Exam!!!
The first big, real exam will be given in the normal class time and place on Friday, October 12. It will cover material up to and including today's lecture (section 3.7, the Chain Rule). The lecturer will have a review session on Thursday evening, October 11. Please do the suggested homework and the workshop problems.

The topic for today is the Chain Rule, which is probably the most important of the differentiation "rules". I want to introduce the subject matter in a way that I think is indirect and natural. So I presented the following economic information.

Chipco data
I then put the chart below on display and we discussed it for a while. Here's a pdf copy which you can print out separately if you like.

	                      CHIPCO
                 INVESTMENT DOLLARS & PRODUCTION

Capital Invested    Chips produced      Marginal chips produced
$ in millions       1,000's of units    1,000's of units per 
                                        millions of $'s

      200               3,000                    .23  
      300               3,040                    .28
      400               3,070                    .42
      500               3,100                    .78 
      600               3,190                    .31 

                         CHIPCO
                    SALES & PROFITS

Chips marketed      Profit gained      Marginal profit
1,000's of units    $'s in millions    Millions of $'s per
                                       1,000's of units

    3,000                1.2                  .03 
    3,050                2.8                  .02 
    3,100                3.6                  .05 
    3,150                4.9                 -.01
    3,200                5.1                  .02 

The first table
I don't think many people were familiar with the economic terms used above. In particular, the word "marginal" is used in a fairly technical sense. In the first table above, it refers to the approximate amount would increase per each million dollars of increase in capital investment. Therefore, for example, if $302 million were invested, then (according to this model) chip production would be 3,040,000 (chip production at the $300 million level) plus .28(2)(1,000) chips. The 2 comes from the additional millions of dollars of capital. The 1,000 comes from the units I use for chip production. The .28 is this "marginal" quantity. In the first table, the marginal quantity is therefore the approximate amount C/M, relating the change in chip production to the change in money invested. It is sort of a slope, or, more likely, sort of a derivative: indeed, the use of "marginal" in economics usually means a derivative. In this model, if $297 million were invested, the approximate expected chip production would be 3,040,000 (again, chip production at the $300 million level) plus .28(-3)(1,000). The novelty here is the use of the minus sign, since here we are decreasing the capital investment rather than increasing it.

Comment
The ratio written as C/M is very "classical" and sometimes can be misunderstood. I don't mean here that has an independent meaning as a variable, so you can't just divide by 's top and bottom. The abbreviation is an abbreviation of complicated ideas, a change in chip production compared to a change in money invested.

The second table
The second table describes a similar phenomenon, here connecting the chip amount, C, with the profit derived from these marketing and sale of these chips. For example, the profit derived from the sale of 3,000,000 chips (the first line of the second table) is $1.2 million. If we now look at the third column, the model predicts a marginal profit of .03 (in the given units). Using this, if 3,010,000 chips are marketed (that's 10 more 1,000 units of chips) the additional profit would be .03(10) million dollars, or $300,000. And if only 2,970,000 chips were marketed, then the profit would be 1.2 million+(.03)(-30)(1,000)million. (I think I got all the units correct.) The third column gives P/C for various amounts of chip marketing: the change in profits compared to the change in chips marketed. Of course the validity of such models can certainly be criticized, but I really wanted to show this to explain what's in the next paragraph.

Linking the tables
The two tables linked together describe a complicated phenomenon. First we "input" capital, M (M is for money), which produces C, a certain number of chips. Then the chips are marketed (and sold, hopefully!) to obtain a certain amount of profit, P. Here we have a composition of functions. For example, suppose we were asked how much profit there is if we put in M=$500 million. From the first table we read off C=3,100,000 chips, and from the second table we can then see that P will be 3.6 million dollars.

I hoped that this was all fairly clear. Now I asked what I thought was a difficult question. Suppose we increase M from 500 million dollars to, say, 503 million dollars. What does the model predict the profit will be? We can trace this if we are sufficiently alert. The first marginal quantity we need to consider is C/M. For M=$500 million, this is .78. So the new chip production is old chip production + increase in chip production, and this will be 3,100,000+(.78)(3)(1,000 chips). Now let us consider the chip/profit table. With C=3,100,000, we see that profit is supposed to be 3.6 million. But we are changing C by adding on the (relatively small) amount of .78(3)(in units of 1,000's of chips). The relevant marginal quantity here is P/C on the row where C is 3,100(,000). The marginal amount here is .05, so that the new profit will be the old profit (3.6 million) plus (.05)(.78)(3) million dollars. The 3 comes from perturbing the capital investment. The really interesting stuff is (.05)(.78): indeed, this represents the marginal profit as capital invested changes, when the capital investment is 500 million dollars. Symbolically, it might make sense written this way:
P/M=P/C·C/M.
So the C's just seem to cancel out. Of course, this is more complicated than just multiplying fractions, since the fractions (the marginal stuff, the derivatives) need to be "evaluated" on the appropriate rows of the tables. So I hope that my discussion has supported:

The Chain Rule
Suppose that f and g are differentiable functions. The F(x)=fog(x)=f(g(x)) is differentiable, and F´(x)=f´(g(x))·g´(x).

Here o is supposed to be a little circle, and the little circle indicates composition. I will more frequently write F(x)=f(g(x)). The tables above sort of indicated that chip production was a function of capital investment, and then that profit was a function of the chips marketed, so that profit as a function of capital investment was a composition of the two functions.

Function diagrams
I tried to make another supporting argument for the Chain Rule, using function diagrams. If x is the input to a differentiable function, g, then the output is g(x). If we perturb or change the input to g a bit, the output, g(x+h), can be thought of in several parts: g(x)+g´(x)h+(Err_g)h. Here g(x) is the old output, g´(x)h is a change in output which is directly proportional to h, and there's other, higher order stuff, which-->0 faster than h, so Err_g-->0 as h-->0.

Now we could take another differentiable function, f, with input w and input perturbation k. The output will be f(w+k), and we can think of it as f(w)+f´(w)k+(Err_f)k. If we wire up the output of g to the input of f, then we can try to think this way:

x+h--> g(x)+g´(x)h+(Errg)h
        w  \___________/
    THIS IS k WHEN CONSIDERING f's INPUT  
          w   +   k  -->  f(w)   +  f´(w)k     +  (Errf)k 
                       f(g(x)) + f´(g(x))g´(x)h  +  (STUFF)h

Why do this? The more ways I can try to explain may allow you more ways to understand. At the basement level, what you must have is the ability to compute with the Chain Rule correctly in straightforward, formulaic circumstances. Examples of this are shown below. But I do think you will be more appropriately equipped to face the demands of a technical career if you understand how this formula works, and what the numbers mean. I hope that the varied approaches will let more of you into the first floor, the second floor, etc., even into the penthouse (sigh, bad metaphor) of understanding the Chain Rule.

Basement example #1
If F(x)=(x²+7)³⁰⁰, what will F´(x) be? I don't need the Chain Rule, not really (?), to compute F´(x) because, after all, F(x) is "just" a polynomial (although here F(x) is a polynomial of degree 600, and this polynomial is not presented in standard fashion). Success (rapid, accurate computation) here probably will result from recognizing that the chain rule applies.
If F(x)=f(g(x)), then g(x) is x²+7 so g´(x)=2x, and f(x) is x³⁰⁰ so f´(x)=300x²⁹⁹. Thus F´(x)=f´(g(x))g´(x)=f´(x²+7)(2x)=300(x²+7)²⁹⁹(2x). Whew!

But now comes the realistic comment. Hardly ever does anyone bother writing down all of these intermediate steps. That is, in practice very few f's and g's are actually identified. What happens is that people see and differentiate the outside most function (f above), put in the inner function (g) in that derivative, and then multiply by g´. For example, consider sin(e^x+x²). What is its derivative? The outside function is sine, whose derivative is cosine. So I begin by writing cos(what's inside)·the derivative of what's inside. The result is cos(e^x+x²)·(e^x+2x). This expression is a formula for the derivative of sin(e^x+x²). Again, I urge you to consider the significance and necessity (!) of appropriate parentheses in these expressions. The "argument" of cosine is e^x+x² and the cosine expression is then multiplied by the expression (e^x+2x).

#2
I think I did another example, something like computing the derivative of F(x)=e^{5x2+7sqrt(sin(x))}. The outside most function is e-to-the- (?) whose derivative is e-to-the-. Therefore the derivative, F´(x), will begin with e^{5x2+7sqrt(sin(x))} and continue by multiplying by the derivative of 5x²+7sqrt(sin(x)). That's a sum, so its derivative is 10x+7·(the derivative of sqrt(sin(x)). What's the derivative of sqrt(sin(x))? That itself is a composition, with the "outside" function sqrt or thing-raised-to-the-half-power. The outside derivative is (1/2)(thing-raised-to-the-minus-half-power). So the derivative of sqrt(sin(x)) is (1/2)(sin(x))^-1/2(cos(x)). Now put it all together:
if F(x)=e^{5x2+7sqrt(sin(x))} then
F´(x)=(e^{5x2+7sqrt(sin(x))})(10x+7((1/2)(sin(x))^-1/2)).

I strongly recommend that you use lots of parentheses when applying the Chain Rule.

Finding a low point -- left over from earlier
In lecture #3, we considered the graph of f(x)=5e^-2x+3e^4x. We decided that it probably was shaped like a big U, with one lowest point. We can now accurately and exactly find this lowest point. Look:
f´(x)=5e^-2x(-2)+4e^4x(4). Here the Chain Rule is being used twice, with the exponential function as the outside function each time, and with multiplication by -2 (respectively by 4) as the inside function.

The lowest point is where f´(x)=0. Here's the algebra:
    5e^-2x(-2)+3e^4x(4)=0
    3e^4x4=5e^-2x2
    12e^4x=10e^-2x
    e^6x=10/12
    6x=ln(10/12)
    x=(1/6)ln(10/12)
By the way, this is about -0.03038692614, which is sort of close to what we guessed way back near the beginning of the course.

QotD
I gave the following information: F(x)=f(3x-x²), and that
    f(1)=4;  f´(1)=8;  f´´(1)=16.
    f(2)=3;  f´(2)=5;  f´´(2)=7.
I asked students to compute F(1) and F´(1) and F´´(1).

F(1)
Since F(x)=f(3x-x²), I know that F(1)=f(3·1-1²). Here the parentheses mean function evaluation, and F(1)=f(3-1)=f(2)=3.

F´(1)
Since F(x)=f(3x-x²), this is a composition of two functions. The inside function is 3x-x² and the outside function is f. Therefore the Chain Rules tells us that the derivative is F´(x)=f´(3x-x²)(3-2x). There is no simpler expression which is valid for all x. We have no information which allows us to write anything else for f´. To get F´(1) we plug x=1 into the equation F´(x)=f´(3x-x²)(3-2x) and we get F´(1)=f´(3·1-1²)(3-2·)=f´(2)(3-2)=5·1=5.
We need to use the formula for F(x) to get a formula for F´(x).

F´´(1)
We need to begin with information about F´(x) and differentiate that. Knowing F´(1) is not enough: we need to know how F´(x) changes.
We do know F´(x)=f´(3x-x²)(3-2x). Let me rewrite this with some emphasis.
    F´(x)=(f´(3x-x²))(3-2x).
If you look carefully at the logic of the right-hand side of this equation, I hope that you will see it is a product of f´(3x-x²) and (3-2x). I will use the Product Rule. The first factor, f´(3x-x²), is a composition. The outside function is f´ and the inside function is 3x-x². I will use the Chain Rule on this.
    F´´(x)= =(f´´(3x-x²)(3-2x))(3-2x)+(f´(3x-x²))(-2).
Now insert x=1. Here is the result:
    F´´(1)=(f´´(3·1-1²)(3-2·1))(3-2·1)+(f´(3·1-1²))(-2)=(f´´(2)(1))(1)+(f´(2))(-2)=7·1+5·(-2)=7-10=-3.

Certainly, if x is very very large positive, the fraction =(1x+2)/(3x²+4) will become small. (The limit as x-->+ of f(x) is 0, we will later say.) A similar thing is also true as x-->-. Both of these limits or asymptotic behaviors or whatever you want to say occur because the bottom is a degree 2 polynomial, and the top is a degree 1 polynomial, and the high degree polynomial "dominates". I'll be more precise about this in the future.

But what happens in between? What sorts of numbers can we get out of f? Well, f(1)=3/7, so 3/7 is in the range. And f(2)=4/16=1/4, so 1/4 (sigh!) is in the range. And ... and ... I think just listing numbers is not even simple fun, and merely is pointless. One nice suggestion for an output, though, was 0. The only x which gives 0 is -2. This is good because now I know that the only sign change is at -2, and (Intermediate Value Theorem) this is where f's outputs change from positive to negative or negative to positive.

How can we systematically learn about f? Well, I do know (Quotient Rule) that f´(x)=[(1)(3x²+4)-(1x+2)(6x)]/(3x²+4)². When can this be 0? Only when the top is 0, so let me "simplify" the top:
(1)(3x²+4)-(1x+2)(6x)=3x²+4-6x²-12x=-3x²-12x+4.
So the top is 0 when -3x²-12x+4. We need to use the quadratic formula (hey, very few "random" degree 2 polynomials with integer coefficients actually can be written as a product of two degree 1 polynomials with integer coefficients!). The roots are -2-(4/3)sqrt(3) and -2+(4/3)sqrt(3). (I wrote something much more horrible in class, and this is what results after some arithmetic.)

What do I know about the SIGN of f´? This derivative is a quotient, and the bottom is (3x²+4)². The bottom is always positive. So the SIGN of f´ in this case is determined by the sign of the top. But the top is -3x²-12x+4. This is a parabola, and because the coefficient of x² is -3<0, the parabola opens down. From this and from knowing the two roots of the top, we see:

• If x<-2-(4/3)sqrt(3) then f´(x)<0.
• If -2-(4/3)sqrt(3)<x<-2+(4/3)sqrt(3) then f´(x)>0.
• If x>-2+(4/3)sqrt(3) then f´(x)<0.

So what will the range of f be, exactly? It seems "clear" looking even at the approximate graph shown to the right that the range will be all numbers between f's values at r_L and r_R. That is, the range is the interval [f(r_L),f(r_R)]. I can be more precise, of course. The range is [f(-2-(4/3)sqrt(3)),f(-2+(4/3)sqrt(3))], and this is (after some work which I would not inflict on people in class!) [(1/4)-(1/6)sqrt(3),(1/4)+(1/6)sqrt(3)].

What this is, and what this is not
I do not claim that the computations just done are wonderful. Maybe they are not even interesting. But if you absolutely need to know the answer to questions of that type, and you need to know with great accuracy, and great certainty, then the analysis we've done is probably the best way. It really allows us to get the correct numbers, and the procedure is understandable. It is also inherently more precise than using a graphing device, although I probably would try to get an approximate answer using a graph first.

By the way, to the right is, of course, a graph (in a weird window, check the axes!) of y=f(x). I hope you can "see" the range. Incidentally, [(1/4)-(1/6)sqrt(3),(1/4)+(1/6)sqrt(3)] is approximately [-0.039,0.539].

Building a BIG cylinder inside a sphere
One of the recent workshop problems asked students to analyze the problem of a cylinder "inscribed" inside a sphere. So the cylinder touches the sphere as mucl as possible. The task is to understand how to write a formula for the volume of the cylinder as a function of the cylinder height, and to also tell what the domain of that formula is, considering the origin of the problem. Here we will look at the formula obtained when the radius of the sphere is 3. The height of the cylinder is related to the cylinder's radius using Pythagoras, if we look at a cross-section of the cylinder. If x is the height of the cylinder, then V(x)=Pi(9-x²/4)x. The domain of V(x), when considering this problem, is 0<x<6. I don't think a cylinder can have a negative height, and I don't think a cylinder insider a sphere of radius 3 can have a height bigger than the diameter of the sphere, which is 6.

We can graph V as a function of the height, x. When x is close to 0, the cylinder is short and wide, but the shortness (a factor of x) makes V(x) quite small. When x is near 6, the cylinder is tall, but the radius is very small (36/4 is 9!) so the V(x) is also quite small.
How can we find the cylinder of largest volume inside this sphere? That will be at the "top" of the graph, and I will locate the top by compute V´(x). So this is (Product Rule) Pi(-2x/4)x+Pi(9-x²/4). This will be 0 when (divide by Pi, collect x²'s): 9-(3/4)x²=0, so x=2sqrt(3) (the negative root is not in the domain for this problem!). Indeed, if you look at the graph, the top of the graph seems to be at about 3.4. This specifies the cylinder. The radius and the volume can then be computed.

What this is, and what this is not
Again: this is not a profound problem! But it does show a systematic way to solving such problems. There may be computational details which can be irritating, but at least we have some method to work with. And I assure you that we will go into great detail about the general method later in this course.

Problem 26 of section 3.4
Section 3.4 has many interesting problems, and I strongly urge students to read it. Here is a lovely problem. When I first looked at it, I thought that it was very complicated, and that we didn't have the necessary technical base to do it. So here's the problem: look at a clock face at 3 o'clock. The minute hand points up, at 12, and the minute hand points to the right, at 3. Fine: a right angle. Here's the question: how fast is the angle changing at that time? Or, phrased more technically, what is the angular velocity (?!) of the angle at 3?
The picture to the right may be lovely, but it is "frozen" and the right angle only occurs at 3 o'clock. We can't study the rate of change of the angle by considering that picture any more than we could deduce the speed of a car by looking at a picture of it racing.

Well, the way I finally thought of doing this is to "decouple" (?) or separate the actions of the two "hands" of the clock.

Let's call 1 the angle between the segment connecting the center of the clock with "12" and the minute hand. I know that 1 varies from 0 to 2Pi in one hour, and that the minute hand moves steadily. Hey: I bet that d1/dt is constant, and that it is 2Pi (in radians per hour).

On the other hand (little joke, sorry) we can analyze the motion of the hour hand. Call 2 the angle between the segment connecting the center of the clock with "12" and the hour hand. The hour hand travels completely around the clock in 12 hours, and it also moves steadily. Therefore its angular velocity, in radians per hour, is 2Pi/12, and that's d2/dt.

Let me call the angle between the minute hand and the hour hand . Then surely (at least, as time progresses after 3 o'clock) I know that =₂-₁. Now we can figure out how changes:
d/dt=d₂/dt-d₁/dt=2Pi/12-2Pi=-11Pi/6. To me, as I confessed in class, this is all not "intuitively" clear, and I do need to think about it. The angle decreases (hey: the minute hand does move faster than the hour hand!) and the rate of decrease is steady. Interesting: not profound, but we needed to think a bit, and I believe derivatives help.

Leibniz notation The derivative of f at x is f´(x). But historically there has been another system used. The difference quotient defining the derivaitve is the slope of a secant line for the curve y=f(x). It is a difference in y's (second coordinates of the points) divided by a different in x's, first coordinates of the points. Traditional notation for the quotient looks like this: (y)/(x). And we want limx-->0(y)/(x). People like remembering the definition, because, as you will see, it helps intuition and applications. The notation that is used is dy/dx for "the derivative of y with respect to x". Use notation that helps you and that doesn't mislead you. Mostly I think Leibniz notation is helpful.

The derivative of sine: a guess
I began by (trying to) draw an accurate picture of sine and then discussing the slopes of the tangent lines to this curve. The derivative of sine, is, of course, exactly those slopes.

Where would the derivative of sine be 0? Well, where the tangent lines are horizontal. That should be at the tops and bottoms of the sine curve. These occur at Pi/2, 3Pi/2, -Pi/2, etc.: lots of places because sine repeats every 2Pi.

Let's look at sine between, say, x=-Pi/2 and x=Pi/2. The derivative, the slope of the tangent line, would start out at -Pi/2 at 0, then it would increase (as the tangent line began to tilt up). Then it would tilt up more (the slope would be more positive) until it would begin to tilt "down": here the language gets complicated. I am not asserting that the derivative is negative, but I am merely asserting that the slope, which stays positive, begins to decrease. Eventually the slope becomes 0 again when x=Pi/2.

What should happen between x=Pi/2 and x=3Pi/2? In that interval the sine curve is sort of a downwards reflection of the behavior in the interval [-Pi/2,Pi/2]. The derivative starts at 0, then becomes negative. It gets more negative, then gets less negative, and ends up at 0. The shape of the derivative exactly reflect the shape in the earlier interval, since the sine curve's shape is a flip of the earlier behavior.

The derivative repeats every 2Pi, since y=sin(x) repeats every 2Pi. I drew sort of what is shown below. The derivative of sine qualitatively looks like cosine, except maybe we don't know the scaling factor: how high the curve is. Things are wonderful: the scaling factor is 1.

The derivative of sine: via limits, algebra, etc.
In a calculus textbook, something like the following is done when f(x)=sin(x):

 sin(x+h)-sin(x)     sin(x)cos(h)+sin(h)cos(x)-sin(x)
----------------- = ---------------------------------- = PIECE #1 + PIECE #2
        h                          h

                   sin(h)
PIECE #1 = cos(x) --------
                     h

                   cos(h)-1
PIECE #2 = sin(x) ----------
                       h

                    [sin(h)]2             sin(h)               1
PIECE #2 = sin(x) ------------- = sin(x) -------- sin(h) ------------
                   h [cos(h)+1]             h             [cos(h)+1]

                 sin(h)                         1
sin(x)-->sin(x); ------ -->1;  sin(h)-->0; ----------- --> 1/2.
Nothing happens!      h                       [cos(h)+1]

The derivative of cosine
Look again at the shape of the graphs drawn just above. If I move the coordinate axes Pi/2 to the right, the graph of sine becomes the graph of cosine. The candidate graph for the derivative needs to be recognized. It is actually minus the graph of sine. This minus sign is slightly annoying, and sometimes I screw up and forget it or I put it where it shouldn't be. The derivative of cos(x) is &ndashsin(x). (There is a shift of Pi/2 in both graphs.)

The derivative of tangent
A specific example of the quotient rule is to compute the derivative of f(x)=sin(x)/cos(x). Here top=sin(x) and bottom=cos(x), and the derivative will be (top'·bottom-bottom'·top)/(bottom)². Notice that because of the minus sign in the quotient rule and the minus sign in the derivative of cosine, there is an asymmetry in the result, and this can lead to errors. f´(x)=[cos(x)·cos(x)-(-sin(x))·sin(x)]/[cos(x)]². Here is an answer which demands simplification. The top is [cos(x)]²+[sin(x)]² which is 1. The whole result is therefore 1/[cos(x)]² or [sec(x)]². Of course, this f(x) is tan(x), so we now know that the derivative of tan(x) is [sec(x)]². Please note the parentheses. I almost always use lots of parentheses because I make mistakes.

There are 6 trig functions. I know the derivatives of sine and cosine and tangent, as I just showed you. Also sometimes useful is the derivative of secant. Notice: since sec=1/cos, sec´=-(-sin)/(cos)²=(1/cos)(sin/cos)=(sec)(tan). I don't offhand know the derivatives of csc and cot. Sigh. Now, the updated table:

Function	Derivative
xn	nxn-1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)
CONSTANT(f(x))	CONSTANT(f´(x))
1/f(x)	-f´(x)/[f(x)]2
f(x)/g(x)	[f´(x)g(x)-g´(x)f(x)]/[g(x)]2
sin(x)	cos(x)
cos(x)	-sin(x)
tan(x)	[sec(x)]2=1/[cos(x)]2
sec(x)	sec(x)tan(x)
In mathematics you don't understand things. You just get used to them.

QotD
Here is the question, and here is its solution, as done by my silicon pal, Si:

> time();
                                     0.004

> f:=(3*exp(x)-4*cos(x))*(x^4-x^2+1)/((5/x^3)+7*sqrt(x));
                                                4    2
                        (3 exp(x) - 4 cos(x)) (x  - x  + 1)
                   f := -----------------------------------
                                    5        1/2
                                   ---- + 7 x
                                     3
                                    x

> time();
                                     0.004

> diff(f,x);
                        4    2                                  3
(3 exp(x) + 4 sin(x)) (x  - x  + 1)   (3 exp(x) - 4 cos(x)) (4 x  - 2 x)
----------------------------------- + ----------------------------------
            5        1/2                         5        1/2
           ---- + 7 x                           ---- + 7 x
             3                                    3
            x                                    x

                               4    2      /   15      7   \
       (3 exp(x) - 4 cos(x)) (x  - x  + 1) |- ---- + ------|
                                           |    4       1/2|
                                           \   x     2 x   /
     - -----------------------------------------------------
                         / 5        1/2\2
                         |---- + 7 x   |
                         |  3          |
                         \ x           /

> time();
                                     0.004

Exam warning!
The first exam of the course will be given two weeks from this lecture, on Friday, October 12. More information about the exam, including review material, sections to be covered, etc., will be available soon.

Function	Derivative
xn	nxn-1

I actually proved this in the last lecture when n is a positive integer. It is, in fact, true for any constant n. So examples would be:

• x¹⁷ whose derivative is 17x¹⁶.
• sqrt(x), which you might see as x^1/2, and its derivative is (1/2)x^-1/2, which can also be written as 1/(2sqrt(x)).
• 1/x⁹, which is x^-9, and its derivative is -9x^-10, which can also be written as -9/x¹⁰.

Constants
I want f´(x)=lim_h-->0[f(x+h)-f(x)]/h. What if f is a CONSTANT function, so its values are all the same? Well then the top of the difference quotient, f(x+h)-f(x), will be CONSTANT-CONSTANT, and it will be 0. So the derivative will be 0.

Function	Derivative
xn	nxn-1
CONSTANT	0

e^x
Let's consider the derivative of an exponential function, say a^x, where a is a constant. Then the difference quotient, [f(x+h)-f(x)]/h becomes [a^x+h-a^x]/h. As I mentioned in class, just plugging in h=0 yields 0/0, and this doesn't help. We can try the algebra that's available:
[a^x+h-a^x]/h=[a^xa^h-a^x]/h=a^x((a^h-1)/h).
So we need to consider (a^h-1)/h as h-->0.

We actually analyzed this limit graphically and numerically in Lecture #3 for several values of a. When a=2, it seems that the limit exists and equals .693, while if a=3, 1.109 was the approximate value of the limit. Since f´(x) will be equal to a^x multiplied by whatever the value of lim_h-->0(a^h-1)/h is, I want to choose a value of a so that the formulas are as simple as possible. So maybe we can get a value of a so the limit of (a^h-1)/h as h-->0 is 1. This can be done. In fact, here is a fairly irritating way of thinking about this special number, which people call e. If h is small, we would like (a^h-1)/h to be approximately 1 when h is small, well, look at the following sequence of ideas:

n	(1+{1/n})n
1	2.000000000
2	2.250000000
3	2.370370369
4	2.441406250
5	2.488320000
10	2.593742460
100	2.704813829
1,000	2.716923932
10,000	2.718145926
100,000	2.718268237
1,000,000	2.718280469

• Since (e^1/100-1)/{1/100} should be approximately 1 we could multiply by 1/100 and see that (e^1/100-1) is approximately 1/100.
• Now add 1 to both sides and see that e^1/100 is approximately 1+{1/100}.
• If we took the 100^th power (raised things to the 100) then the 1/100 and 100 in the exponent of e cancel (repeated exponentiations multiply) so that e is about (1+{1/100})¹⁰⁰.

Well, the numerical value of (1+{1/100})¹⁰⁰ about 2.704813829. To the right is a table of values of (1+{1/n})ⁿ. This method of "computing" or approximating e is actually very very slow. As previously written, here are several million digits of e, if you need them. As I mentioned in class, it is certainly possible to prove that the formula converges as n gets large. This can be done with "only" high school algebra. It is quite tedious, and, to me at least, has little redeeming social value. (I may have the wrong attitude here, so please forgive me.)

Function	Derivative
xn	nxn-1
CONSTANT	0
ex	ex

What does that limit statement mean?
The definition f´(x)=lim_h-->0[f(x+h)-f(x)]/h is sometimes a bit difficult to understand. What if I just throw out "lim_h-->0"? Well, certainly f´(x) is NOT the same as [f(x+h)-f(x)]/h (hey: one of them has an h and the other doesn't even mention h!). So it really means f´(x)=[f(x+h)-f(x)]/h+ERR, where ERR (stands for "ERROR", of course) is some "mess", and all I know about it (and care about it, at this time!) is that it goes to 0 as h goes to 0. I don't like division, so let me multiply by h. Here's the result:
f´(x)h=f(x+h)-f(x)+ERR·h.
I don't like subtraction, so let me add f(x). f(x)+f´(x)h=f(x+h)+ERR·h.
People usually put the "ERR" term on the other side, so let me do that. It is irritating, but I won't change the sign on the ERR term, because right now I am interested in the qualitative aspect. So what I have is:

f(x+h)=f(x)+f´(x)h+Err·h

So what the heck do we have? If we think of the function as taking an input variable, x, and smooshing (?!) it around to get an output value, f(x), then f(x+h) is what results if we "kick" the input value by a little bit, h. If f is differentiable, then the output seems to split up into several pieces.

1. The old value, f(x). If h is very very very small, we should expect to see only something that looks like f(x). This is continuity -- there will be no big jumps if the input is changed just a little.
2. Suppose h is small, but not (somehow!) very very very small. Then we should see some change in the output. If f is differentiable, the major change will be a CONSTANT multiplying the input change. The multiplier is the value of the derivative.
3. So the two terms before aren't all. There's also "ERR·h". This is stuff that is much smaller than h. In reality, I would not expect to observe this for small h. It is "higher order" than h (people think of h³, for example: when h is small, higher order stuff is much smaller).

Sums
Suppose f and g are differentiable functions. Then I know:
     f(x+h)=f(x)+f´(x)h+Err_f·h
     g(x+h)=g(x)+g´(x)h+Err_g·h
I used different notation for the error terms for f and g to keep track of stuff better than I did in class. I can add these equations. Here's the result:
    f(x+h)+g(x+h)=f(x)+f´(x)h+Err_f·h+g(x)+g´(x)h+Err_g·h
but this is an awfully silly way to write the result. I should write it in such a way that the "structure" of the equation is shown. Here:
    (f(x+h)+g(x+h))=(f(x)+g(x))+(f´(x)+g´(x))h+([Err_f+Err_g]h)
The left-hand side is the function f+g at the input value x+h. The first piece, (f(x)+g(x)), is the unperturbed value of the function f+g at x. The second piece, (f´(x)+g´(x))h, is a multiplier not involving h, multiplied by h. The third piece is h multiplied by some stuff: [Err_f+Err_g], and all this stuff-->0 as h-->0. Hey, this is the higher order vanishing. So I know the next line in the table.

Function	Derivative
xn	nxn-1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)

Products
This is harder. Suppose I know
     f(x+h)=f(x)+f´(x)h+Err_f·h
     g(x+h)=g(x)+g´(x)h+Err_g·h
and now I multiply the equations. Well, the left-hand side isn't bad, but there are three terms on the right-hand side of each equation, so there will be NINE terms if I distribute out the product/sums. I am really considering the product function of f and g. Here is how to organize the result. In class we thought a bit about how to organize this. In "print" such discussion is more difficult to write.

Please: I don't want all this stuff to be memorized. I haven't memorized it. But I do know the general idea, and that's what I'd like you to get used to. Here is the left-hand side:
     (f(x+h)·g(x+h)). This is the function f·g's value at x+h.
Here are the pieces of the product of the two right-hand sides:
     (f(x)·g(x)). This is the function f·g's value at x: the "old", unperturbed value of f·g.
    (f´(x)·g(x)+f(x)·g´(x))h. This is the first-order term, stuff (no h's!) multiplied by one h.
Here is all the rest of the stuff. There are (good grief!) six different terms. But I can "pull out" one h from all of these terms, and what is left in all six of these terms are things that -->0 as h-->0. Wow!
     (f(x)·Err_g+f´(x)·g´(x)h+f´(x)·Err_gh+f(x)·Err_f+Err_f·g´(x)+Err_f·Err_gh)h.
If you see this, sort of, then you can see the next line of the table.

Function	Derivative
xn	nxn-1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)

A quote from von Neumann
John von Neumann (1903-1957) was a mathematician who was raised in Hungary and spent most of his career in the United States. He worked in many areas of pure and applied mathematics. His ideas were influential in quantum mechanics, the development of nuclear weapons, game theory, and the theory and construction of digital computers.

In mathematics you don't understand things. You just get used to them.

Another way of looking at products
Well, first let's reconsider addition. We could imagine f and g being functions that somehow model intervals which are growing in length. At a certain time, the intervals have some length (the lengths labeled just f and g to the right). If we increase time ("+h") then each interval grows. I used the Greek capital Delta () to indicate how much each would grow. If we combine the functions with addition, then the growth just adds. To me this is sort of straightforward. I hope it is to you. Since the growths add, the average growths add, and the "instantaneous growths" (the derivatives!) also just add.

A simple physical (?) model of multiplication in this setting would be to just make the intervals into sides of a rectangle. Then the area of the rectangle will be the product function, f·g. When we allow the intervals to grow, the area grows in a more complicated manner. Look at the picture: the increase of the area has three "chunks". One of them is an increase in f multiplied by g. Another is f, multiplied by the increase in g. Then there's the corner piece, which is increase in f multiplied by increase in g. When the increments get small, the corner piece's decrease is much faster than the two other pieces' decreases. (Sorry: my use of language is not very good here.) So I think that the instantaneous increase in f·g will be given by f´(x)g(x)+f(x)g´(x): these are the first order terms of the increase. Of course, this is the product rule.

Examples
The derivative of 37x⁸⁸: well, this is a product. It is 37·x⁸⁸. Here f(x)=37 and g(x)=x⁸⁸. So f´(x)g(x)+f(x)g´(x) becomes 0·x⁸⁸+37·88x⁸⁷. Most people think that multiplication by a constant deserves an entry of its own. So back to the table:

Function	Derivative
xn	nxn-1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)
CONSTANT(f(x))	CONSTANT(f´(x))

• 47x³²+sqrt(Pi)x¹²-.007x²+99 has derivative 47·32x³¹+sqrt(Pi)·12x¹¹-.007·2x+0. So now we can differentiable polynomials. People who aren't as neat as I am might leave out the +0. :-)
• 8e^x-9x^1/3 has derivative 8e^x-9·(1/3)x^-2/3.
• Consider (5e^x-9x)(x¹²-12/x³). This is a product with f(x)=5e^x-9x and g(x)=x¹²-12/x³ (but also recognize that -12/x³ is -12x^-3). This has derivative (5e^x-9)(x¹²-12/x³)+(5e^x-9x)(12x¹¹-12(-3)/x⁴)
• Look at, say, x¹². This can also be written x⁷·x⁵. If we use the Product Rule, we will get 7x⁶·x⁵+x⁷·5x⁴. If we "clean up" a bit, we seem to get 7x⁶⁺⁵+5x⁷⁺⁴ which is 7x¹¹+5x¹¹ which is 12x¹¹. That is the correct and expected answer, so the weird Product Rule does reinforce what we already know.

Reciprocal
Now I want to consider the reciprocal ("one-over"). This in not the inverse function, which is much more complicated. Here is just want the multiplicative inverse of the result of the function. For example, if f(x)=3x²+5, then f(2)=3·4+5=17. If g(x)=1/f(x), g(2) would be 1/17.

We could for a second pretend that the graph of f looks roughly ("qualitatively") like what is shown to the right. I haven't included much information, but as we travel from left to right on the curve y=f(x), the graph goes up. The slope of the tangent line is a geometric representation of the derivative. I bet that the slopes of the tangent lines to y=f(x) are positive. Now look at y=1/f(x). Since the values of f got bigger (yeah, I made it simple, everything is positive) the values of y=1/f(x) get smaller. That graph is decreasing and the tangent lines have negative slope. So I bet there should be some negative sign appearing somewhere in some formula. Let's get the formula.

Let me assume that g(x)=1/f(x), and that g(x) is differentiable. Then f(x)·g(x)=1. We could differentiate this equation. As I did in class, I will be very adventurous (?) and differentiate the right-hand side first. That gives me 0. I will use the product rule on the left-hand side and then this equation results:
    f´(x)·g(x)+f(x)·g´(x)=1.
But I want a "formula" for the derivative of g(x), so I will "solve" for the derivative of g. Here is the result:
     g´(x)=–f´(x)·g(x)/f(x)=–f´(x)/[f(x)]² (because g(x) is 1/f(x) and I substituted).

Function	Derivative
xn	nxn-1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)
CONSTANT(f(x))	CONSTANT(f´(x))
1/f(x)	–f´(x)/[f(x)]2

just one crummy Example
We did get our minus sign. The derivative of 1/(3x²+5) must be -(6x)/(3x²+5)².

Quotient rule
Finally, I will state the Quotient Rule. This is a formula which allows you to differentiate f(x)/g(x). If you think of this as f(x)·[1/g(x)], you could differentiate by using the Product Rule and the Reciprocal Rule (no one calls it that!) in succession. But people rarely do that. They just remember another formula. Here it is, the last line in the table for today.

Function	Derivative
xn	nxn-1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)
CONSTANT(f(x))	CONSTANT(f´(x))
1/f(x)	-f´(x)/[f(x)]2
f(x)/g(x)	[f´(x)g(x)-g´(x)f(x)]/[g(x)]2
In mathematics you don't understand things. You just get used to them.

Actually, von Neumann was the sort of person who liked to say shocking things. I believe you and I should try to understand things (he certainly did, in spite of the quote), in order to increase the amount of structure in what we know. Such structure makes the incredible accumulation of facts more tolerable.

The derivative of (7e^x+9x-11)/(x⁵-sqrt(x)) is [(7e^x+9)(x⁵-sqrt(x))-(7e^x+9x-11)(5x⁴-1/{2sqrt(x)})/(x⁵-sqrt(x))².
What a mess. Remember that sqrt(x)=x^1/2, whose derivative is (1/2)x^-1/2 or, if you wish, 1/{2sqrt(x)}. Also we now can officially differentiate rational functions, which are quotients of polynomials.

What grows faster?
I wanted to compare x¹⁰ and e^x. When x=2, the first function is 2¹⁰=1,024, and the second function is e² which is certainly less than 9 (e is less than 3). In fact, e² is about 7.389. Which one grows faster? I asked people to graph this, say from 0 to 5. A graph is shown to the right. My "tool" has lots more computing power and pixels than the typical hand-held device, but please notice the differing scales on the vertical and horizontal axes. The red curve is y=x¹⁰ since it sort of exits the window at the top. The lazy (?), apparently more slowly increasing green curve is y=e^x.

Comparing the asymptotics as x gets large of these functions turns out to be extremely important in practice. For example, similar discrepancies or differences occur in descriptions of different types of chemical reactions, so knowing what's fast and what's slow is valuable. Consider the function F(x)=x¹⁰/e^x. The Quotient Rule tells me that its derivative is F´(x)=(10x⁹e^x-x¹⁰e^x)/[e^x]². Some algebra (cancel e^x top and bottom, factor the top) gives me
     F´(x)=x⁹·(10-x)/e^x.
Let me concentrate on x>0. The first factor, x⁹, is positive. The second factor, 10-x, is positive for x between 0 and 10. It is 0 at 10, and afterwards, for all x's bigger than 10, it is negative. So:
x¹⁰/e^x begins with the top increasing faster than the bottom. Things sort of are even at 10, but forever afterwards, the bottom is increasing faster than the top.

The graph shown is very deceptive. To the right is a graph of y=x10/ex more truthfully displayed. The window should be examined carefully. The x's go from 0 to 30. The y's go from 0 to (don't miss the little number in the top left!) 5·105: that's 500,000, five hundred thousand. So it looks like x10 is winning, winning, winnning ...until after x=10, it loses, loses, loses. To give you some idea of how badly it loses, 8310 is about 1.55·1019, big, but e83 is about 1.11·1036, much, much, much bigger. Some chemical reactions occur in nanoseconds. Others just loaf along.

QotD
Suppose f(x)=(1x²+2)/(3x²+4). Find an equation for the line tangent to y=f(x) when x=1.

A solution For this we need a POINT and a SLOPE.
POINT Since f(1)=(1+2)/(3+4)=3/7. the tangent line passes through (1,3/7).
SLOPE Since (Quotient Rule) f´(x)= [(2x)(3^x2+4)-(6x)(x²+2)]/(3x²+4)², f´(1)=[2·(7)-6·3]/(3+4)²=-4/49.
An equation for the line is therefore (y-3/7)=-(4/49)(x-1). I am lazy, and unless I am told that the form I just wrote is unacceptable or unless I needed another form to work more with the equation, I would leave the equation exactly that way. (Please!)
Another (?) answer: y=-(4/49)x+(25/49).

A nature note
Adult Winter: Birds that molt twice per year have two adult plumages: adult winter and adult breeding. Adult winter plumage is also called adult nonbreeding plumage.

We discussed various curves which could represent the position of Francine on the parkway in terms of miles from the start of the parkway at time t, in terms of hours elapsed from 7 AM. I tried to show that our everyday intuition lead to the graph being increasing (as you travel from left to right, the points on the graph go up). The graph can have level spots, where Francine pulls over for a rest stop. Legally Francine isn't supposed to drive backwards, though.

If we believe that motion is continuous (so Francine does not have a Star Trek transporter or other device) then the graph of Francine's position goes from (7 AM, 0 miles) to (10 AM, 172 miles) and therefore the graph must have on it at least one point with coordinate description (*,135). All of this, by the way, rests on some complicated assumptions, some of them philosophical (why should motion be continuous?). Today, though, I believe that motion is continuous, and therefore at sometime Francine must be at Mile 135. This is all the idea behind the following important result.

Intermediate Value Theorem
Suppose that the function f is defined and continuous on the interval [a,b]. Then the equation f(x)=y has at least one solution for every y which is between f(a) and f(b).

In mathematics, the word theorem is applied to results that are deduced from basic principles, and usually the term is used for more important conclusions in the subject. In this case, the Intermediate Value Theorem follows from basic principles governing the real numbers. A particular basic principle which is used in the proof of the theorem is the "least upper bound" property of the real numbers. This essentially declares that there are "no holes" in the real numbers. A precise statement is fairly delicate, and this property essentially shows that the reals and the rationals are distinct. Several upper-level math courses spend quite a bit of time exploring the statement. You can read about it in Wikipedia but I do add that detailed knowledge of such foundational material is not needed for success in Math 151 (or, for that matter, for successful careers in almost all of science and engineering!).

Problem #5 in section 2.7
Here's the problem:
     Show that cos(x)=x has a solution in the interval [0,1].
Since I am a picture person, I frequently try to draw a graph or two to understand what the problem is about. In this case, certainly the graph of y=x is easy to imagine on [0,1]. What about y=cos(x)? Well, it does help, if you don't have access to a graphing device, to know that cos(0)=1. I know that cosine drops down as x increases from 0. What do we know about cos(1)? As one student declared, cos(1) is both less than 1 and greater than 0. That's because Pi/2 is about 1.57 and cosine decreases between 0 and Pi/2, and does not reach 0 until Pi/2. So the graph of y=cos(x) decreases from (0,1) to (1,cos(1)), and the end point is somewhere above the x-axis. The mental picture I have built is shown to the right. I deliberately did not have a graphing device create an "accurate" display of the situation -- I wanted to show what we should be able to do inside our own heads.

The picture now encourages me to believe the assertion of the problem. The text supplies this hint:
     Show that f(x)=x-cos(x) has a zero in [0,1].
The phrase "has a zero in [0,1]" means that there is some root of f(x)=0 in the interval [0,1]. Well, f(0)=0-cos(0)=-1<0 and f(1)=1-cos(1). Since cos(1) is between 0 and 1, 1-cos(1) will be between 1 and 0. So we know that f(1)>0. Since the values at the endpoints are both positive and negative, I know that 0 is between the endpoint values. The Intermediate Value Theorem then applies to show that f(x)=0 has a solution between the endpoints, which are 0 and 1.

Can we locate this root more precisely?
So far what we know is that there is a root of cos(x)=x in [0,1]. Such equations occur quite frequently in applications. It is rare that solutions to such equations can be written exactly in terms of simple operations (roots, logs, etc.) and classical constants, such as e and Pi. But it may be (usually is!) important to know them accurately. Can we get better information?

If f(x)=x-cos(x), we know f(0)<0 and f(1)>0. I've indicated this with the + and - labels on the ends of the unit interval to the right.
f(.5)=–.3775..., so we now know there is a root in the interval [.5,1].
f(.75)=+.0183..., so we now know there is a root in the interval [.5,.75].
f(.625)=–.1859..., so we now know there is a root in the interval [.625,.75}
Etc. By this I mean we can continue chopping the interval, looking for the sign of f's value at the center, making the length of the interval where a root is located as small as we like. This is the key idea of the Bisection Algorithm. The weird entry condition below, f(a)·f(b)<0, means that f has different signs at the two ends of the interval.

The Bisection Algorithm
Entry conditions
A continuous function f(x) defined on an interval [a,b], with f(a)·f(b)<0;
a positive tolerance E for the error.

Output
An interval [c,d] so that d-c<E and f(c)·f(d)<0. This identifies an interval of length less than Ein which must contain a root of f(x)=0.

"Loop" structure
Given [p,q] with f(p)·f(q)<0: let m=(p+q)/2. Compute f(m). If f(m)·f(p)<=0, then replace q by m else replace p by m.
Exit check If q-p<E then return p and q as c and d in the output else go to loop.

The very alert student might notice a slight difference from what I described in class. Here I wrote f(m)·f(p)<=0 inside the loop. In class I omitted the = possibility thinking that the program would just work anyway. Mr. Sloane asked me about this and I dismissed his worry. I was wrong. If you "test" the program on a fuctnion like f(x)=2x-1 on the interval [0,1], whose only root is at the midpoint, 1/2, then without the =, the program goes from [0,1] to [.5,1] to [.75,1], and loses the root entirely! With this change, the algorithm properly goes from [0,1] to [0,.5] to [.25,.5] to [.375,.5], etc. This is probably the way it should perform. My programming and logical skills have suffered another crushing insult. A simple program implementing the bisection method Here's a bisection program program written in the Maple programming language. I wouldn't use this for serious, real applications. Such numerical programs must have much more careful analysis of the kinds of errors which can occur and should be carefully tested. I hope the logic is clear. bisection := proc (f, a, b, E) local p, q, m; p := a; q := b; while E <= q-p do print(p, q); m := (1/2)*p+(1/2)*q; if f(p)*f(m) <= 0 then q := m else p := m end if end do end proc; The function f has been defined by the formula f(x)=x-cos(x) in another statement: f:=x->x-cos(x);. This bisection program prints out the intermediate stages, so you can see the program focusing on the interval in which the root sits. > bisection(f, 0., 1., 0.001); 0., 1. 0.5000000000, 1. 0.5000000000, 0.7500000000 0.6250000000, 0.7500000000 0.6875000000, 0.7500000000 0.7187500000, 0.7500000000 0.7343750000, 0.7500000000 0.7343750000, 0.7421875000 0.7382812500, 0.7421875000 0.7382812500, 0.7402343750 0.7392578125 When the program ends, it reports the last variable's value, the middle of the subinterval. We will get more sophisticated root-finding methods, but this one is simple to understand and to use. Algorithm? I mentioned the word "algorithm". Let me give some further information about this word in the form of quotes from The Art of Computer Programming by D. E. Knuth: The modern meaning for algorithm is quite similar to that of recipe, process, method, technique, procedure, routine, except that the word "algorithm" connotes something just a little different. Besides merely being a finite set of rules which gives a sequence of operations for solving a specific type of problem, an algorithm has five important features: Finiteness An algorithm must always terminate after a finite number of steps. Definiteness Each step of an algorithm must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case. Input An algorithm has zero or more inputs, i.e., quantities which are given to it initially before the algorithm begins. These inputs are taken from specified sets of objects. Output An algorithm has one or more outputs, i.e., quantities which have a specified relation to the inputs. Effectiveness An algorithm is also generally expected to be effective. This means that all of the operations to be performed in the algorithm must be sufficiently basic that they can in principle be done exactly and in a finite length of time. Knuth continues on the same page to contrast his definition of algorithm with what could be found in a cookbook: Let us try to compare the concept of an algorithm with that of a cookbook recipe: A recipe presumably has the qualities of finiteness (although it is said that a watched pot never boils), input (eggs, flour, etc.) and output (TV dinner, etc.) but notoriously lacks definiteness. There are frequently cases in which the definiteness is missing, e.g., "Add a dash of salt." A "dash" is defined as "less than 1/8 teaspoon"; salt is perhaps well enough defined; but where should the salt be added (on top, side, etc.)? ... a computer programmer can learn much by studying a good recipe book

Average and Instantaneous Rates of Change
A few lectures ago I tried to analyze a number of real phenomena. I hope that background should help you accept the following definitions:
The average rate of change of f in the interval [x₀,x₁] is (f(x₁)-f(x₀)/(x₁-x₀). Geometrically, this is the slope of a secant line connecting the two points (x₀,f(x₀)) and (x₁,f(x₁)) on the graph of y=f(x). The instantaneous rate of change of f at x₂ is a stranger thing, that probably can't be physically measured in most cases. It is the slope of the tangent line at (x₂,f(x₂)). The instantaneous rate of change of f at x₂ is better and better approximated by the average rate of change if the numbers x₀ and x₁ are close to f.

The definition
Consider lim_h-->0(f(x+h)-f(x))/h. If this limit exists, then the value of the limit is f´(x), the derivative of f at x, and we say that f is differentiable at x.

Most of this course will be devoted to studying the derivative of a function and its uses. Actually, the next few lectures will show that, for familair functions defined by formulas, the derivative can be computed fairly easily. So computation of derivatives, while both necessary and useful, is not the ultimate aim of the course (hey, such computation can be described carefully enough so that [capable] programmers can create differentiation programs!). We will spend most of the time investigating how to use derivatives.

Example
A traditional first example is f(x)=x². Then we need to consider (f(x+h)-f(x))/h. Let's look at the top of the fraction.
     f(x+h)-f(x)=(x+h)²-x²=x²+2xh+h²-x²=2xh+h²=h(2x+h).
In the last step, I factored out an h because I was thinking ahead:
     (f(x+h)-f(x))/h=(h(2x+h))/h=2x+h.
Now lim_h-->02x+h=2x. We're done.
Conclusion The function f(x)=x² is differentiable at all x's, and its derivative is given by f´(x)=2x.

A tangent line
What is an equation of a line tangent to y=x² at x=3? Here we need a POINT and a SLOPE.
     (3,9). If x=3, y=3³=9.
     f´(3)=2·3=6. The derivative's value at x=3 is the slope of the tangent line at x=3.
An equation for the tangent line at x=3 is therefore y-9=6(x-3). I'd probably leave the equation this way unless there was a reason to change it (if I were requested to provide the answer in a different form, or if I needed to compute with it more).

Is it correct?
Here I wanted to consider the formula for the derivative, f´(x)=2x and consider if the answer were reasonable. The graph of y=x² to the left of the y-axis is decreasing, and the tangent lines should be tilted "down". Their slopes should be negative. And the algebraic candidate we have for the slopes of tangent lines, 2x, is negative when x<0. On the other side of the y-axis, the tangent lines tilt "up", and their slopes seem to be positive. Of course, for x>0, 2x>0.
In this case, considering the answer and seeing that it is reasonable and consistant with other information is easy. Certainly in more complicated situations, such checks are more difficult. But if at all possible, within the limits of time and effort, please try to make such a check. Everyone makes mistakes: humans, computers, humans using computers, etc. A few seconds "thought" can catch errors that can be very irritating later.

And now for xⁿ
Everyone knows the answer. O.k., but why? (Why is "the answer" actually the answer, not why does everyone know it.) If f(x)=xⁿ where n is a positive integer, then:
     f(x+h)-f(x)=(x+h)ⁿ-xⁿ.
We need to consider (x+h)ⁿ. It is possible, using the Binomial Theorem, to write an explicit exact expanded form of this object. I don't need that. I need much less information. So:
     (x+h)ⁿ=(x+h)(x+h)···( times)···(x+h).
There are lots of ways to multiply things out here: you need to choose in each factor either the left (x) term or the right (h) term. But how may ways are there which would get only x's? For this, we would need to make only the x choice each time. There is exactly one way to get all x's, so xⁿ would only come out one time. How many ways would result in one h and all the rest (n-1 of them) x's? Well, we could choose the h from the first term and choose all x's from the other terms. Or we could choose an h from the second term and all the others (the first term and the terms after the second term) x's. Etc. Here by "Etc." I mean that we could take an h from exactly one of n factors, with all the other choices being x's. So since there are n factors, there are n ways to get a product with one h and all the rest x's. So in the result, there's nhx^n-1. What about the rest? We took care of the "no h" term and the "one h" terms. So all the rest has at least two factors of h. So, actually, we now know:
     (x+h)ⁿ=xⁿ+nhx^n-1+h²JUNK.
In this expression "JUNK" is not something bad. It represents terms I don't need to care about at this time. In fact, later in the course, we will try to understand some aspects of JUNK and how they can be useful. Anyway, let's continue:
     (x+h)ⁿ-xⁿ=xⁿ+nhx^n-1+h²JUNK-xⁿ=nhx^n-1+h²JUNK=h(nx^n-1+hJUNK)
where again I factored the h out because I was thinking ahead. Now the limit:
     lim_h-->0(f(x+h)-f(x))/h=lim_h-->(h(nx^n-1+hJUNK))/h=lim_h-->0nx^n-1+hJUNK=nx^n-1.
Therefore, f(x)=xⁿ is differentiable, and its derivative, f´(x), is nx^n-1.

QotD
If f(x)=1/x, use the definition of derivative to find f´(x).

So f(x+h)-f(x)=1/(x+h)-1/x=(x-(x+h))/((x+h)x)=-h/((x+h)x) and lim_h-->0(f(x+h)-f(x))/h=lim_h-->0[-h/((x+h)x)]/h=lim_h-->0-1/((x+h)x)=-1/(x·x)=-1/x².
Interesting aspects of this computation: if f(x)=1/x, then f(x+h)=1/(x+h). You must understand the grammar (?) of functions to do this. And combining the fractions and converting to a simple fraction: you must know how to do this sort of algebra.

Limit test!!!
Please see here and be prepared for a short exam about limits next Wednesday.

          1        1 
        ----  -  ------
        3x+4      x2+6
 lim  -------------------
x-->2        x-2

"Plugging in" x=2 gets 0 on the bottom, and on top, if we substitute correctly, 1/(3x+4) becomes 1/10 and 1/(x²+6) becomes 1/10. So the top is (1/10)-(1/10) which is 0. Surprise (not!): this is a 0/0 situation. We use some algebra. My feeling is great dislike for compound fractions, that is, fractions within fractions. I find them difficult to understand and difficult to manipulate. My advice is to change them into "simple" fractions, where only one division sign will appear. But we need to do this carefully.

The top is 1/(3x+4)- 1/(x²+6) which is
(x²+6)-(3x+4)
-------------------
 (x²+6)(3x+4)
Notice that this fraction is sitting "on top" of (x-2). We have something like this:

   A         A      1            A
 -----     ----- · ---        -------  
   B         B      C           B·C          A
------- = --------------  =  ---------- =  -----    
   C              1                           B·C
             C · ---             1
                  C

Limits and algebra
If lim_x-->af(x)=L₁ and If lim_x-->ag(x)=L₂ then
    •  lim_x-->af(x)+g(x) exists and equals L₁+L₂;
    • lim_x-->af(x)·g(x) exists and equals L₁·L₂;
    • (when L₂ is not 0) lim_x-->af(x)/g(x) exists and equals L₁/L₂.

QotD (definitely less absurd)
(Please accept my guarantee that this is not absurb, at least for a little while [2 lectures].)What is

       1     1     
      --- - --- 
       x2    w2 
 lim  --------- ?
x-->w    x-w

   w2-x2    (w-x)(w+x)       w+x
--------- = ---------- = - ------
x2w2(x-w)    x2w2(x-w)       x2w2

A weird but not absurd limit
f(x)=|x+5|/(2x+10). So what happens as x-->7, for example? The x+5-->7+5=12. The absolute value for x's near 7 give x+5 near 12. The results are all positive, so |x+5|-->12. The bottom, 2x+10, approaches 2·7+10, which is 24. The resulting limit value is 12/24.

There's trouble near -5, though. So what about lim_x-->-5|x+5|/(2x+10)?

Here there's a real difficulty with absolute value. Recall that |w| is w if w>0 and |w| is -w if w<0. So |x+5| is x+5 if x+5>0. and |x+5| is -(x+5) if x+5<0. But x+5>0 is the same as x>-5, and x+5<0 is the same as x<-5. So put all this together:
suppose f(x)=|x+5|/(2x+10).
    If x>-5, f(x)=(x+5)/(2x+10)=1/2;
    if x<-5, f(x)=-(x+5)/(2x+10)=-1/2.

To the right is a graph of f. I don't "know" what the value of f(-5) should be (computer programs handle this differently, but many report that trying to evaluate the function at -5 is an error).

Such functions occur when trying to model real situations (hey: "hit" a plate -- that will be a shock! Or drop a chunk of salt into a container of water -- that's a shock to the salt concentration!). So getting some language to describe the behavior is useful.

As x gets close to -5 from the left, the values of f(x) are all -1/2. So people say that lim_x-->-5^-f(x), the left-hand limit of f at -5, exists, and they say that the value of this limit is -1/2.

As x gets close to -5 from the left, the values of f(x) are all 1/2. So people say that lim_x-->-5⁺f(x), the right-hand limit of f at -5, exists, and they say that the value of this limit is 1/2.

You may be confused at first with the superscript/exponent + and -. The "+" up there always means "from the right" and the "-" up there will always mean from the left. In the expression x-->-5⁺ the minus before the 5 means look 5 units to the left of 0. The plus sign in the exponent means see what happens to the right of that number (that is, -5), but look close to it.

A piecewise linear graph and some limits

I drew a graph similar to what's shown to the right. The idea was to use the geometric information on the graph to find out what we know about function values and limits. I also wanted to urge students to understnad the possibilities of our algebraic "language". So to the right is a graph of y=f(x). Some questions:

• What is f(1)? What is lim_x-->1f(x)? (This is the easy one!)
  I see that the point (1,3) is on the graph of y=f(x). A point is on the graph if it has the form (x,f(x)), so f(1)=3. Also, as x gets close to 1, the heights on the graph, the f(x)'s, get close to 3. So lim_x-->1f(x)=3.
  The function y=f(x) is continuous at 1. This is because the limit as x approaches 1 of f(x) exists, and 1 is in the domain of f, and the value of f at 1 is equal to the limiting value of f.
• What is f(4)? What is lim_x-->4f(x)? (This is not so easy.)
  Well, f(4)=1 since there a big dot at (4,1). What about the limiting behavior as x-->4? A key part of the idea of limit is that what matters is behavior near the point but not the behavior at the point. Now as x-->4 from either the right or the left, the height, f(x), gets close to 2. I think that lim_x-->3f(x) exists, and that the limit is 2.
  Since the limiting value is not equal to f's value at 4, f is not continuous at 1.
• What is f(2)? What is lim_x-->2f(x)? (A bit stranger.)
  If I consider points whose first coordinate is 2, points with coordinates (2,?), there are no such points on the graph of y=f(x). So 2 is not in the domain of f. What about lim_x-->2f(x)? Here the heights of the function depend on which side x gets close to 2. If x gets close from the left, then the f(x) values are close to 1. So I know lim_x-->2^-f(x) exists, and its value is 1. But if x gets close to 1 from the right, the heights, the f(x) values, seem to get close to 0. So now lim_x-->2⁺f(x) exists, and its value is 0. Since the left and right handed limits do not agree, the (two-sided) limit, lim_x-->2f(x) does not exist.
  Since the limit does not exist, the function can't be continuous at 2.
• What is f(-1)? What is lim_x-->-1f(x)?
  Look at the graph. There's a nice thick dot at (-1,2). Therefore I conclude that f(-1)=2. The two-sided limit at -1 of f does not exist. As x gets close to -1 from the left, the f values (f(x) is the height of the curve) get close to 1. So lim_x-->-1^-f(x)=1. Look now at the right side of x=1. The f(x) values drop go to 2, so I conclude that lim_x-->-1^-f(x)=2.
  f is not continuous at -1. To be continuous at -1, all these must exist and be the same number: f(-1) and the left-hand limit at -1 and the right-hand limit at -1.

Lots of oscillations, I
I asked people to consider the function f(x)=sin(1/x). The range of this function is certainly easy enough to understand. The outputs are outputs from sine, and therefore they are numbers from -1 to 1. The domain is all x's except 0 (division by 0 is bad).

What makes this function interesting is its wild behavior near 0. When x is very small, 1/x is very large. So a short interval of x's near 0 gets changed to a long interval of numbers away from 0, and then these numbers are "fed into" sine. For example, look at the interval of x's between 1/5Pi and 1/4Pi. These numbers work out to 0.064 and 0.079. If x is between 1/5Pi and 1/4Pi, then 1/x is betweeb 4Pi and 5Pi. The sine curve oscillates once, from 0 to 1 and back to 0, between 4Pi and 5Pi. Therefore the function f(x)=sin(1/x) goes from 0 to 1 to 0 in the interval [0.064,0.079]. Hey! Look at the x's between, say, 1/(100Pi) and 1/(101Pi). Now 1/x goes from 100Pi to 101Pi. And, again sine's values go from 0 to 1 to 0. So there is another "blip" in the graph. Of course I've been looking at stuff from even multiples of Pi to odd multiples of Pi. If I had looked from odd multiples to evens, I would have seen blips down, from 0 to -1 to 0. Hey, the sin(1/x) curves oscillates infinitely often from 0 to 1 to 0 to -1 and back to 0 in every small interval centered around 0. Certainly lim_x->0sin(1/x) does not exist, and neither do the left- and right-handed limits.

Computing machines (calculators and other graphical devices) have a finite number of pixels and a finite ability to consider numbers. They usually can't show too much detail of the curve y=sin(1/x), and sometimes very strange results occur. You should check what your calculator does. Here is a graph drawn by Maple on [-1,1]. Maybe it is slightly better than a calculator-drawn graph, but it still can't cope with all the oscillations.

Lots of oscillations, II Here I looked at f(x)=x·sin(1/x). This function takes the previous result and multiplies it by x. So the heights will vary. The sine outputs are between -1 and +1, but then we multiply by x. If x is very small, then the function result is very small. Actually, since -1<=sin(1/x)<=+1, so -x<=x·sin(1/x)<=+x. This graph will always be between the lines y=x and y=-x. It will touch the lines each time sin(1/x) is +1 or -1, and this happens (look at the preceding analysis) infinitely often. So what happens here to the limit as x-->0? The heights are being squeezed down to 0. I claim that the outputs all get close to 0, and that limx-->0f(x) exists, and the value of the limit is 0.

Now another function ...
Here is one of very important limits of calculus. It will be built into many of the methods we will use. We should understand it. So:
What can we say about lim_x-->0[sin(x)]/x? Please be very careful. Here the division is outside of the function evaluation. If we try to "plug in" we again get the equivocal 0/0, and that is not helpful. I suggested that we plot f(x)=[sin(x)]/x.

Here is a plot of y=[sin(x)]/x. It doesn't look too strange. Certainly, this seems to suggest that limx-->0[sin(x)]/x=1. In fact, this is true, when x is measured in radians.

If you insist on using degrees, a calculator, with window properly adjusted, will show you something like the graph to the right for a plot of y=[sin(x)]/x near 0. This graph seems almost flat, and it seems to show that IF YOU MEASURE ANGLES WITH DEGREES then limx-->0[sin(x)]/x=.017, which is also true. Please note that the vertical and horizontal scales are not the same in this graph.

Degrees or radians?
Choose between the numbers .017 and 1: which is better to compute with? Everyone I know likes 1 better. Therefore everyone analyzing periodicity with calculus will use the trig functions with radian measure.

What is the limit?
The limit of [sin(x)]/x as x-->0 is 1. Better, because it is so important:

         sin(x)
 lim  --------  = 1
x-->0    x

Hey, engineers, it is sometimes handy to know that if |x| is very small, then sin(x) and x are about the same. For example: Values of sine Difference between sin(x) & x sin(.12345)=0.1231366779 sin(.12345)-.12345=-0.0025380486 sin(.00678)=0.006779948056 sin(.00678)-.00678=-0.51944·10-7 The differences get very, very small. In fact, we can predict how small the differences get using calculus --later in the course.

The Squeeze Theorem
The Squeeze Theorem is a result which is occasionally useful. Here are the hypotheses:

IF f(x)<=g(x)<=h(x) and
IF lim_x-->af(x) and lim_x-->ah(x) both exist and
IF the two limits are equal,
THEN lim_x-->ag(x) exists, and equals the common value of the other two limits.

Now for this course ...
I mentioned the following ideas.
COMPETITION The course is not a struggle between students. The grades are given essentially on an absolute scale, so one student getting a high grade does not imply another student will get a lower grade. Please help each other. This is a good thing.
COMBAT This is also not a struggle between students and the instructor. It is my job and strong desire to help you do well. This means learning calculus and learning how to write about calculus. My comments and grades are part of my effort to do this. Pleae take advantage of the interaction given by the course. It isn't as much as we would ideally like, due to economic factors, but try to take advantage of what there is.

I returned workshops and QotD's to students who attended.

The idea of limit
Today we begin an official discussion of limit. This is a very useful idea. There are lots of computations and manipulations which are related to limits. But I must be very careful about how class time is used, and I must rely on you to read the textbook and do homework problems. I will choose to emphasize ideas in the lectures. I will try to do a sufficient number of examples to illustrate these ideas, but your own efforts will be key to getting enough experience and sufficient familiarity.

Tree growth
Catalogs sell trees with short descriptions which frequently include such information as "The eventual height of this tree is 40 ft and its growth rate is 2 ft/yr." What do these numbers mean? We discussed this for a while.

First approach
If one took the 2 ft/yr literally, then in a century the tree would be 200 ft high, and a bit later (well, quite a bit later!) the tree would knock down the moon. This is probably not realistic.

Another attempt to understand ...
We made a second attempt to understand. The tree would grow at 2 ft/yr, and then, after 20 years (40/2=20) the tree would stop growing. Well, if we plotted the growth curve, it might look like what is shown to the right. To me, a simple piecewise linear curve is really neat and simple. But I don't think this represents behavior of complex organic objects. For example, I don't think a really tiny tree would actually grow two feet taller in its first year. I think it would begin by growing rather slowly. I also don't think that trees grow at a steady rate, and then tree growth would suddenly and totally stop at some specific height. That's probably not what happens.

Maybe more real ...
Reality is sloppier. I bet that the tree begins its growth fairly slowly. Then the growth gets more rapid and steadier, and persists for quite a while until the tree gets near its "ultimate height". The tree growth slows then, but I would expect more growth, at a slower rate, for a while. I think that one more reasonable possibility for a growth curve is shown to the right.

In fact, I think what's shown to the right is, itself, only an imitation of real tree growth. Growth will depend on climate and weather (moisture, temperature), and nourishment (how much of what kind of minerals, etc.), and competition (in the shadow of something else?), and such factors as where is the lumber harvester. Each tree will likely be somewhat different, and what's presented and described is an average of many observations.

How could tree growth around, say, year 13, be calculated? Well, suppose H(t) be the height of the tree in feet, at time t years after the tree seed germinates. Then we could measure the height of the tree once a year. So we'd have values for H(12) and H(13) and H(14), etc. The growth rate in feet per year for the year [13,14] would be (H(14)-H(13))/(14-13). This number would represent the average growth rate of the tree during that year. Of course, we could want more precise information. I could imagine someone measuring the tree height every 3 months, a quarter of a year. The interval [13,13.25] would be closer to 13. And I would think that the quantity (H(13.25)-H(13))/(13.25-13) would be a better idea of what might be the tree growth rate "at" 13. Of course, this fraction is what's called the average growth rate in the interval [13,13.25]. We could imagine an enthusiast desiring better information. This person could want 10,000 measurements of the tree height during the year, etc. This information could be used to give more precise information about tree growth rate. Now think about real life. I don't believe that 10,000 tree measurements are likely. I think that approximate average rates of growth are all we can expect, and the idea of some totally precise "growth rate" is fictional. But, even though it is fictional, such an ideal growth rate may be useful.

Diary entry in progress! Rock dropping
I used feet and pounds, an antique system of measurements in this discussion. The units confused people. I am sorry.
Suppose we drop a rock. Then the rock will fall approximately s(t)=16t² ft after t seconds. If we wanted to know the average velocity that the rock fell in the one second after t=3, we could compute:

s(4)-s(3)  16(42)-16(32) 
-------- = ------------= 16(16-9) = 112 
 4-3             1

s(3.07)-s(3)  16(9.042049)-16(9)   16(.042049)
----------- = ------------------ = ----------- = 96.112
  3.007-3           .007               .007

Let's now use algebra. Algebra allows almost everyone to be clever. So we can ask: what is the average velocity of the rock during the time interval [3,3+h] if h is a small positive number? The computation would be something like this:

s(3+h)-s(3)  16(3+h)2-16(32)   16(9+6h+h2)-16·9    16(6h+h2)  
---------- = --------------- = ---------------- = ---------- =16(6+h)
    h              h                  h               h

This is actually a rather remarkable computation, and just because you may have seen such things before is no reason to ignore the wonder: you must be like a child .... Let's check the formula. If h=.007, then 16(6+h) is 16(6.007)=96.112 (that's what we got before). To me, the most remarkable part of the computation is that we have a quotient, a tiny distance change on top, and a tiny time change on the bottom. Amazing, somehow the "tiny"'s sort of cancel, and we are left with an average velocity, 16(6+h), which has a sort of stability property as h gets smaller. In fact, if you don't think too hard, as h gets smaller ("h-->0") then 16(6+h) gets close to 16(6)=96 ("16(6+h)-->96").

So the average velocity over the interval [3,3+h] approaches 96 as the length of the time interval goes to 0. People abbreviate this by declaring that 96 is the instantaneous velocity of the rock at time 3. As far as I know, human beings don't actually measure or observe instantaneous velocities. These instantaneous things are lies, o.k., we shouldn't call them lies, they are useful fictions or mental constructions. Most of this course will be about various useful fictions.

Local linearity
Last time we considered the curve y=3^x and zoomed in on it around the point (0,1). We observed that the curve seemed more and more to appear like a straight line. Hey: I even gave this phenomenon a name: local linearity. The line goes through the point (0,1), and therefore complete information about it only depends on knowing its slope. Well, if we think that the line is about the same as the curve, on a very very small scale, then maybe another point on the line would approximately be (h,3^h) if h is a small positive number. The slope is then the difference in the second coordinates divided by the difference in the first coordinates: (3^h-1)/h (on the bottom, h-0 is just h). In this case, unlike the low-degree polynomial investigated with the rock, the h's don't obviously somehow fade away. We guessed using numerical evidence that as h-->0, this quotient also "stabilizes" near the value 1.1. This behavior is not obvious to me algebraically, and, indeed, if I wanted to verify it for you right now, showing the limiting behavior would take a great deal of effort.

A function which isn't locally linear
Look at the graph to the right, and specifically look "near" (0,0). Here I am deliberately omitting the scale marks on the axes, because they don't matter for the purposes of this discussion. If we zoom in around (0,0), we get something which looks much the same. This is a self-similar object and scale doesn't matter. (Self-simlarity is a key part of what's called fractals, one natural (?) appearance of which is here).

There's no magnification which will make the graph to the right look like a straight line through the point (0,0). So this is a graph of a function which is not locally linear. In this course, most of the functions we will consider will be locally linear, so you should see that some rather simple graphs will just be thrown out of consideration.

Please realize that the graph to the right is associated with |x|, and is not an abstract, weird invention.

Observations and ideals
We, humans, can observe average rates of change. We can measure how balls roll down inclined planes. We can, with considerable more ingenuity and equipment (go over to the Chem Department!) make observations of chemical reactions on a near-nanosecond scale. As far as I know, we look at average rates of change. A nice invention is the instantaneous rate of change, which will be a principal object of study in this course. As far as I know, we do not observe instantaneous rates of change, we can only approximate them by real measurements. This course studies the useful fictions related to instantaneous rates of change.

What is a limit?
The idea of a limit took several hundred years to evolve and understand. We try to recaptulate ("summarize briefly") this development today in about 70 minutes!

What is a limit?
"Real" functions are not very exact. A chemical engineer might have (I'm simplifying hugely) some process which produces the correct kind of plastic if, say, a certain amount of benzene is used in the mix. So a certain percent of benzene, say 3%, may produce a plastic which has the desired amount of, say, translucence. But in reality, even very precise measurements may not create an input with exactly 3% benzene. Maybe some days we get 3.5, or 4.2. And maybe the desired output measurement is not necessary -- we can deal with some error in the output. Real measurements imply that we need to contemplate error, not as a moral defect (at least in this case), but as part of our mathematical model which we must deal with.

Output tolerance as controlled by input tolerance
So maybe we need to understand the following idea: we want a certain output tolerance: that is what we can "live with" in the material created. Is there an input tolerance describing inputs in an interval around the ideal input which makes the corresponding outputs close enough to the desired output? The existence of such a relationship is what's meant by limit.
lim_x-->af(x)=c means that if some output tolerance is specified, so we want |f(x)-c|<some acceptable output error, then there is some input error or wiggle that will force that acceptable output error.

Very precisely, here is what is meant by limx-->af(x)=c: Given some positive OTE (Output Tolerance Error), there is a positive number AIE (Acceptable Input Error) so that: If 0<|x-a|<AIE then |f(x)-c|<OTE. Maybe this is too darn abstract. Let me consider a numerical example, with a very simple function. x2 near x=3 If the input to f(x)=x2 is 3, the output will be 9. Suppose we are willing to live with a +/- error of 1 in the output. That is, we can tolerate |f(x)-9|<1. Is there some simple specification of an interval of inputs near 3 which will guarantee this? In this case, almost everyone thinks that the simplest kind of input specification would be an interval of x's centered at 3 (your feelings may differ here, but this is what is usually done). So I ask if there is some convenient number ("CN") so that if |x-3|<CN then |x2-9|<9. Most people don't want an exact, precise, most perfect (?) value of CN. It may be difficult to get something like that. They are willing to settle for something convenient. In this case, the graph to the right shows the needed output tolerance using horizontal red lines. The next graph, shown here to the right, adds part of the vertical lines corresponding to x=2.9 and x=3.1. The part of f(x)=x2 shown which is between the vertical blue lines is "clearly" forced to be between the red horizontal lines. This geometry reflects the following algebraic statement:      If |x-3|<.1 then |x2-9|<1. So an input tolerance of .1 around 3 will guarantee that the output tolerance of 1 around 9 is true. Change the output tolerance to .2 What if we changed the desired output tolerance to .2? That is, we want to get |x2-9|<.2 by controlling the size of |x-3|. Here's the picture of f(x)=x2 with the two horizontal red lines indicating the output tolerance needed. Now here are the "old" input tolerance lines which we just used. Please note that this input tolerance won't work: there is part of the graph of x2 between 2.9 and 3.1 which is not between y=9-.2 and y=9+.2.1. Here is a display showing a satisfactory input tolerance. I used .1 as the input tolerance (yes, I experimented a bit). I repeat that in reality people rarely care about the "best possible" input tolerances. They usually just want to get some number that works. To me this graph doesn't show very much. So let's try a different scale. I hope that you can see here a graphical "verification" of the algebraic implication:      If |x-3|<.01 then |x2-9|<.2 An input tolerance of .01 satisfies the output tolerance of .2.

I do think that a few of the students in Math 151 may need to get this more intimate (?) view of limits, but not more than a few of the students. Most students will need to understand how to compute limits for familiar functions and familiar combinations of functions. There are some standard techniques. The first one is almost ludicrous, but I recommend it strongly.

Plug in
Suppose you want to understand lim_x-->af(x). If the function is defined by some formula you understand, plug in. That is, evaluate f(a).
Example What's lim_x-->3x²? I think it is 3².

This "trick" is important enough so that there is a label which goes with it. A function f is continuous at x=a if the domain of f includes an interval surrounding a, and if lim_x-->af(x) exists, and if the value of that limit is f(a).
In other words, the function is continuous at x=a if the limit of the function as x "approaches" a is gotten just by plugging in a to f: computing f(a). Due to many facts about limits that we will mention next time, most familiar functions are continuous in their domains.

Many of the most interesting limits we need to deal with (those involving rates of change, for example) can't be evaluated by just plugging in. Here are two examples: lim_x-->0(3^x-1)/x and lim_x-->0(sin(x))/x. In both cases, plugging in gives 0/0. In fact, both limits do exist. The value of the first one is ln(3) and the value of the second one is 1. Neither fact is obvious, and some effort is needed. But sometimes limits can be seen by (relatively!) simple algebraic manipulation. We will try to change them into forms which can be evaluated by "plugging in".

Algebraic transformation, with the goal being ...
I think we looked at something like lim_x-->2(x²-3x+2)/(x-2). First plug in. The top becomes 2²-3·2+2=4-6+2=0. The bottom is also 0. I don't know what 0/0 means. But we can realize that the top can be analyzed:
x²-3x+2=(x-2)(x-1). Therefore, if x is not equal to 0, (x²-3x+2)/(x-2)=[(x-2)(x-1)]/(x-2)=x-1.
We are dividing the top and bottom of the fraction by x-2. This would not be valid if x were equal to 0.
Two comments: first, I'll rarely be so careful as to mention restrictions like "x not equal to 0" again. I'm just not that pedantic ("marked by a narrow focus on or display of learning especially its trivial aspects"). Second, the "x not equal to 2" is actually part of the precise definition of limit, if you look either in the text or what's written above. We don't need to know what happens exactly at x=2. We want to know what happens when x is close to 2. Back to the limit:

Therefore lim_x-->2(x²-3x+2)/(x-2)=lim_x-->2x-1=1 (just by plugging in).

Another example Graphical example
Suppose we have a piecewise-defined function:

     1-x2 for x<2
f(x)=I don't care (!) for x=2
     2x+1 x>2

What can we say about lim_x-->2f(x)? If we first restrict attention to x<2, we can use one formula:
For x<2, lim_x-->2f(x)=lim_x-->21-x²=1-2²=-3. This limit is just "plugging in".
Now if we consider x>2, use the other formula:
For x>2, lim_x-->2f(x)=lim_x-->22x+1=2·2+1=5. Again, the answer uses "plugging in".

I hope that if you look at the picture you can see these two results. But what about lim_x-->2f(x)? There is no unique one number which occurs as output when x gets close to (but not equal to) 2. The outputs get close to -3 (if inputs are from the left) and they get close to 5 (if the inputs are from the right). So we say that "this limit does not exist". If a limit exists (without any further description, so without the qualifications of left or right) then there should only be one value.

One further algebraic example
I think we looked at the more complicated
lim_x-->w(x-w)/(sqrt(x)-sqrt(w)).
Here plugging x=w in just gets you 0/0, which I don't understand. I can see two ways of analyzing the fraction inside the limit.

Method 1 Multiply by the conjugate. Here this means:

     (x-w)                  (x-w)((sqrt(x)+sqrt(w))       (x-w)((sqrt(x)+sqrt(w))  
----------------- = ----------------------------------- = ------------------------ = sqrt(x)+sqrt(w)
(sqrt(x)-sqrt(w))   (sqrt(x)-sqrt(w))((sqrt(x)+sqrt(w))           x-w

      (x-w)         (sqrt(x)-sqrt(w))(sqrt(x)+sqrt(w))  
----------------- = ---------------------------------- =(sqrt(x)+sqrt(w))
(sqrt(x)-sqrt(w))           (sqrt(x)-sqrt(w))

Use any valid method that works for you, please. Here's another view of conjugation

QotD
What is lim_x-->3(x²-4x+3)(x²-x-6)? Plugging in gets
(3²-4·3+3)(3²-3-6)=(9-12+3)/(9-3-6)=0/0
so that doesn't work. But:

(x2-4x+3)   (x-3)(x-1)   x-1
--------- = ---------- = ---
(x2-x-6)    (x-3)(x+2)   x+2

3⁶ means 3·3·3·3·3·3. That is, 3's repeated and multiplied 6 times. Therefore
     RULE 3^{positive integer} means 3's multiplied together that "positive integer" number of times.
Then, surely, people noticed things like 3⁵⁺² is 3⁷, seven 3's multiplied, and this is the same (associativity) as 3⁵·3²: just line things up and multiply them.
     RULE If n and m are positive integers, then 3^n+m=3ⁿ·3^m.
Well, then, this "rule" is neat and easy to remember. But consider the following "equation" (quotes because until you decide what some of the symbols mean, the equation is silly): 3⁶=3⁶⁺⁰=3⁶·3⁰. If this equation is valid, then 3⁰ should mean 1. Making this equation correct would mean that the simple template just above would be true for n and m non-negative integers, and then I (and other people) would have less to remember.
     RULE 3⁰=1.
If we want to extend the n+m equation more, consider the following "equation" (quotes again because we are exploring how the symbols work): 3⁰=3^7-7=3^7+(-7)=3⁷·3^-7. Then since we want 3⁰ to be 1, 1=3⁷·3^-7 so that 3^-7=1/3⁷. Well, then, here's another definition:
     RULE If n is a positive integer, then 3^-n=1/3ⁿ.
So, for example, 3^-2 must be 1/9. Finally, what happens if we repeat exponentiation? Let's consider (3⁴)². What is this? Well, 3⁴ is 3·3·3·3 (four 3's multiplied). Then squaring this gets us (3·3·3·3)·(3·3·3·3). If you count this is eight 3's multiplied: 3⁸. So therefore (3⁴)²=3⁸. Therefore we get another:
     RULE If n and m are positive integers, then (3ⁿ)^m=3^n·m.
But what about, say, 3^7/2? If we believe in the previous result, or, better, we want to use it as a simple template for computation, then consider (3^1/5)⁵. This "should" be 3^(1/5)·5=3¹=3. So therefore 3^1/5 should mean the positive number which is the fifth root of 3, the unique positive number which solves x⁵=3.
     RULE If n is a positive integer, then 3^1/n is the n^th root of 3, that is, the positive solution of xⁿ=3.
People have adapted these rules so that exponentiation is easy to handle.

If b is a positive number, then: bn=b·b·...·b n times. b0=1. bp+q=bp·bq. b-p=1/bp. (bp)q=bp·q. Although maybe these rules are all "obvious" to you, in fact I hope that you see they are designed so people and numbers live together comfortably. To a certain extent, math is a human construction. Certainly mathematical notation is human. The obviousness is partly due to your constant exposure to such equations (since maybe fifth or sixth grade?). I remarked that the repeated exponentiation equation, obvious and clear, is an important part of RSA encryption, a very widely used method for secure electronic communication (for example, commercial web transactions). Google has two and quarter million links to the phrase "rsa encryption".

The definition (?) of 3^sqrt(2)
I mentioned this in class, although I suspect it seemed weird and totally useless to most students. Very briefly, the problem is how to define and compute such things as 3^sqrt(2). The previous discussion showed how to compute, say, 3^77/17 First find the 17^th root of 3 and then take the 77^th power of that root. But sqrt(2) is not rational. It can't be written as a quotient of two integers. Numbers like 3^sqrt(2) will occur in calculus, and we may want to get some approximation of them. So what's done? First, get an approximation of sqrt(2): 1.414213562, for example (!). This is rational (you may not like it, but it is 1414213562/100000000. Then compute 3^1.414213562 which is about 4.728804386. And this is what is done.

The curve y=3^x investigated
I asked people to graph y=3^x on their graphing calculators. Below are some results.

To the right is a graph of y=3x in the window -1<=x<=1, 0<=y<=2. I think the curve looks sort of the way everyone thinks it should. This window has the point (0,30), which is (0,1), in the middle of it.
Please keep track of the dimensions of the window. Here I've "zoomed in". The window is -.25<=x<=.25, .75<=y<=1.25. This is again a square window. The zoom is a factor of 4. The bend in the curve is still visible to me, but the amount of curviness is less.
The zoom is higher. Now the square window has these dimensions: -.01<=x<=.01, .99<=y<=1.01. The curve is, to my weak human eyes, just about flat. The phrase "locally linear" is used for functions like 3x. This phrase means that if the graph of the function is zoomed at enough, the result looks very much like a straight line. In fact, the functions of interest in this course will be exactly the locally linear functions. The straight line, if you look at it carefully (some students did) has slope approximately equal to 1.1. (Hey, (3.05-30)/(.05-0) is about 1.12).
We then went through the same sequence of graphs for y=2x. I'll skip to the third graph. So what's shown to the left is the third result, which is y=2x in the window -.01<=x<=.01, .99<=y<=1.01. Again, this curve is locally linear, and the window is sufficiently small so that the piece of graph visible looks to me like a segment of a straight line. The slope of this line is about .69. (A computed value of (2.05-30)/(.05-0) is about .705.)

Exponential functions are very commonly used to model such phenomena as radioactive decay and bacterial growth and lots of other things. The local linear approximations are things we will work with. I don't find 1.1 ("slope" of 3^x near (0,1)) and .69 ("slope" of 2^x near (0,1)) particularly nice numbers to work with. I would rather work with 1. We can adjust the local slope of b^x by varying b. There is a unique number b (between 2 and 3) so that the local slope of b^x at (0,1) is 1.
This number is called e. To the right is a graph of 2^x in blue and 3^x in green and e^x in red. The graph does show e^x "sandwiched" between the other two exponentials. It doesn't clearly show that e^x has the desired local slope of 1, but this is actually true. We will later see how to compute e with as much accuracy as desired. e^x is widely known as the exponential function, and frequently referred to as exp(x).

e is approximately 2.71828, and I sort of doubt that more digits will be needed by almost anyone in the class for practical computation. e is not rational, and its decimal expansion does not repeat. If you wish, here are about the first 2 million decimal places of e. People have computed billions of decimal places of both e and Pi. What else is there to do?

Graphing 5e^-2x+3e^4x
I asked people what the graph of f(x)=5e^-2x+3e^4x would look like. I admire "technology" (that is in this case, graphing devices) but to some extent one should learn what to expect. So here is sort of a thinking process, a way we go through the graphing step-by-step. The pictures in the process won't be totally accurate, but I will try to draw them as I see them. I hope this helps.

Start with y=ex. We know what this looks like. The domain is all reals and the range is all positives. The point (0,1) is on the curve. The curve goes down to 0 as x-->- and goes up to "+" as x-->+. BACTERIAL GROWTH!
Now modify to y=e4x. Multiplying the x by 4 inside the exponential function makes the function increase faster as x gets large positive, and also makes it increase faster as x gets large negative. Another way to think of this is by the formula e4x=(ex)4. The fourth power means that numbers bigger than 1 get bigger (the outputs of exp to the right of 0) and numbers between 0 and 1 get smaller (the outputs of exp to the left of 0). So a point whose second coordinate is 2, say, gets changed to one whose second coordinate is 24=16, and one with second coordinate is 7 gets changed to a point with second coordinate 74=2401. The point (0,1) has height 1, and since 14=1, (0,1) is still on this curve. Or if you wish, think this way: e4x=(e4)x. Since e4 is about 54.598, this curve is like y=54.598x, and the exp curve is about y=2.718x. Notice the 54.5980=1, so (0,1) is on the curve. BACTERIAL GROWTH!
And change it to y=3e4x. This stretches stuff by a factor of 3 in the output. That means that vertical distances get mutliplied by 3. A point with second coordinate 2 gets changed to one with second coordinate 6, and a point with second coordinate 7 gets changed to one with second coordinate 21. Please note that this is very different from the previous change, which was not a uniform change of length. The point (0,1) becomes (0,3). BACTERIAL GROWTH!
Now let's try the first piece of the function. What is e-x? Here the minus sign inside reverses x, so the graph gets flipped over the y axis. The domain is still all reals, the range is still all positive numbers, but the function is decreasing. (0,1) is on the curve. RADIOACTIVE DECAY!
Here we have e-2x. As you travel left on this curve, the curve moves up more. As you travel right, the curve gets closer to the x-axes. If you wish, think of e-2x=(e-x)2. Squaring makes numbers bigger than 1 bigger, and makes numbers between 0 and 1 smaller. (0,1) is on the curve since 12=1. Or, you could think this way: e-2x=(e-2)2, and this is about (.135)x, so certainly things shrink as x increases. And since .135 is smaller than 2.718 (e's approximate value) you get the behavior shown. (0,1) is on the curve since .1350=1. RADIOACTIVE DECAY!
We finish preparing this piece by looking at the graph of y=5e-2x. This is a constant factor stretch of 5 in vertical direction of the plane, and the result is shown. The point (0,1) has become (0,5). RADIOACTIVE DECAY!
Now we take both of the previous red curves and add them up. This picture is intentionally drawn a bit screwy. The constants involved (5, -2, 3, and 4) were also intentionally chosen to have no simple relationship or obvious symmetry. The result, the adding of the two curves, shouldn't be symmetric. I just wanted to get a rough idea of what was going on. This will allow me to check what a machine does. (Machines can make mistakes, and so can human beings, as the human beings use machines.)

Here is a Maple graph of f(x)=5e-2x+3e4x. Please consider the ridiculous difference in the vertical and horizontal scales. The total width of the bottom is 6 units, and the height is a bit more than 2,000 units. The exponential function gets really big really quick.

I also asked people where the coordinates of the "bottom" of the curve were. Several people responded, who were quick with their calculators. If we plot the curve with a different range of x's, we see what is to the right. Be careful again with the scale of the vertical and horizontal axes. A bit more work should shows that the bottom is fairly near (-.03,7.97). We will learn how to compute the coordinates of such points. Note The first graph I put here was wrong. I screwed up. Sorry. Human beings can make mistakes, and here I am an average representative of the species. Mr. Sloane spotted the error, and I thank him.

Inverting exp to get ln
The exponential function has domain all real numbers and range all positive numbers. It is increasing and 1-to-1. It has an inverse called ln for "natural log". I may sometimes slip and write log instead of ln. Almost the only logarithm function I use is ln. Since the point (10,e¹⁰) on on the graph of exp, we know that (e¹⁰,10) is on the graph of ln. Therefore ln(e¹⁰)=10. So, in words, ln(a number) is the number of powers of e "in" the number.
The picture to the right has a part of exp in red and a part of ln in blue. The dashed line in green is the "main diagonal" y=x. The graphs of the curves are obtained by flipping each over the main diagonal.

The exp graph increases very rapidly and the ln graph increases very slowly. Again, we will discuss this later in detail.

Logarithm "laws"
Some other logs are used in practice (base 10 for pH and decibels [sound intensity] and Richter scale [earthquakes] and base 2 for some computer science/engineering applications). But in this course almost surely we'll only refer to ln. The algebraic equations which we constructed (?) for exponentiation have counterparts for ln:

• ln(a·b)=ln(a)+ln(b)
• ln(1)=0
• ln(1/a)=-ln(a)
• ln(a/b)=ln(a)-ln(b)
• ln(a^b)=b ln(a)

How many digits does 2⁴⁰ have?
As one amusing exercise, I mentioned that ln(2) is about .7 and ln(10) is about 2.3. I asked how to use these values to see how many decimal digits the number 2⁴⁰ has.

Here is how this problem can be done. If 10^x=2⁴⁰, we can "ln" both sides. Now ln(10^x)=x ln(10) and ln(2⁴⁰)=40 ln(2). Therefore x is 40 times ln(2)/ln(10). This is about 40 times .7/(2.3), so about 40/3 which is near the integer 13. Indeed, 2⁴⁰ is exactly 1099511627776. Better: it is 109 95116 27776 (I put spaces between groups of 5 digits). So there are 13 decimal digits in 2⁴⁰.

I don't think this is a vital computation, but I do think that applied scientists and engineers should be able to get approximate sizes of reasonable numbers without exact computation.

QotD: solve exactly -10e^-2x+12e^4x=0
Here I expected something like this: 12e^4x=10e^-2x
e^4xe^2x=(10/12)e^-2xe^2x because I want to use 1/e^-2x=e^2x
e^6x=10/12 since e^ae^b=e^a+b and e⁰=1.
e^6x=10/12
6x=ln(10/12) since ln is the inverse of exp. x=(1/6)ln(10/12).

All of these answers are also correct:     ln({10/12}^1/6)     -(1/6)ln(12/10)     (ln(10)-ln(12))/6
and others (!) because they are all different names for the same number (properties of logs!).

As I mentioned to students who worked on this, I would like you to make mistakes while doing the QotD, and get rid of them now, rather than on exams, when such errors will be more painful and costly. By the way, the approximate numerical value of the answer is -.03 which is not a coincidence. (1/6)ln(10/12) turns out to be the first coordinate of the "bottom" of the curve sketched earlier in the lecture. Stick with the course and I will explain how this all works.

The second pre-class problem
Again, some happy volunteers, Mr. N. Patel and Ms. Rana, worked on this. The function f was defined by the formula f(x)=sqrt(x)+sqrt(3-2x). I first asked what the domain of this function was. Since sqrt(x) occurs, I know that x>=0. But sqrt(3-2x) is also present, so 3-2x>=0, which is equivalent to 3>=2x which is equivalent to 3/2>=x. Both of the restrictions must be satisfied, so that the domain is the closed interval [0,3/2] (both endpoints are included).
I also asked that a graph of the function be given. Here I expected and wanted people to use a graphing calculator!
I don't know the exact range of this function. We'll come back to this question later.

Why?
I discussed these pre-class problems because the answers to the first QotD were not correct. People gave graphs which were not graphs of a function, and did not seem familiar with the idea of composing functions. I hope these examples help.

The hierarchy of trig
There are several ways of looking at the trig functions. I tried to give a rapid review of the different conceptual levels and uses.

Level 0: measurement of sides of triangles
Several thousand years ago some smart people noticed that trigonometry can be used to make indirect measurements. For example, if you had a very tall tree whose height might be difficult to get directly, then you could more easily measure the shadow and the estimate the angle from the tip of the shadow to the tree top.

The idea is to notice that two right triangles which have one acute angle equal then have their other acute angle equal, and so they are similar. But ratios of corresponding sides of similar triangles are equal. So in the second diagram, we know that ?/!=??/!!. It may be that ?, for example, is difficult to measure directly. If we know the ratio ??/!! for the angle shown, and if we can measure ! directly, then the similarity equation will allow us to get ?. This is actually a very clever idea. People who used this idea actually collected data about the ratio of sizes of right triangles (they began to build a "knowledge base").

Ratios and special numbers
There are 3 sides to a right triangle. If you choose an acute angle, the standard English names for the sides are as Adjacent (ADJ) and Opposite (OPP) and Hypotenuse (HYP). These are shown to the right. Then there are 6 possible ratios to study. I'll refer almost always to three of these.
    sine of the angle=OPP/HYP;
    cosine of the angle=ADJ/HYP;
    tangent of the angle=OPP/HYP.
I'll rarely mention or use the other three ratios (secant, cosecant, and cotangent).

Fairly early in history some special right triangles were "discovered" and used frequently.

The triangle with sides labeled	Sine of the acute angle	Cosine of the acute angle	Tangent of the acute angle
	1/2	sqrt(3)/2	1/sqrt(3)
	1/sqrt(2)2	1/sqrt(2)	1
	sqrt(3)/2	1/2	sqrt(3)

Level 1: triangles in a circle
Consider the unit circle. In Math 151, "the unit circle" will mean a circle of radius 1 centered at (0,0). Take a point (x,y) on the circle. In the drawing to the right, the point is somewhere in the first quadrant. Draw a right triangle with one corner at the origin and one corner at (x,y), and the right angle sitting on the x-axis. Since the hypotenuse has length 1, the ratios defining the sine and cosine of the acute angle with vertex at (0,0) get simpler: the bottoms are both 1. The legs of the triangle are x and y long, so that the x and y coordinates are themselves equal to cosine and sine of that angle. The text and most other sources usually will call the acute angle at the origin , and I will try to do that also. Then the coordinates of the point on the unit circle are cos() and sin(). In fact, we could define those coordinates to be cos() and sin(), and then we'd be able to get more angles and more values of sine and cosine. For example, if =0, then the corresponding point on the unit circle is at (1,0), so that cos(0) should be 1 and sin(0) should be 0. And if =Pi/2 (a right angle), the point would be "at" (0,1), and then cos(Pi/2)=0 and sin(Pi/2)=1. Certainly tan(0)=0 and tan(Pi/2) is not defined.

Level 2: a moving point
Very few folks in this course likely will often use trig to find heights of trees. But the functions of trigonometry will be used to model periodic phenomena (body temperatures varying during the day, seasonal changes of weather, electrical and magnetic fields, etc.). These applications need a slightly different view. Consider a particle moving around the unit circle. The slightly obnoxious picture to the right is supposed to show such motion. A particle "starts" moving from (1,0) at time 0, and moves with unit speed around the circle in the positive (always, counterclockwise) direction. The coordinates of the particle dynamically change. You can almost read them off from the particle motion. The first, x-coordinate, is the left/right position of the point. It goes from 1 to 0 to -1 to 0 to ... This is cosine of the time it takes the particle to get to that position. The other coordinate, in one circular movement, starts at 0, goes up to 1, down to 0, then down to -1, and back up to 0. In fact, if we really think of the particle as moving for all time, the positions constantly cycle back and forth. The simplest way to measure angles, if we are going to consistently use this dynamic view, is with the time lable. Of course, this measurement is what is called radian measure. The preceding table then can be augmented as follows:

Angle measurement	The triangle with sides labeled	Sine of the angle	Cosine of the angle	Tangent of the angle
0, and 0o		1	0	1
Pi/6, and 30o		1/2	sqrt(3)/2	1/sqrt(3)
Pi/4, and 45o		1/sqrt(2)2	1/sqrt(2)	1
Pi/3, and 60o		sqrt(3)/2	1/2	sqrt(3)
Pi/2, and 90o		0	1	Not defined

It is not true that Babylonians had 360 fingers. Not everything you hear in class is totally correct.Dynamical view.

Graphs of trig functions
Now we can look at the graphs of the trig functions. Sine and cosine are geometrically the same, only starting at different "times" (after all, the circle is round and looks the same left/right as it does up/down).

The graph of sine
To the right is part of the graph of y=sin(x). The curve is what happens to the second coordinate of the imaginary particle. The sine function has domain all real numbers. The range is just the closed unit interval, [-1,1]. Since the imaginary particle described above travels around the circle forever, the sine function is periodic, with period equal to 2Pi. That is, for all numbers x, sin(x+2Pi)=sin(x).
The graph of cosine
To the right is part of the graph of y=cos(x). The curve is what happens to the first coordinate of the imaginary particle. The sine and cosine curves are congruent -- you can move one exactly onto the other. Sine begins at height 0 and goes up to 1. Cosine begins at height 1 and goes down to 0. The cosine function has domain all real numbers. The range is just the closed unit interval, [-1,1]. Since the imaginary particle described above travels around the circle forever, the cosine function is periodic, with period equal to 2Pi. That is, for all numbers x, cos(x+2Pi)=cos(x).
The graph of tangent
To the right is part of the graph of y=tan(x). It is very different from the graphs of sine and cosine. In one way, it is simpler. It consists of repeated "chunks" of horizontal width Pi, and each chunk when considered by iteself just goes up: it is increasing. But the graph also has breaks (where the cosine function is 0). The function has domain all real numbers except odd multiples of Pi/2 (such at -3Pi/2, -Pi/2, Pi/2, 3Pi/2, etc.). The range is all real numbers. Amusingly, tangent is periodic but its minimal (smallest) period is Pi, so that tan(x+Pi)=tan(x) (if x is in the domain of tangent).

It is neat to notice that most people (include me here!) usually draw sine and cosine incorrectly. One total wiggle of each of them occurs in an interval of length 2Pi, about 6.28 or six and a quarter. But the up/down is -1 to +1. So the darn curves aren't really as bumpy as many people think.

Addition formulas
The only one I could remember is
sin(₁+₂)=sin(₁)cos(₂)+cos(₁)sin(₂)
and this was embarrassing. You should look at the cosine formula, which is about the same order of complexity, and try not to look at the tangent formula, because it is slightly ferocious.

Amplitude and frequency Let me show some more graphs. I should have drawn more in class, because this comes up in many applications. Here is a graph of f(x)=3sin(2x). Please look carefully at the horizontal and vertical scale. Since values of sine fill up [-1,1], multiplying by 3 leads the height of the graph to fill up [-3,3]. The 3 in front affects the height, the amplitude, of the graph. What about the 2 inside? Multiplying by 2 there makes changes in x occur twice as quickly. Since sine repeats when the input increases by 2Pi, this will make the f(x) here repeat every Pi. Look carefully at the horizontal scale and check this. The frequency has changed. It has doubled.   And now f(x)=-1+4cos({1/2}x). Again I have asked the computer drawing this picture to make careful choices of the x and y scale. There is a {1/2} (that's "one-half") multiplying what is input to cosine. So changes will occur half as fast. Since cosine repeats every 2Pi, the effect here is to make the graph of this f(x) repeat every 4Pi. The frequency is halved. The 4 multiplying cosine on the outside makes a curve which natively would go from -1 to +1 stretch from -4 to +4. So the amplitude of the wiggling has changed. Last, the -1 pushes the whole curve down 1 unit. I hope you can see this. There are three changes, and it may be difficult to see all of them. The curve has range [-5,3]. Comment Thanks to Mr. Sloane who noticed an earlier misprint here. Sigh. I frequently (pun intended) confuse frequency and wavelength. They vary inversely. So when one doubles, the other one halves. Function Frequency Wavelength 3sin(2x) Twice as fast Pi -1+4cos({1/2}x) Half as fast 4Pi This, along with other reasons, is why I am not an engineer. Radios I'd design would pop and hiss, power lines would shake themselves to fragments, etc.

Functions and their inverses
Consider the function f(x)=x³+7. A graph of this function is shown to the right. I hope you recognize this as x³ moved up 7 units. Certainly f(2)=2³+7=15: the output corresponding to the input 2 is 15. Since f is a function, 15 is the unique, the only output corresponding to the input 2. But the function f has an interestping property. Suppose f(x)=15. This means x³=7=15 so x³=15-7=8. And x=8^1/3=2. The only input corresponding to the output 15 is the input 2. The geometric interpretation of this is gotten by considering the horizontal line y=15. This line intersects the curve y=x³+7 at exactly one point, (2,15). In fact something more general is happening.

The specific function f considered here is one-to-one (1-1). This means there is exactly one distinct input for each output. How can we verify this? If you believe the graph, the fact means that any horizontal line will intersect the curve at most one time. Algebraically, we can do this: y=x³+7 implies that x=(y-7)^1/3. If y is the output, we have just "constructed" the input, x, corresponding to the output, y.

One-to-one
If we have a 1-1 function, we can think about reversing the process, and the reversed process is the inverse function to the original function. Right now the idea is what I'd like to make sure of. There are certainly implementational issues to worry about, and these aren't easy. But I think the idea should precede computations in this case. The ludicrous diagram to the right is supposed to make clear (?) what's going on. If there is exactly one output value corresponding to each input value, then we could think about reversing this association, and creating an inverse function.

In the case of f(x)=x³+7 I know that f is 1-1 because I can solve f(x)=y and get exactly one value of x for each y. The function g(y)=(y-7_^1/3 is a function defined by the algebraic formula I got while solving. It is a formula for the function inverse to f.

If the point (x,y) which is (f's input, f's output) is on the graph of f, then (y,x) will be on the graph of the inverse to f. The switch (x,y) to (y,x) is geometrically done with a flip across the "main diagonal", y=x.

Part of the graph of the function inverse to this f(x)=x³+7 is shown to the right. It is the cubic curve flipped over the main diagonal.

Another example, not so easy
Piecewise linear functions are used in many computational applications. They look rather simple, but because they can wiggle a lot and they can be computed easily, they are very useful. I first "defined" the function with its graph. The function had domain (-,3]. Let's start at the top, at 3. The graph was a straight line segment connecting (3,2) to (1,-2). Then there was a line segment connecting (1,-2) to (0,1). Finally, the graph went left as a half-line, starting at (0,1), passing through (-1,0) and just traveling on left.

Some people feel uneasy about pictures, so I request that the class provide the relevant formulas. Only a few moments later (with the instructor roaming the class and urging on student efforts, like a wild slug), we got:

     x+1 for x<=0
f(x)=-3x+1 for 0<x<1
     2x-4 for x>=1 and x<=2

This function is not 1-1. For example, f(-1)=0 and f(1/3)=0 (I solved -3x+1=0) and f(2)=0 (I solved 2x-4=0). Geometrically the horizontal line y=0 (the x-axis) hits the "curve" three distinct times.

Therefore if we flipped the graph over the main diagonal, we would not get the graph of a function. Well, what can we do? If we insist on trying to undo (?) this function f, we could somehow break the function into pieces (as indicated by the colors on the flipped graph).

The left-most piece of f Consider f with domain restricted to x<=0. There f(x)=x+1, and, if y=x+1, then x=y-1. The function g1(y)=y-1 is inverse to f. Function Formula Domain Range f f(x)=x+1 (-,0] (-,1] g1 g1(y)=y-1 (-,1] (-,0] Please notice how the domain and range are interchanged for the function f (suitably restricted) and its inverse, g1. That is the way it should be, since the inputs and outputs are being interchanged. The middle piece of f Consider f with domain restricted to [0,1]. There f(x)=-3x+1, and, if y=-3x+1, then x=-(1/3)(y-1). The function g2(y)=-(1/3)(y-1) is inverse to f on this domain. Function Formula there Domain Range f f(x)=-3x+1 [0,1] [-2,1] g2 g2(y)=-(1/3)(y-1) [-2,1] [0,1] Again notice how the domain and range are interchanged for the function f (suitably restricted) and its inverse, g2. That is the way it should be, since the inputs and outputs are being interchanged. The right-most piece of f Consider f with domain restricted to [1,2]. There f(x)=2x-4$, and, if y=2x-4, then x=(1/2)(y+4). The function g3(y)=(1/2)(y+4) is inverse to f on this domain. Function Formula there Domain Range f f(x)=2x-4 [1,2] [-1,3] g3 g3(y)=(1/2)(y+4) [-2,3] [1,2] And in this final piece also, the domain and range are interchanged for the function f (suitably restricted) and its inverse, g3: inputs and outputs are being interchanged.

Diary entry in progress!

Squaring
I believe you could, with some seriousness, declare the the previous rather elaborate example was silly. It was technical, and overly caring about "logic". I would like to argue against this. My argument would consist first of discussion of a rather obvious function.

Consider squaring: f(x)=x². We know about this function. It is perhaps the most obvious (once the negative numbers are learned) example of a function which is generally (except for 0) not 1-1. Look: f(2)=2²=4 and f(-2)=(-2)²=4 also. Take the graph, a parabola whose vertex is at (0,0), and flip it over the main diagonal. The result is a curve which is not the graph of a function. "Everyone" knows this. Yet at the same time everyone would like to indicate square roots, and would like to tell that the square root of 4 is ... well, it could be -2 or +2. If a unique output is desired, the unique output which people have chosen is the non-negative answer (a classical word for this was "branch", really, apparantly because the flipped over curve might look like it has several slanting tree things). So sqrt(x) (really, better written with that strange almost-division sign) means the non-negative number whose square is x. So there has been a choice, in terms of the graph shown to the right. "Everyone" has decided to take the top, green, piece of the graph, and if ever the bottom part has to be used, it will be designated -sqrt(x).

Choosing an inverse
In this case the choice of a part of the function's domain and the inverse has been done in our culture years and years ago, and I think it is likely we have all grown up with it so nicely that we haven't thought about it!

The trig functions and their inverses
The six trig functions are all periodic. Therefore the difficulties of squaring actually become much more obvious. If sin()=1/2, well, then, actually, could be Pi/6 or 5Pi/6 or Pi/6-2pi or 5Pi/6-2Pi or Pi/6+2PI or 5Pi/6+2Pi or ... actually infinitely many other possibilities. Yet if you are interested in numbers, solving equations like sin() =some number for might be important (hey, I mentioned in class and meant the following: construct a robot arm, and "watch" its motion -- you'd better be able to understand how the values of the input and output of sine and cosine work!).

arcsine People have agreed to "restrict" the domain of sine to [-Pi/2,Pi/2]. Then the function is 1-1, changing the inputs [-Pi/2,Pi/2} to putputs [-1,1} (sine is increasing in that domain, and we will develop techniques to verifty such claims easily). So there is an inverse function. The text calls it sin-1 and I will try not to use that notation since a -1 in the exponent makes me think of reciprocal. The notation I will use is arcsin. Also I should mention that most of the computer languages I know use the designation arcsin to mean what I am describing here (since superscripts are difficult to type!). The graph to the right is arcsin. The endpoints are (-1,-Pi/2) and (1,Pi/2). I labeled these wrong in class and I am sorry. Everyone agrees that the domain of arcsin is [-1,1] and the range is [-Pi/2.Pi/2].

arctan
Tangent, from the inverse function point of view, is in some silly sense twice as bad as sine. ! This is because sine is 2Pi periodic and tangent is Pi periodic. The tangent domain restriction that people use almost always (99.999% of the time!) is (-Pi/2,Pi/2). The open interval results from the fact than tangent is not defined at odd multiples of Pi/2 (hey, cosine is 0 there). So flip the graph of tangent. The result is arctan (or, in the text, tan^-1).

The domain of arctan is all real numbers. The graph is increasing, and when x gets very large positive, then arctan(x) gets close to Pi/2. On the other side, when x gets very large negative, then arctan(x) gets close to -Pi/2. There are two (horizontal) asymptotes, y=Pi/2 and y=-Pi/2. As I mentioned, arctan is used to take big positive and negative numbers and output numbers which are more controlled, between -Pi/2 and Pi/2. arctan has been more used by me than tan. It is a neat function.

QotD
To the right is a graph of a function, f. It is piecewise linear, and has period 5. Values of f can be read from the graph. For example, f(0)=1 because (0,1) is on the graph, and f(3)=0 because (3,0) is on the graph. So:

Suppose g(x)=3f(2x)+4.
Compute g(0), g(1), and g(2). Graph g(x). What is the range of g?

g(0)=3f(2·0)+4=3f(0)+4=3·1+4=7.
g(1)=3f(2·1)+4=3f(2)+4=3·1+4=7.
g(2)=3f(2·2)+4=3f(4)+4=3·0+4=4.

We can do the graph of g in "stages".

y=f(x) The original graph: period 5 and range [0,1].
y=f(2x) The 2 is inside the parentheses. It affects numbers before they are inputs to f. It stretches the inputs, so they appear to change twice as fast. The frequency changes. This graph: period 2.5 and range [0,1].
y=3f(2x) The 3 is outside of f. It changes the values or outputs. It multiplies them by 3. The amplitude changes. This graph: period 2.5 and range [0,3].
y=3f(2x)+4=g(x) The "+4" moves the graph up 4 units. This graph: period 2.5 and range [4,7]. Please note that the three values found for g are verified on this graph.

Introduction
What's calculus about? The simple answer, which we will use as initial motivation, is to learn how to find and compute lines tangent to curves and areas of regions with curved boundaries. These problems are easy to state and most people like pictures. But truthfully very few of you will have any reason to compute tangent lines and areas after calculus courses. So here's the goal of the course:

We want to teach you how to model, analyze, and, to the extent possible, solve problems involving all sorts of rates of change and accumulation.

• For example, stretch a rubber band. Hooke's Law, which describes some of what happens, declares that the amount of stretching is directly proportional to the force. Well, this means if one ounce stretches the rubber band, say, 1/3^rds of an inch, then two ounces should stretch the band 2/3^rds of an inch. That's easy. But I don't think that putting 10 tons on the rubber band would make it many miles long. The simple model has restricted validity.
• Later in the course I will show you a picture of the solubility of sodium sulfate in water as the temperature varies. The result is quite (unexpectedly to me!) complicated. What model will describe this?
• Disease spread (for example, influenza) has been extensively studied and is quite important (how much vaccine should be made, and what types, etc.). The models are fairly accurate, but ... sometimes are complicated.
• Enzyme-catalyzed chemical reactions can be fairly accurately modeled using techniques of this course. We may see some simple examples later.

A simple toy model

Equal-sized squares are cut from the corners of an 8 inch by 11 inch rectangular piece of paper. The flaps in the resulting piece of paper are folded up. What is the resulting volume which the paper object encloses?

Althought this is a rather simple problem, already some interesting features are available. Notice that x=-50 or x=200 makes no sense in this problem. Also notice that V (in units of inches³) could not be -11 or 206,358 (I think!).

New
Math 151 has a new textbook this semester. The text will be used through the three semester sequence (151-152-251). It is cheaper and slightly shorter that the previous text, and perhaps it offers better exposition. Look out for errors, and please report them to me. Although the book was extravagantly proofread, almost all first editions have errors.

Is the course important?
This course teaches the basic "language" used in virtually all fields of engineering, technology, and science. It is likely that the grade in the first college calculus course is a very good indicator of eventual success in such majors.

Who teaches what? How is student work assessed and how are grades determined? Please see here.

A rough comparison
In New Jersey, a high school calculus course has 50 minute periods, nominally 180 student days: let's say 160. So the time in-class is about 8,000 minutes.
Math 151 has two lectures each week. There are 14 weeks, and the lectures are 80 minutes long. Two lectures are lost to exams. So 13 times 80 is about 1,040, then doubled is 2,080. I will add about half the recitation time (more about this below): that's another 14·40=560 minutes. So 151 has 2,640 educational minutes. That's about one-third of the high school regimen!
Therefore YOU will be your most important teacher! Working in groups is good. Asking questions is good. Taking advantage of the LRC and the MSLC is good.

Modes of instruction
Two meetings each week are likely to be classical lectures. There's usually not much interaction in the lectures, and (1600<8,000!) there is not much time for interaction. The additional "recitation" meeting will have more varied activities. Part of the time will be question/answer about course material and textbook homework problems. Maybe there will be a quiz. Much of the time will be spent doing workshops. Students will discuss somewhat non-routine problems in groups. Students will be required to write up a detailed solution of a selected problem and hand this in the next week. Why?
• Essentially all engineering students will participate in a group project in one or more upper-level courses, and extended written solutions will be required. Practice now, when the stakes are lower.
• Some of the workshop problems will be intentionally (!) ill-defined, and the methods needed may not be obvious. Real-world problems frequently are not given with the methods of solution, and sometimes what a solution is may not be clear!
• Many job environments have people working in teams. So get used to it now, when the stakes are lower. And maybe use these workshops to begin study sessions outside of class.

Let's really begin ...
Solutions to two textbook homework problems should be handed in tomorrow (1.1: 18 and 68). Consider numbers:

I'll use this background color if I write things which I don't have time to cover in class. Or I just forgot! You should know how to count: the positive integers. Then how to add and therefore the additive identity, 0. Additive inverses enlarge our number universe to all integers. Multiplication and division bring us to the collection of all rational numbers. We can "solve" linear equations in one variable: ax+b=c. Now interesting things happen if we try to solve non-linear equations. Even an equation as simple-looking as x2-2=0 has no solution in the rational number universe. So we need to enlarge our collection of numbers to real numbers. There are many ways to understand real numbers and one way is with decimal expansions. Rational numbers have finite or repeating decimal expansions. In this course we will use various sequential approximation processes, and we will need to consider numbers with all sorts of decimal expansions, not just repeating expansions. These weird and "chaotic" numbers are not rational: they are irrational. In various ways which can be made quite precise, there are many more irrational numbers than there are rational numbers: a typical number is irrational. Any physical experiment or process when measured gives only an approximation to some sort of "true" answer, and this sequence of approximations, like the decimal expansion, is likely to be an approximation of an irrational number. The decimal expansion of a real number provides an "address" to locate the number geometrically. For example, 23.4680472... means we first move right from 0 by 23 steps, each 1 long. Then look between 23 and 24, and divide that length in tenths. The number we're looking for resides (?) in the 4th subinterval. Now divide that subinterval into tenths. The number is in the 6th subinterval of that collection. Etc. The "Etc." of course conceals an approximation process which will locate the real number with the specified decimal expansion. Some real numbers have two valid decimal expansions. This doesn't bother me, since, say, a house on a corner of a grid system of streets could also possibly have two valid addresses (if it is on the corner of Cedar Avenue and Third Street, the house could have the address 12 Cedar Avenue and also 42 Third Street). So, for example, the real number with decimal address 23.45999... (with the 9's repeating) also have the decimal address 23.46, a terminating expansion (there are a string of 0's which are usually not written.

Geometry
The real numbers are usually thought of as corresponding to a specific geometric object, the real line. I usually think of this line as horizontal with 0 sitting "in the middle". 1 is to the right of 0. And this geometric picture brings up the idea of order. In addition to the algebraic structure, there is order: a<b. To me this means (in my picture of the line) that a is to the left of b. Negative numbers are to the left of 0 and positive numbers, including all the positive integers, are to the right of 0. Here are some interesting aspects to note.

Distance including a discussion of | |
The distance between two points is a non-negative real number whose size expresses how far apart the numbers are. This will be important when we study approximation schemes. We'd like to know that the approximation gets "close" to the correct answer, and the closeness will be measure by the distance. Algebraically, if the points correspond to the real numbers a and b, the distance between them is |a-b| and this is the same as |b-a|, so that distance has some symmetry. But I just used absolute value, and here is the piecewise definition of absolute value:
|x|=x if x≥0 and |x|=-x if x<0

Therefore absolute value is always a non-negative number. The absolute value of a number is 0 only when the number itself is 0. And absolute value of a product is the product of the absolute values (this actually is not totally obvious, and needs a bit of thought, I believe).

Intervals
Suppose we want to discover what numbers x are closer to 9 than a distance of 4. Algebraically this requirement translates to |x-9|<4. We can sort of "unroll" the inequality. The absolute value will be less than 4 if the number itself is both less than 4 and greater than -4. The two inequalities can be compactly written as follows:
-4<x-9<4 which implies 5<x<13.
This is an interval and an interval which does not contain either endpoint is called an open interval. The notation for this interval is (5,13). Intervals which contain both endpoints are called closed. An examples of such an interval is [-4,6], which means the numbers x satisfying -4≤x≤6. There are also half-open intervals, unbounded intervals (with notation using + or - ), etc. Please see the textbook.

Warning! If you wanted to "solve" (better: understand!) the inequality |x-9|>4 you can't just "unroll" it to -4>x-9>4. This inequality has no solutions. There is no number which is simultaneously less than -4 and greater than 4. You can't write this so compactly and using such implications represents an invalid (wrong!) method of solution. A valid method of solution would involve separately solving the inequalities: -4<x-9 or x-9>4 which gives 5<x or x>13. This is actually, therefore, two intervals: (-,5) and (13,). So the inequality |x-9|>4 has a solution set which is two intervals.

The plane
The conventional way to describe the plane algebraically is to drop down two lines perpendicular to each other: coordinate axes. A point in the plane will then be described by an ordered pair of real numbers. The first coordinate will usually be called the x-coordinate and the second, the y-coordinate. This pair describes ordered distances from the horizontal (the x-axis) and vertical (the y-axis) lines. Please see the text for more about this.

The embarrassment of all this, especially with "new" students, is that (3,8) could describe both a point in R², the plane, and could also describe an open interval (with [missing!] endpoints 3 and 8). The context is supposed to help, but still the notational confusion is possible, and this is lousy.

A (non-vertical) line and its algebraic description
Suppose we wanted an algebraic description of points with coordinates (x,y) which lie on the straight line which goes through (4,3) and (8,13). If (x,y) is such a point, then look at the picture: two right triangles indicated are similar, so the corresponding sides have the same ratios:

13-y   13-3
---- = ----
 8-x    8-4

Distance in the plane, R2 Look, please, at the diagram to the right. In the plane, points correspond to ordered pairs of numbers. So a point p might correspond to an ordered pair, (x1,y1), and q might correspond to (x2,y2). Then the point (x1,y2) is the vertex of a right triangle whose hypotenuse is a line segment connecting p and q. One leg of the right triangle is on a line where all the first coordinates are x1, and the length of that leg is given by the one dimensional formula, |y1-y2|. The other leg is on the line where all the second coordinates are y2, and the length of that leg is |x1-x2|. Then by Pythagoras, the hypotenuse has length sqrt(|x1-x2|2+|y1-y2|2). And usually the absolute values are discarded since we are squaring the quantities. Therefore we officially define: dist(p,q)=sqrt((x1-x2)2+(y1-y2)2) if p has coordinates (x1,y1) and q has coordinates (x2,y2).

Example
The distance between between (3,-2) and (6,4) is sqrt([3-6]²+[4-(-2)]²)=sqrt(3²+2²)=sqrt(13).

A circle and its algebraic description Suppose we wanted to describe algebraically the collection of all points which have distance sqrt(13) from (6,4). This is, of course, a circle of radius sqrt(13) and center (6,4). Well, if (x,y) is such a point, then sqrt([x-6]2+[y-4]2)=sqrt(13). This does describe the circle. Of course, if you must, all sorts of algebraic things could be done to the equation. But I am the laziest person in the room, and therefore ...

Functions
The word function is used in a technical sense in calculus, and is one of the most important vocabulary words. It is the logical setting for how things are transformed to other things. In the case of Math 151, the "things" are numbers. So functions will change numbers to numbers. A function is a rule changing numbers to numbers, with the important restriction that each "input" number is assigned a unique output number. You could think of a function as a machine with an input and a unique output associated to each input. The collection of all valid inputs to the machine, those inputs which don't cause the machine to break, is called the domain. The collection of all of the outputs for these valid inputs is called the range.

One example, from the toy model, already tricky!
In the toy model we got the equation V=x(11-2x)(8-2x). Well, the formula on the right side of the equation defines a function. Usually we write V(x) to show the dependence of V on the input value, x. So V(x)=x(11-2x)(8-2x). What is the domain of this function, V? If we just consider the formula, we'll get one answer: the polynomial has a value for all x's. The "natural domain" of a function defined by the formula x(11-2x)(8-2x) is all x. But if the formula is being used to model the physical situation, then there are definite restrictions on x. Certainly x should not be negative (cut out a square with negative sides?). And x can't be bigger than half of the smaller side length (remember that the paper's original size was 8 by 11 inches). So the domain is (0,4). Some people (many people, actually) argue that it makes more sense for the domain in this model to be [0,4]. I won't continue this discussion now, but we will come back to this. But certainly the physically reasonable domain for V in this problem is not the same as the "natural domain".
The range of V is interesting. If we allow x=0, the range "begins" at 0. How big does it get? I don't know now. So the range of V is [0, I don't know now].

Student-specified domain -- even weirder!
Before class Mr. Green "gave" me the interval (0,5) and Mr. Sloane "gave" me the interval [7,8]. Below is a picture of the union of these intervals, with the open interval indicted with a small circle at the end, and the closed interval shown with a large dot at each end.

I would like to "construct" a formula whose natural domain of definition is these two intervals. Familiar restrictions include division (we're not supposed to divide by 0) and square roots (in this course we deal with real numbers, so we should only take square roots of non-negative numbers.

• I want to throw away things bigger than 8. Well, if x>8, then 0>8-x. So if I include sqrt(8-x) in my formula, then those x's bigger than 8 can't be inputted (?): that is, they won't be in the domain.
  
  The picture is supposed to indicate that when this term is included, the red stuff, the open interval (8,), is thrown out of the domain.
• Now I want to exclude numbers less than or equal to 0. So I think of sqrt(x). That will certainly exclude negative numbers. But I want to throw out 0, also. Hey: just write 1/sqrt(x). Now if x=0, the result would imply dividing by 0 which is not legitimate.
  
  Now the red has touched (-,0].
• Let's see: I need to do something with numbers between 5 and 7. So I want something which will be negative between 5 and 7 and positive elsewhere. If we put such an expression inside a square root, the numbers between 5 and 7 will not be in the domain. The expression (x-7)(x-5) is positive if x>7 (then x-7>0 and also x>5 so x-5>0). Similarly, (x-7)(x-5) will be positive if x<5, since both of the factors will be negative (check me!). But Mr. Green and Mr. Sloane want me to include 7 in the domain and throw out 5. So I should look at sqrt([x-7]/[x-5]). That will keep 7 and throw out 5. The signs will be the same, but x-7 on top allows x=7 and x-5 on the bottom throws out x=5.
  
  

If you don't believe me, the graph to the right of the function f is produced by Maple, a program available to all Rutgers students (on Eden, other systems, and in computer labs). The instruction I used is below. The discont=true simply advised the program to expected "discontinuities", and not to try to connect the dots (it is a program, and will try to connect the dots!). I hope that you could get something similar on a graphing calculator. Notice that there is a graph exactly over the intervals that were named, and the boundary "behavior" over (0,5) sort of indicates that the endpoints are not in the domain.

plot(sqrt(8-x)+1/sqrt(x)+sqrt((x-7)/(x-5)),x=-1..9,y=0..10,thickness=2,color=black,discont=true)

The graph of a function
The collection of all points in the plane which correspond to the ordered pairs (x,f(x)), when x is in the domain of the function, is called the graph of the function. I like pictures, so I like graphs. Substantial evidence exists (allocation of neural resources -- brain power!) suggesting that humans can process lots of visual clues efficiently, much more, say, than many numbers or algebraic formulas. It is possible to specify a function with its graph. So that's what I did.

An example
I drew a curve much like what is displayed to the right. What was drawn? It was a nice ("continuous", technical word to be defined later) curve which "interpolated" (was drawn to connect) the five "data points" (-2,4), (-1,0), (0,-2), (1,-3), and (2,-1).

This is the graph of a function, since it passes the vertical line test: every vertical line intersects the graph at most one time. This is logically the same as requiring that every input yields exactly one output.

What is the domain of this function? That's the set of x's for which there is a point (x,something) on the graph. Look at the collection of vertical lines which touch this graph. The x's which touch or cross this graph are all x's in the closed interval [-2,2].

What is the range of this function? That's the set of y's for which the point (something,y) is on the graph. You can find these by considering all of the horizontal lines which touch or cross the graph. Notice, please, that there are horizontal lines which cross the graph more than once. This is definitely permitted by the definition of function. There can be distinct inputs which have the same outputs. The range is [-3,4].

Creating some new functions

Define the function A by the equation A(x)=f(x)+2. What does the graph of A look like? What is the domain of A? What is the range of A? Here the "+2" pushes the graph up two units. The resulting graph is shown to the right. The second coordinates of the data points are increased by 2. The domain, the input values, are unchanged. So the domain is [-2,2]. The output values, the range, are all increased by 2, and therefore the range is [-1,6].
Define the function B by the equation B(x)=f(x+2). What does the graph of B look like? What is the domain of B? What is the range of B? This is a bit trickier. The "+2" is on the inside of the function. The values that will emerge certainly will be values of f. This "+2" affects what are the legal inputs to B. The inputs to f are numbers between -2 and 2. Numbers which will be inputs to B are numbers which when 2 is added will be numbers between -2 and 2. If you can unwind this, you will see that such numbers are between -4 and 0. This graph will be the graph of f translated two units to the left. The domain is [-4,0] and the range is [-3,4].
Define the function C by the equation C(x)=|f(x)|. What does the graph of C look like? What is the domain of C? What is the range of C? This is definitely more fun than the other two. The absolute value is on the outside of f. So the numbers which are valid for inputs to f are going to be o.k. for the function C. If an output is negative, the absolute value flips the sign. Geometrically the numbers make the point on the graph appear symmetrically flipped over the x-axis. The result is what is shown. The domain of this function is [-2,2]. The range changes, and is now [0,4]. To me, the most interesting geometric feature of this graph is the sharp corner (?) at (-1,0). We will analyze such behavior and misbehavior in detail as part of this course.
Define the function D by the equation D(x)=f(|x|). What does the graph of D look like? What is the domain of D? What is the range of D? This was the Question of the Day (see the discussion which follows for more information about the QotD). This is a rather tricky question. It shows that elaborate technical information is not needed to generate an intricate situation. A hint that I gave students was to compute D(1) and D(0) and D(-1). Well, let's try to do this:
We know that D(x)=f(|x|). Look below, please. x |x| What value of f? The value of D(x) 1 1 f(1) D(1)=f(1)=-3 0 0 f(0) D(0)=f(0)=-2 -1 1 f(1) D(-1)=f(1)=-3 These data points are shown on the graph to the right.
You've got to get D's values very carefully. The absolute value is on the "inside" of the function. The values of the function are values of f, and these may be (actually, they are!) negative! With these data points, I hoped that students could indeed sketch a graph of D. Such a graph is to the right. The domain of D is [-2,2]. The range of D is [-3,-1]. The graph of D has several interesting qualitative aspects. It has a corner at (0,-2). And it is symmetric with respect to the y-axis, and this is because |x| is inside D.

About the QotD QotD=the Question of the Day. I will try (unless I get tired or busy or bored ...) to ask a question at every lecture. Every student who hands in an answer gets full credit, whatever the answer is! Why do I do this? Here are some possible reasons: Take attendance. (Yeah, one reason, but not the most important reason.) Give me a chance to learn if students understand what I am talking about. Give students a chance (I will read the answers, which will be returned at a future recitation) to learn if what students think they know is correct. As far as I can see, this is a win-win for everyone. People can work together on the QotD, and I want to encourage working together (which is why I'll have a list of students in these sections available soon). Further questions I wanted to explore more about the poor function f which was given graphically above. Here is what you could do, and let me know if you have questions: A new function What is the domain? What is the range? What does a graph of the function look like? E(x)=1/f(x)       F(x)=f(1/x)       Some answers Here are the answers to the domain and range questions for E and F, but not in the correct order. [1/4,)union(-,-1/3]    [-3,4]    [-2,-1)union(-1,2]    (-,-1/2]union[1/2,)

Maintained by greenfie@math.rutgers.edu and last modified 9/3/2007.