The Right Answers to The Rong Questions:

Solutions to the Review Questions for Final Exam Math 151, Fall 2003, Rutgers Univ

Doron Zeilberger (“Dr. Z.”)

Disclaimer: I believe the answers to be correct, but I assume no responsibility to any damage due to possible errors and/or typos.

1. For what values of B is the following function continuous at x = 1?

f(x) = \{ \begin{matrix}3x^{3} − x^{2} + Bx, & if x > 1; \\ Bx − 2, & if x ≤ 1. \end{matrix}

Sol to 1.: The limit from the left (at x = 1), lim_{x→1^{−}} f(x) is 3 · (1)^{3} − 1^{2} + B · 1 = 2 + B, whereas the limit from the right lim_{x→1^{+}} f(x) is B · 1 − 2 = B − 2. In order for f(x) to be continuous at x = 1 we set these equal and solve. 2 + B = B − 2 getting 4 = 0. This has no solutions.

Ans. to 1.: None.

2. Find the equation of the tangent line to the curve y = \frac{x−1}{x+1} at each point where the tangent is parallel to the line x − 2y = 2.

Sol. to 2. The line x − 2y = 2, can be written as y = x/2 − 1, hence its slope is 1/2. Now the slope of the tangent to the curve y = \frac{x−1}{x+1} is

\frac{dy}{dx} = \frac{(x − 1)′(x + 1) − (x − 1)(x + 1)′}{(x + 1)^{2}} = \frac{1 · (x + 1) − (x − 1) · 1}{(x + 1)^{2}} = \frac{2}{(x + 1)^{2}} .

Setting this equal to 1/2 yields that the points we are interested in satisfy

\frac{2}{(x + 1)^{2}} = \frac{1}{2} .

Solving this we get (x + 1)^{2} − 4 = 0 hence, x^{2} + 2x − 3 = (x + 3)(x − 1) = 0, whose roots are x = −3 and x = 1. Plugging-in into y = \frac{x−1}{x+1}, gives, when x = −3 y = 2, and when x = 1, y = 0. So the points we need are (−3, 2) and (1, 0).

Ans. to 2: The tangent at the point (−3, 2) is (y − 2) = (1/2)(x + 3), which is x − 2y = −7, and the tangent at the point (1, 0) is (y − 0) = (1/2)(x − 1), which is x − 2y = 1.

3. Find the equation of the tangent line to the curve defined by the equation ln(xy)+2x−y +1 = 0 at the point (\frac{1}{2}, 2).

Sol. to 3. First simplify to get

ln x + ln y + 2x − y + 1 = 0 .

By implicit differentiation

\frac{1}{x} + \frac{y′}{y} + 2 − y′ = 0 .

Now plug-in x = 1/2 and y = 2, to get

\frac{1}{1/2} + \frac{y′}{2} + 2 − y′ = 0 ,

getting 4 = y′/2, and so the slope y′ (at the designated point) is 8. The equation of the tangent line at the designated point (\frac{1}{2}, 2) is (y − 2) = 8(x − \frac{1}{2}), which is y = 8x − 2. Ans.: y = 8x − 2.

4. Find the following limits

a) lim_{x→−1}(x^{2} − 2x + 1) b) lim_{x→∞} \frac{3x^{2} − 7}{\sqrt{x^{2} + 2}} c) lim_{x→0} \frac{sin 3x}{2 sin 5x}

d) lim_{x→∞} \frac{3e^{x} + 4e^{−x}}{5e^{x} + 4e^{−x}} e) lim_{x→8^{−}} \frac{|x − 8|}{x − 8} .

Sol. to 4.

(a) Plugging in we get (−1)^{2} − 2(−1) + 1 = 4. Ans. to 4a): 4.

(b) “forgetting about the little ones” we get

lim_{x→∞} \frac{3x^{2} − 7}{\sqrt{x^{2} + 2}} = lim_{x→∞} \frac{3x^{2}}{\sqrt{x^{2}}} = lim_{x→∞} \frac{3x^{2}}{x} = lim_{x→∞} 3x = ∞ .

Ans. to 4b): ∞.

(c) When the argument of sin goes to 0, you can replace sin w by w, hence

lim_{x→0} \frac{sin 3x}{2 sin 5x} = lim_{x→0} \frac{3x}{2(5x)} = lim_{x→0} \frac{3}{10} = \frac{3}{10} .

Ans to 4c): 3/10.

d) It is better to replace e^{x} by another variable, let’s call it y.

lim_{x→∞} \frac{3e^{x} + 4e^{−x}}{5e^{x} + 4e^{−x}} = lim_{y→∞} \frac{3y + 4/y}{5y + 4/y}

Now ‘forgetting about the little ones’ this equals

lim_{y→∞} \frac{3y}{5y} = \frac{3}{5} .

Ans to 4d): 3/5. e) Recall that |z| = −z for z < 0, so |x − 8| = −(x − 8) for x < 8. We have

lim_{x→8^{−}} \frac{|x − 8|}{x − 8} = lim_{x→8^{−}} \frac{−(x − 8)}{x − 8} = lim_{x→8^{−}}(−1) = −1 .

Ans. to 4e): −1.

5. Find the following limits

a) lim_{x→0^{+}} \sqrt{x} ln x  b) lim_{x→0^{+}} x^{sin x}  c) lim_{x→∞} \frac{(ln x)^{2}}{x} .

Sol. to 5 a) We use L’Hôpital:

lim_{x→0^{+}} \sqrt{x} ln x = lim_{x→0^{+}} \frac{ln x}{x^{−1/2}} = lim_{x→0^{+}} \frac{(ln x)′}{(x^{−1/2})′} = lim_{x→0^{+}} \frac{1/x}{(−1/2)x^{−3/2}} =

lim_{x→0^{+}} \frac{x^{1/2}}{(−1/2)} = lim_{x→0^{+}} −2\sqrt{x} = lim_{x→0^{+}} −2\sqrt{0} = 0 .

Ans. to 5a): 0.

b) We must first take the ln, getting ln(x^{sin x}) = sin x ln x, and writing it as a quotient we get \frac{ln x}{csc x}. Taking the limit of this we have:

lim_{x→0^{+}} \frac{ln x}{csc x} = lim_{x→0^{+}} \frac{(ln x)′}{(csc x)′} = lim_{x→0^{+}} \frac{1/x}{− cot x csc x} =

lim_{x→0^{+}} \frac{tan x sin x}{x} = lim_{x→0^{+}} \frac{sin x sin x}{(cos x)x} = lim_{x→0^{+}} \frac{x^{2}}{x} = lim_{x→0^{+}} x = 0 .

Now the limit of the original is the exponential of this Ans. to 5b): e^{0} = 1. Shortcut: When x is near 0, sin x can be replaced by x, so we get the simpler limit lim_{x→0^{+}} x^{x} whose ln is x ln x, and the limit

lim_{x→0^{+}} x ln x = lim_{x→0^{+}} \frac{ln x}{x^{−1}} = lim_{x→0^{+}} \frac{(ln x)′}{(x^{−1})′} =

lim_{x→0^{+}} \frac{1/x}{−x^{−2}} = lim_{x→0^{+}}(−x) = 0 .

and we get quicker Ans.: e^{0} = 1.

c) By L’Hôpital again:

lim_{x→∞} \frac{(ln x)^{2}}{x} = lim_{x→∞} \frac{((ln x)^{2})′}{x′} = lim_{x→∞} \frac{2(ln x)(1/x)}{1} = lim_{x→∞} \frac{2 ln x}{x} .

Using L’Hôpital once again, we have that this equals

2 lim_{x→∞} \frac{(ln x)′}{x′} = 2 lim_{x→∞} \frac{1/x}{1} = 2 lim_{x→∞} \frac{1}{x} = 0 .

Ans to 5c): 0.

6. Find y′ in each case. Do not simplify your answer.

a) y = x^{7} − 3x + 6 − \frac{1}{x^{4}} b) y = x^{5x} sin(x^{2}) c) y = \frac{2 tan x}{\sqrt{1 − x^{2}}}

d) x^{4}y + 5y^{6}x^{3} = 8 e) xe^{xy+3y} = y f) f(x) = (1 + 2x)^{1/x} .

Sol. to 6. a).

y′ = (x^{7} − 3x + 6 − \frac{1}{x^{4}} )′ = (x^{7} − 3x + 6 − x^{−4})′ = 7x^{6} − 3 − (−4)x^{−5} = 7x^{6} − 3 + \frac{4}{x^{5}} .

Ans. to 6a): 7x^{6} − 3 + \frac{4}{x^{5}} . b)

y′ = (x^{5x} sin(x^{2}))′ = (x^{5x})′ sin(x^{2}) + x^{5x}(sin(x^{2}))′ = (e^{5x ln x})′ sin(x^{2}) + x^{5x}(sin(x^{2}))′ =

(5x ln x)′(e^{5x ln x}) sin(x^{2}) + x^{5x}(2x) cos(x^{2}) =
(5 ln x + 5)(e^{5x ln x}) sin(x^{2}) + x^{5x}(2x) cos(x^{2}) =
(5 ln x + 5)x^{5x} sin(x^{2}) + x^{5x}(2x) cos(x^{2}) .

Ans. to 6b): (5 ln x + 5)x^{5x} sin(x^{2}) + x^{5x}(2x) cos(x^{2}). c)

y′ = (\frac{2 tan x}{\sqrt{1 − x^{2}}} )′ = \frac{(2 tan x)′\sqrt{1 − x^{2}} − (2 tan x)((1 − x^{2})^{1/2})′}{(\sqrt{1 − x^{2}})^{2}} =

\frac{2\sqrt{1 − x^{2}} sec^{2} x − (2 tan x)(1/2)(1 − x^{2})^{−1/2} · (−2x)}{1 − x^{2}} =

\frac{2\sqrt{1 − x^{2}} sec^{2} x + 2x(1 − x^{2})^{−1/2} tan x}{1 − x^{2}} = \frac{2 sec^{2} x}{\sqrt{1 − x^{2}}} + \frac{2x tan x}{\sqrt{1 − x^{2}}^{3}}

Ans, to 6c): \frac{2 sec^{2} x}{\sqrt{1−x^{2}}} + \frac{2x tan x}{\sqrt{1−x^{2}}^{3}} . d)

(x^{4}y + 5y^{6}x^{3})′ = 0 ,

so

(x^{4})′y + x^{4}y′ + 5y^{6}(x^{3})′ + 5(y^{6})′x^{3} = 0 ,
4x^{3}y + x^{4}y′ + 5y^{6}(3x^{2}) + 5(6y^{5}y′)x^{3} = 0 ,
4x^{3}y + x^{4}y′ + 15y^{6}x^{2} + 30y^{5}y′x^{3} = 0 .

Moving all the terms not involving y′ to the right gives x^{4}y′ + 30x^{3}y^{5}y′ = −4x^{3}y − 15y^{6}x^{2} .

(x^{4} + 30x^{3}y^{5})y′ = −4x^{3}y − 15y^{6}x^{2} .

Hence

y′ = \frac{−4x^{3}y − 15y^{6}x^{2}}{x^{4} + 30x^{3}y^{5}}

Since we are not supposed to simplify, we have Ans. to 6d): y′ = \frac{−4x^{3}y−15y^{6}x^{2}}{x^{4}+30x^{3}y^{5}} . e):

(xe^{xy+3y})′ = y′ ,
x′e^{xy+3y} + x(e^{xy+3y})′ = y′ ,
e^{xy+3y} + xe^{xy+3y} · (xy + 3y)′ = y′ ,
e^{xy+3y} + xe^{xy+3y} · (x′y + xy′ + 3y′) = y′ ,
e^{xy+3y} + xe^{xy+3y} · (y + (x + 3)y′) = y′ ,
e^{xy+3y} + xye^{xy+3y} + (x + 3)xy′e^{xy+3y} = y′
(1 + xy)e^{xy+3y} = y′(1 − x(x + 3)e^{xy+3y})

Hence

y′ = \frac{e^{xy+3y}(1 + xy)}{1 − x(x + 3)e^{xy+3y}} .

Ans. to 6e): y′ = \frac{e^{xy+3y}(1+xy)}{1−x(x+3)e^{xy+3y}}. f)

(ln f(x))′ = (\frac{ln(1 + 2x)}{x})′

\frac{f′(x)}{f(x)} = \frac{(ln(1 + 2x))′x − ln(1 + 2x) · x′}{x^{2}} = \frac{(2x/(1 + 2x)) − ln(1 + 2x)}{x^{2}}

Hence

f′(x) = f(x)\frac{(2x/(1 + 2x) − ln(1 + 2x)}{x^{2}} = (1 + 2x)^{1/x} \frac{2x/(1 + 2x) − ln(1 + 2x)}{x^{2}} .

Ans. to 6f): (1 + 2x)^{1/x} \frac{2x/(1+2x)−ln(1+2x)}{x^{2}}.

7. State the formal definition of the derivative of the function f(x). Use the definition to calculate f′(x) for f(x) = \sqrt{3 − 5x}. Sol. to 7.: The formal definition of the derivative of a function f(x) is

f′(x) = lim_{s→x} \frac{f(s) − f(x)}{s − x} .

Note: There is another version with h and lim_{h→0}, but for this problem it is more convenient to use this one.

f′(x) = lim_{s→x} \frac{\sqrt{3 − 5s} − \sqrt{3 − 5x}}{s − x} = lim_{s→x} \frac{(\sqrt{3 − 5s} − \sqrt{3 − 5x})(\sqrt{3 − 5s} + \sqrt{3 − 5x})}{(s − x)(\sqrt{3 − 5s} + \sqrt{3 − 5x})}

= lim_{s→x} \frac{(\sqrt{3 − 5s})^{2} − (\sqrt{3 − 5x})^{2}}{(s − x)(\sqrt{3 − 5s} + \sqrt{3 − 5x})} = lim_{s→x} \frac{(3 − 5s) − (3 − 5x)}{(s − x)(\sqrt{3 − 5s} + \sqrt{3 − 5x})}

= lim_{s→x} \frac{−5(s − x)}{(s − x)(\sqrt{3 − 5s} + \sqrt{3 − 5x})} =

lim_{s→x} \frac{−5}{\sqrt{3 − 5s} + \sqrt{3 − 5x}} = \frac{−5}{\sqrt{3 − 5x} + \sqrt{3 − 5x}} = \frac{−5}{2\sqrt{3 − 5x}} .

Ans. to 7. \frac{−5}{2\sqrt{3−5x}} (but it had to be derived from the definition!).

8. Suppose that S(x) = \sqrt{x} for x ≥ 0, and let f and g be differentiable functions about which the following is known

f(3) = 2, f′(3) = 7, g(3) = 4, g′(3) = 5 .

Compute the following

(f + g)′(3) (f · g)′(3), (\frac{f}{g} )′ (3), (S ∘ g)′(3), (\frac{f · g}{f − g} )′ (3) .

Sol. to 8.

(f + g)′(3) = f′(3) + g′(3) = 7 + 5 = 12 .

(f · g)′(3) = f′(3) · g(3) + f(3) · g′(3) = 7 · 4 + 2 · 5 = 38 .

(\frac{f}{g} )′ (3) = \frac{f′(3)g(3) − f(3)g′(3)}{g(3)^{2}} = \frac{7 · 4 − 2 · 5}{4^{2}} = \frac{18}{16} = \frac{9}{8} .

(S ∘ g)′(3) = S′(g(3))g′(3) = S′(4) · 5 = 5 · (1/2)(4)^{−1/2} = 5/4 .

(\frac{f · g}{f − g} )′ (3) =

(\frac{(f · g)′(f − g) − (f · g)(f − g)′}{(f − g)^{2}} ) (3) =

(\frac{(f′g + fg′)(f − g) − (f · g)(f′ − g′)}{(f − g)^{2}}) (3) =

(\frac{f^{2}g′ − f′g^{2}}{(f − g)^{2}} ) (3) =

\frac{f(3)^{2}g′(3) − f′(3)g(3)^{2}}{(f(3) − g(3))^{2}} =

\frac{2^{2} · 5 − 7 · 4^{2}}{(2 − 4)^{2}} = −23 .

Ans. to 8: 12, 38, 9/8, 5/4, −23.

9. Suppose f′′(x) = −3x + cos(πx) and f(1) = 2 and f′(1) = −1. What is f(5)?

Sol. to 9.: The general antiderivative of −3x + cos(πx) is f′(x) = −3x^{2}/2 + sin(πx)/π + A, where A is a general constant. To find A, we plug-in x = 1, and get f′(1) = −3 · 1^{2}/2 + sin(π)/π + A = −3/2 + 0 + A. On the other hand, in this specific problem, we are told that f′(1) = −1, so we get the equation −1 = A − 3/2, and solving for A, we get A = 1/2. So

f′(x) = −(3/2)x^{2} + sin(πx)/π + 1/2 .

To get f(x), we take the antiderivative of the above, getting f(x) = −(1/2)x^{3}−cos(πx)/π^{2}+x/2+B, where B is a constant yet to be determined. Plugging-in x = 1 into this yields f(1) = −1/2 − cos(π)/π^{2} + 1/2 + B = 1/π^{2} + B. On the other hand, in this specific problem, we are told that f(1) = 2, so we get the equation 2 = 1/π^{2} + B, so B = 2 − 1/π^{2}. Hence

f(x) = −(1/2)x^{3} − cos(πx)/π^{2} + x/2 + 2 − 1/π^{2}

Finally, plugging-in x = 5, we get f(5) = −(1/2)5^{3} − cos(5π)/π^{2} + 5/2 + 2 − 1/π^{2} = −125/2 + 1/π^{2} + 5/2 + 2 − 1/π^{2} = −120/2 + 2 = −60 + 2 = −58.

Ans. to 9: -58.

10. A farmer with 450 feet of fencing wants to enclose the four sides of a rectangular region into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?

Sol. to 10. Let’s call the length of the side parallel to the three internal walls x, and the length of the other side (perpendicular to these walls) y.

We have to maximize the area. So the Goal function is xy. We also have the constraint: 2y + 5x = 450. Solving for y, we get y = (450 − 5x)/2, and plugging this into the goal function, we express it in terms of x alone, getting f(x) = x(450 − 5x)/2 = (450/2)x − (5/2)x^{2}. To find the maximum, we take the derivative f′(x) = 450/2 − 5x, and then set it equal to 0. Solving 450/2 − 5x = 0, we get x = 45. Plugging this into f(x) we get that the maximal total area is f(45) = 45 · (450 − 5 · 45)/2 = 45^{2} · 5/2 = 10125/2.

Ans to 10: 10125/2.

Note: In some problems, they might ask you about the dimensions of the optimal rectangle. Then the answer would be 45 × (225/2).

11. A 15 feet long ladder is leaning against a wall when its base begins to slide along the floor, away from the wall. By the time the base is 12 ft. away from the wall, the base is moving at the rate of 5 ft/sec. How fast is the top of the ladder sliding down the wall then? How fast is the area of the triangle formed by ladder, wall, and floor changing at that time?

Sol. to 11. This is a related rates problem. Let’s call the height of the top of the ladder y, and the distance of the foot of the ladder from the wall x. By Pythagoras,

x^{2} + y^{2} = 15^{2} .

Taking derivatives with respect to time (using the chain rule in its abstract setting, a.k.a. implicit differentiation), we get

2x\frac{dx}{dt} + 2y \frac{dy}{dt} = 0 .

Right now x = 12 and dx/dt = 5. We also need y right now. Using the relation, x^{2} + y^{2} = 15^{2} we plug-in x = 12 and solve for y: 12^{2} + y^{2} = 15^{2}, getting y = 9. Substituting these values into the derived relation we get

2 · 12 · 5 + 2 · 9\frac{dy}{dt} = 0 ,

and solving for dy/dt, we get dy/dt = −20/3 = −6\frac{2}{3}. Since the sign is negative, it is sliding down (which is also obvious from common sense).

Ans. to the first part of 11: The top of the ladder is sliding down at a speed of 20/3 = 6\frac{2}{3} ft/sec.

The area, let’s call it A, is given by the relation A = xy/2. Taking derivative with respect to time (using the product rule), we get

\frac{dA}{dt} = \frac{1}{2} (x\frac{dy}{dt} + y \frac{dx}{dt}) ,

and we plug-in the above numbers, getting

\frac{dA}{dt} = \frac{1}{2} (12 · (−20/3) + 9 · 5) = −35/2 ,

Ans. to second part of 11): The area is decreasing at a rate of 17\frac{1}{2} square feet per second.

12. Let f(x) = \frac{3x}{x^{2}−1}. Find the domain of the function, the intervals where f(x) is increasing or decreasing, maximum and minimum, the concavity and inflection points, horizontal and vertical asymptotes of the graph. Then sketch the graph of f(x).

Sol. to 12. Domain: {x|x ≠ −1, 1}. Even though the horizontal, and vertical asymptotes are mentioned last in the statement of the problem, it is best to start with them, since they contain the most important information.

horizontal asymptotes:

lim_{x→∞} f(x) = lim_{x→∞} \frac{3x}{x^{2} − 1} = lim_{x→∞} \frac{3x}{x^{2}} = 0 .

So y = 0 (alias the x-axis) is a horizontal asymptote.

vertical asymptotes: Set the denominator equal to 0, we get x = −1 and x = 1.

Let’s consider each vertical asymptote on its own. Plugging-in a value of x slightly less than −1, say −1.01, we get −BIG, so this means that lim_{x→−1^{−}} f(x) = −∞, and the curve would drop down to “hell” (−∞) as it approaches the vertical asymptote x = −1. Plugging-in a value of x slightly bigger than −1, say −.99 we get that f(−.99) = BIG, so the curve “comes down from heavens” right after x = −1. Similarly the curve goes down to hell as x approaches x = 1, and emerges from heavens right after x = 1.

It is also a good idea to find the x-intercept (solve y = 0, getting (0, 0)) and the y-intercept, plugging-in x = 0, getting, once again (0, 0). So the curve intersects the x-axis only at the origin (0, 0).

Notice that we haven’t taken derivatives yet, but we can already pretty much draw the function. When x is very negative, it is almost coincident with the x-axis (the horiz. Asymptote) but it is below the x-axis, since f(x) is negative there, and then it is dropping down-down right before x = −1. By inspection you can see that the curve is concave down at the interval −∞ < x ≤ −1

Then from x = −1 to x = 1 it is a steady decrease (since at x = −1 it is coming from y = ∞ dropping to y = −∞ at x = 1. Somewhere between these, there should be an inflection point, that by symmetry is (0, 0). Also for x > 1 it dropping from ∞ to 0. So you can draw the curve without any calculus! Also it is clear that there are no local maxima and minima.

But now we can do some calculus.

By the quotient rule

f′(x) = −3\frac{x^{2} + 1}{(x^{2} − 1)^{2}} ,

this is always negative, so intervals of increase: none.

intervals of decrease: (−∞, −1),(−1, 1), (1, ∞).

To find potential (local) max or min. points, solve f′(x) = 0, which has no roots. Hence max and min. points: none. Using the quotient rule one more time, we get

f′′(x) = \frac{6x(x^{2} + 3)}{(x^{2} − 1)^{3}} .

The inflection point(s) are where f′′(x) = 0. Setting the numerator equal to 0, we get x = 0. The y coordinate is f(0) = 0. So inflection point: (0, 0). The transition points in the life of this function are x = −1,x = 0, and x = 1. Plugging-in random values in each interval we get The interval (−∞, −1) is concave down (since. e.g. f′′(2) < 0). The interval (−1, 0) is concave up (since. e.g. f′′(−\frac{1}{2}) > 0). The interval (0, 1) is concave down (since. e.g. f′′(\frac{1}{2}) < 0). The interval (1, ∞) is concave up (since. e.g. f′′(2) > 0). Ans. to 12.: max. and min. points: none; intervals of increase: none; intervals of decrease: (−∞, −1), (−1, 1), (1, ∞); inflection point: (0, 0); intervals of concave down:(−∞, −1), (0, 1); intervals of concave up:(−1, 0), (1, ∞); horiz. asymptote: y = 0; vertical asymptotes: x = −1 and x = 1.

13. Repeat the same problem for y = \frac{x^{2}}{x^{2}+3}. Sol. to 13.: Horiz. Asymptotes: y = 1. Vertical asymptote: none. By the quotient rule (do it!)

f′(x) = \frac{6x}{x^{2} + 3}

f′′(x) = \frac{−18(x^{2} − 1)}{x^{2} + 3}

So potential max or min.: x = 0. Plugging into the second derivative, we get f′′(0) = 6 > 0 so x = 0 is a minimum. Plugging x = 0 into f(x), we get that when x = 0, y = 0, so the point (0, 0) is a local minimum, and that’s the only one. To get the inflection points solve f′′(x) = 0, and get x = −1 and x = 1. When x < −1, f′′(x) is negative so in the interval −∞ < x < −1 the curve is concave down. In the interval −1 < x < 1 f′′(x) is positive, so the curve is concave up, and in the interval 1 < x < ∞ the curve is concave down since f′′(x) is negative there.

Now draw it!

Ans. to 13: Horiz. asymptote y = 1; Vertical asymptotes: none. local min. (0, 0); inflection points: (−1, 1/4) and (1, 1/4); interval of decrease: (−∞, 0); interval of increase: (0, ∞); intervals of concave down:(−∞, −1), (1, ∞); interval of concave up: (−1, 1).

14. Sketch the graph of f(x) which satisfies the following conditions f′(1) = f′(−1) = 0, f′(x) < 0 if |x| < 1, f′(x) > 0 if 1 < |x| < 2, f′(x) = −1 if |x| > 2, f′′(x) < 0 if −2 < x < 0, inflection point (0, 1).

Sol. to 14. First let’s spell everything out w/o the annoying absolute value symbol ||.

For −∞ < x < −2 : f′(x) = −1.

For −2 < x < −1 : f′(x) > 0.

For −1 < x < 1 : f′(x) < 0.

For 1 < x < 2 : f′(x) > 0.

For 2 < x < ∞ : f′(x) = −1.

Now we are almost ready to sketch, noting that at x = −2 and x = 1 we have local minima, since the derivative changes sign from negative to positive, and at x = −1 and x = 2 we have local maxima, since f′(x) changes sign from positive to negative. We also have an inflection point at (0, 1) and since f′′(x) < 0 between x = −2 and x = 0, it is concave down there, and hence between x = 0 and x = 2 it is concave up.

To actually draw a curve fitting all these specifications, draw a straight line with slope −1 coming from −∞ to the point (−2, 0), then draw a hat-like curve that peaks at x = −1 and inflects at the point (0, 1) turning into a cup-like arc that bottoms at x = 1 and goes up to x = 2 and then it suddenly becomes a straight line with slope −1.

15. Find the linearization of f(x) = \sqrt{x + 1} at a = 15 and use it to find an approximation to \sqrt{15} and \sqrt{17} . Use the second derivative to determine whether the estimate is greater or less than the actual value.

Sol to 15. The linearization of a function f(x) at x = a is L(x) = f(a) + f′(a)(x − a). In this problem f(x) = (x + 1)^{1/2}, so

f′(x) = (1/2)(x + 1)^{−1/2} = \frac{1}{2\sqrt{x + 1}} .

We have f(a) = f(15) = \sqrt{1 + 15} = \sqrt{16} = 4, and f′(a) = f′(15) = \frac{1}{2\sqrt{15+1}} = 1/8. Hence Ans. to first part of 15: L(x) = 4 + \frac{1}{8}(x − 15) . Now \sqrt{15} = f(14), and plugging x = 14, we get L(14) = 4 + \frac{1}{8}(14 − 15) = \frac{31}{8} = 3\frac{7}{8}, hence Ans. to second part of 15: \sqrt{15} is approximately 3\frac{7}{8}. Also \sqrt{17} = f(16), and plugging x = 16, we get L(16) = 4 + \frac{1}{8}(16 − 15) = \frac{33}{8} = 4\frac{1}{8}, hence Ans. to third part of 15: \sqrt{17} is approximately 4\frac{1}{8}.

f′′(x) = (−1/4)(x + 1)^{−3/2} = \frac{−1}{4\sqrt{x + 1}^{3}} .

Since f′′(15) = −1/256 is negative, the above approximations to \sqrt{15} and \sqrt{17} are more than the actual value. Ans. to fourth part of 15: both are more than the actual values.

16. Find the derivative of the following functions:

a) F (x) = ∫_{π}^{x} tan(s^{2})ds b) g(x) = ∫_{1}^{cos x} \sqrt{1 − t^{2}}dt .

Sol to 16.a) Since the upper-limit of integration is identical to the argument of F, (both are x), the derivative is simply the integrand with s replaced by x. Ans. to 16a): F′(x) = tan(x^{2}). Sol to 16b) Now the upper-limit of integration, cos x is not identical with the argument of g, which is x. We must first use the chain rule, multiplying top and bottom by d cos x (so to speak!).

g′(x) = \frac{d}{x} ∫_{1}^{cos x} \sqrt{1 − t^{2}}dt = \frac{d cos x}{dx} · \frac{d}{cos x} ∫_{1}^{cos x} \sqrt{1 − t^{2}}dt .

The first piece is a straightforward differentiation, while the second is done exactly as in a): replace t by cos x in the integrand. Hence F′(x) = − sin x\sqrt{1 − cos x^{2}}. Using the famous trig. identity this is − sin^{2} x. Ans. to 16b): g′(x) = − sin^{2} x.

17. Suppose that f is continuous on [0, 4], f(0) = 1 and 2 ≤ f′(x) ≤ 5 for all x ∈ (0, 4). Show that 9 ≤ f(4) ≤ 21. Sol. to 17. We must use the Mean Value Theorem. If f(x) is continuous in [a, b] and differentiable in (a, b) then there exists a c in (a, b) such that

f′(c) = \frac{f(b) − f(a)}{b − a} .

In this problem a = 0 and b = 4 so, there exists a c in (0, 4) such that

f′(c) = \frac{f(4) − f(0)}{4 − 0} = \frac{f(4) − 1}{4 − 0}

But we know that 2 ≤ f′(x) ≤ 5 for every x ∈ (0, 4), in particular for the c, whatever it is. Hence

2 ≤ \frac{f(4) − 1}{4} ≤ 5 .

Multiplying by 4, we get (since 4 is positive, the inequalities are preserved) 8 ≤ f(4) − 1 ≤ 20, and adding 1 everywhere, we get 9 ≤ f(4) ≤ 21.

18. Let f(x) = 3x^{3} − 2x^{2} + x − 1. Show that f(x) must have real roots in [0, 1].

Sol. to 18. The given function f(x) is continuous (because it is a polynomial), Plug-in the leftmost end, 0, into f(x), getting f(0) = −1. Plug-in the rightmost end, 1, into f(x), getting f(1) = 3 · 1^{3} − 2 · 1^{2} + 1 − 1 = 1. Since these two numbers −1 and 1 have opposite signs (one is positive and the other is negative), it follows by the intermediate value theorem, that there must be a c ∈ (0, 1) such that f(c) = 0. In other words, f(x) = 0 must have a real root in [0, 1].

19. Show that the equation x^{101} + x^{51} + x − 1 = 0 has exactly one real root.

Sol. to 19: Let f(x) = x^{101} + x^{51} + x − 1. Since lim_{x→−∞} f(x) = −∞ and lim_{x→∞} f(x) = ∞ (because the polynomial f(x) has odd degree), and these have opposite signs (like in problem 18), there must be at least one root in (−∞, ∞).

Taking derivative, f′(x) = 101x^{100} + 51x^{50} + 1. Since all the powers are even, and the coefficients are positive and the constant term (1) is strictly positive, f′(x) is strictly positive. If there would have been more than one root, by Rolle’s theorem there would have been a c such that f′(c) = 0, contradicting the fact that f′(x) is always positive. Hence there is exactly one real root.

20. Use two steps of Newton’s method to approximate the root of the equation x^{4} + x − 4 = 0 in the interval [1, 2].

Sol. to 20, Recall that Newton’s method is:

x_{n} = x_{o} − \frac{f(x_{o})}{f′(x_{o})} ,

where x_{0} is the old approximation and x_{n} is the new approximation. Implementing this here, we get f(x) = x^{4} + x − 4, and f′(x) = 4x^{3} + 1, so for this problem we have

x_{n} = x_{o} − \frac{x_{o}^{4} + x_{o} − 4}{4x_{o}^{3} + 1} .

f(1) = −2 and f(2) = 14. Since −2 is so much closer to 0 then 14, it is reasonable to take as the initial approximation x_{o} = 1. We get

x_{n} = 1 − \frac{1^{4} + 1 − 4}{4 · 1^{3} + 1} = 1 − \frac{−2}{5} = 7/5 .

Now let x_{0} = 7/5 and we get

x_{n} = 7/5 − \frac{(7/5)^{4} + 7/5 − 4}{4 · (7/5)^{3} + 1} .

Since you don’t have a calculator, you leave it like this. (It turns out to be 9703/7485). Ans. to 20: 7/5 − \frac{(7/5)^{4}+7/5−4}{4·(7/5)^{3}+1} = 9703/7485.

21. [Note: slightly changed from previous version]. If a stone is thrown vertically upward from the surface of the moon with a velocity of 10m/s, its height after t seconds is h(t) = 10t − 0.83t^{2}.

(a) What is the velocity of the stone after 3s?

(b) What is the velocity of the stone after it has risen 26.72m? Sol. to 21.: The velocity is the derivative of the position (in this problem height). So

v(t) = h′(t) = 10 − 1.66t .

The velocity after 3 seconds is v(3) = 10 − 1.66 · 3 = 10 − 4.98 = 5.02. Ans. to 21a): 5.02m/s. To solve b), we need to find at what time the height is 26.72, so we have to solve the equation

10t − .83t^{2} = 26.72 ,

getting t = 4 and t = 8.8048 (you need a calculator for this) . Since t = 4 comes before, we have h(4) = 10 − 1.66 · 4 = 3.36 Ans. to 21b): 3.36.

22. Find the absolute maximum and absolute minimum values of f(x) = \frac{x}{x^{2}+1} in [0, 2]. Sol. to 22. Using the quotient rule, we have

f′(x) = \frac{x′(x^{2} + 1) − x(x^{2} + 1)′}{(x^{2} + 1)^{2}} = \frac{1 · (x^{2} + 1) − x · 2x}{(x^{2} + 1)^{2}} = \frac{x^{2} + 1 − 2x^{2}}{(x^{2} + 1)^{2}} = \frac{1 − x^{2}}{(x^{2} + 1)^{2}} .

We have to solve f′(x) = 0. Setting the numerator equal to 0, we get 1−x^{2} = 0, so (1−x)(1+x) = 0 and we get the two roots x = −1 and x = 1. But x = −1 does not satisfy the residency requirement, i.e. it does not belong to the interval [0, 2], so we have the three finalists: 0, 1, 2. (The endpoints of the interval always participate in the final contest). Now let’s plug these numbers in f(x) = \frac{x}{x^{2}+1}:

f(0) = \frac{0}{0^{2} + 1} = 0 f(1) = \frac{1}{1^{2} + 1} = \frac{1}{2} = .5 f(2) = \frac{2}{2^{2} + 1} = \frac{2}{5} = .4 .

Ans. to 22: Absolute minimum value is 0, Absolute maximum value is 1/2.

23. Find the most general antiderivative of the functions

a) f(x) = 1 − x^{3} + 5x^{5} − 3x^{7} b) g(x) = \frac{5 − 4x^{3} + 3x^{6}}{x^{6}}

c) h(x) = 3e^{x} + 7 sec^{2} x + 5(1 − x^{2})^{− \frac{1}{2}} .

Sol. to 23. Recall the formula

∫ x^{n} = \frac{x^{n+1}}{n + 1} + C .

a)

∫ f(x)dx = ∫ (1−x^{3} +5x^{5} −3x^{7})dx = x− \frac{x^{4}}{4} +5\frac{x^{6}}{6} −3\frac{x^{8}}{8} = x−(1/4)x^{4} +(5/6)x^{6} −(3/8)x^{8} +C .

Ans. to 23a:x − (1/4)x^{4} + (5/6)x^{6} − (3/8)x^{8} + C .

b) First break it up, then integrate term by term.

∫ g(x)dx = ∫ \frac{5 − 4x^{3} + 3x^{6}}{x^{6}} dx = ∫ (5 · x^{−6} − 4x^{−3} + 3)dx = 5x^{−5}/(−5) − 4(x^{−2}/(−2)) + 3x =

−x^{−5} + 2x^{−2} + 3x = \frac{−1}{x^{5}} + \frac{2}{x^{2}} + 3x + C .

Ans. to 23b): \frac{−1}{x^{5}} + \frac{2}{x^{2}} + 3x + C.

c) Recall (or look up in the formula sheet) that (tan x)′ = sec^{2} x, hence the antiderivative of sec^{2} x is tan x + C. Also (sin^{−1} x)′ = (1 − x^{2})^{− \frac{1}{2}} , hence the antiderivative of (1 − x^{2})^{− \frac{1}{2}} is sin^{−1} x + C. Of course the antiderivative of e^{x} is e^{x} + C. So

∫ f(x)dx = ∫ (3e^{x} + 7 sec^{2} x + 5(1 − x^{2})^{− \frac{1}{2}} )dx = 3e^{x} + 7 tan x + 5 sin^{−1} x + C .

Ans. to 23c): 3e^{x} + 7 tan x + 5 sin^{−1} x + C.

24. Evaluate the integral, if it exists.

a) ∫_{1}^{9} \frac{\sqrt{u} − 2u^{2}}{u} du. b) ∫_{0}^{2} y^{2}\sqrt{1 + y^{3}}dy. c) ∫_{1}^{5} \frac{dt}{(t − 4)^{2}} .

d) ∫_{0}^{1} t^{2} cos(t^{3})dt. e) ∫ \frac{e^{\sqrt{x}}}{\sqrt{x}} dx.

Sol. to 24. a)

∫_{1}^{9} \frac{\sqrt{u} − 2u^{2}}{u} du = ∫_{1}^{9}(u^{− \frac{1}{2}} − 2u)du = u^{\frac{1}{2}} /(\frac{1}{2}) − \frac{2u^{2}}{2} |_{1}^{9} = 2\sqrt{u} − u^{2}|_{1}^{9} =

[2\sqrt{9} − 9^{2}] − [2\sqrt{1} − 1^{2}] = −76 .

Ans. to 24a): −76. b): We make the substitution u = 1 + y^{3} which entails du/dy = 3y^{2} hence dy = du/(3y^{2}). Also the limits of the new integral must be computed. When y = 0, u = 1, and when y = 2, u = 9. So

b) ∫_{0}^{2} y^{2}\sqrt{1 + y^{3}}dy = ∫_{1}^{9} y^{2}\sqrt{u}\frac{du}{3y^{2}} = \frac{1}{3} ∫_{1}^{9} \sqrt{u}du = \frac{1}{3} ∫_{1}^{9} u^{\frac{1}{2}} du = \frac{1}{3} \frac{1}{3/2}u^{\frac{3}{2}} |_{1}^{9} =

\frac{2}{9}\sqrt{u}^{3}|_{1}^{9} = \frac{2}{9}(\sqrt{9}^{3} − \sqrt{1}^{3}) = \frac{2}{9}(3^{3} − 1^{3}) = \frac{52}{9} .

Ans. to 24b): 52/9.

c) Beware! The integrand blows up (and hence is undefined) at t = 4, that lies within the interval of integration. Ans. to 24c: Undefined!

d) Here the substitution is u = t^{3} giving du/dt = 3t^{2}, and so dt = du/(3t^{2}). As for the limits of integration, when t = 0 u = 0 and when t = 1, u = 1.

∫_{0}^{1} t^{2} cos(t^{3})dt = ∫_{0}^{1} t^{2} cos(u)\frac{du}{3t^{2}} = \frac{1}{3} ∫_{0}^{1} cos(u)du = \frac{1}{3} sin u|_{0}^{1} = \frac{1}{3}(sin 1 − sin 0) = (sin 1)/3 .

Ans. to 24d): (sin 1)/3. e): This is an indefinite integral since there are no limits of integration. The answer should be an expression in x. In the method of substitution, we must translate back at the end. Take u = \sqrt{x} = x^{\frac{1}{2}} , so du/dx = (1/2)x^{− \frac{1}{2}} and dx = 2x^{\frac{1}{2}} du = 2\sqrt{x}du. We have

e) ∫ \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = ∫ \frac{e^{u}}{\sqrt{x}}2\sqrt{x}du = 2 ∫ e^{u}du = 2e^{u} = 2e^{\sqrt{x}} + C .

Ans. to 24e): 2e^{\sqrt{x}} + C.

25. If ∫_{1}^{5} f(x)dx = 12 and ∫_{4}^{5} f(x)dx = 3.6 , find ∫_{1}^{4} f(x). Sol. to 25. We use the obvious relation

∫_{a}^{b} f(x)dx = ∫_{a}^{c} f(x)dx + ∫_{c}^{b} f(x)dx , or rather its companion

∫_{a}^{c} f(x)dx = ∫_{a}^{b} f(x)dx − ∫_{c}^{b} f(x)dx .

So

∫_{1}^{4} f(x)dx = ∫_{1}^{5} f(x)dx − ∫_{4}^{5} f(x)dx = 12 − 3.6 = 8.4 .

Ans. to 25: 8.4.

26. Find the area bounded by the two curves (a) y_{1} = 2x^{2} and y_{2} = 8x. (b) y_{1} = sin x and y_{2} = cos x, 0 ≤ x ≤ \frac{π}{2}

Sol. to 26. Recall that the area is (if the two curves meet only at two points)

∫_{a}^{b}(T OP − BOT T OM)dx ,

where a and b are the two roots of y_{1} = y_{2}. Solving 2x^{2} = 8x yields 2x(x − 4) = 0 so x = 0 and x = 4, and indeed there are only two meeting points. To find who’s on top just plug-in any x in the interval (0, 4). x = 1 is a good choice. When x = 1, y_{1} = 2 and y_{2} = 8, since 2 < 8 we know that y_{2} = 8x is on top and y_{1} = 2x^{2} is at the bottom. So the area is

∫_{0}^{4}(8x − 2x^{2})dx = 4x^{2} − (2/3)x^{3}|_{0}^{4} = 4 · 4^{2} − (2/3) · 4^{3} = \frac{64}{3} = 21\frac{1}{3} .

Ans. to 26a): \frac{64}{3}.

b): (Note, in class I solved a different problem).

y_{1} = sin x and y_{2} = cos x meet when sin x = cos x, and that happens when x = π/4. There are two parts (equal by symmetry). The one bounded by y = sin x, y = cos x, and the y-axis (alias x = 0), and the one bounded by y = sin x, y = cos x, and x = π/2. The first one happens between x = 0 and x = π/4 and the second between x = π/4 and x = π/2. In the first half y = cos x is on top and y = sin x is at the bottom, and at the second half it is the other way. Hence the area is

∫_{0}^{π/4}(cos x − sin x)dx + ∫_{π/4}^{π/2}(sin x − cos x)dx = (sin x + cos x)|_{0}^{π/4} + (− cos x − sin x)|_{π/4}^{π/2} =

(sin(π/4)+cos(π/4))−(sin(0)+cos(0))+(− cos(π/2)−sin(π/2))−(− cos(π/4)−sin(π/4)) = 2\sqrt{2}−2 .

Ans. to 26b): 2\sqrt{2} − 2.

27. Evaluate the limit

lim_{n→∞} ∑_{i=1}^{n} sin(\frac{iπ}{n})\frac{π}{n} .

This is an instance of the formal definition of the definite integral

∫_{a}^{b} f(x)dx = lim_{n→∞} ∑_{i=1}^{n} f(x_{i})h ,

where h = (b − a)/n and x_{i} = a + ih, i = 1, 2, …, n.

The name of the game is to first decide on the function f(x), which in this problem is obviously f(x) = sin x. The argument of sin is iπ/n so x_{i} = iπ/n. To get a we plug-in i = 0 getting a = 0, and to get b we plug in i = n, getting b = π. Hence h = π/n, which is exactly as in the problem. Hence this complicated-looking limit is nothing but

∫_{0}^{π} sin x = − cos x|_{0}^{π} = − cos(π) − (−cos(0)) = −(−1) + 1 = 2 .

Ans. to 27.: 2.

28. a) Use the definition of the natural logarithm as an integral and compare areas to prove that ln 2 < 1 < ln 3.

b) Use a) to deduce that 2 < e < 3.

Sol. to 28: a) Recall that

ln x = ∫_{1}^{x} \frac{1}{t}dt .

Hence ln 2 is the area under the curve y = 1/t between t = 1 and t = 2. To get an upper bound for ln 2 note that it is enclosed in the square bounded by the t-axis, the horizontal line t = 1, and the vertical lines t = 1 and t = 2. Its area is 1 · 1 = 1, so ln 2 < 1.

The lower bound 1 < ln 3 is more tedious. Take the inscribed rectangles with n = 8, getting that ln 3, the area under the curve y = 1/t between t = 1 and t = 3 is definitely more than

\frac{1}{4} (\frac{1}{5/4} + \frac{1}{6/4} + \frac{1}{7/4} + \frac{1}{8/4} + \frac{1}{9/4} + \frac{1}{10/4} + \frac{1}{11/4} + \frac{1}{12/4}) = \frac{28271}{27720} > 1 .

b): Since e^{x} is an increasing function we can exponentiate the inequality ln 2 < 1 < ln 3 getting

e^{ln 2} < e^{1} < e^{ln 3} .

But e^{ln w} = w, so

2 < e < 3 .