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Please read these answers after you work on the problems. Reading the answers without working on the problems is not a useful strategy!
Notation e^{Frog} will be called exp(Frog) here, mostly because it is easier to type and read. Also, sqrt(Toad) will denote the square root of Toad (so sqrt(4) is 2).
Comment Answers will generally not be "simplified" unnecessarily. It is the philosophy of the management that formulas are delicate, and there is some risk of damaging them any time they are touched, so ... unless you really need to change them, don't bother.
Problem 1 B, the amplitude, is half of the difference between the maximum and the minimum, so B=2. The vertical shift A is the average of the maximum and minimum, so A= 7. Substituting x=0 will yield a value for C: algebraically y = A+Bsin(0+C) = 7+2sin(C). But (0,7.6) seems to be the y-intercept of the graph (at least approximately!) so that 7.6= 7+2sin(C). Solve this with your calculator (be sure to use radians!) and .3 will be the approximate value of arcsin(.3).
Problem 2 The two triangles ABF and FDE are similar so pairs of corresponding sides are proportional. In ABF, we choose the pair of sides AB and BF, and in FDE, we choose the pair of sides FD and DE. Then 12/x=x/5, so that x=sqrt(60). The hypotenuse AE can be obtained from the Pythagorean Theorem: it is just sqrt((12+sqrt(60))^{2}+(5+sqrt(60))^{2}), which is approximately 22.834. There are many equivalent ways to write this quantity, such as sqrt(289+34sqrt(60)) or (sqrt(17))(2sqrt(3)+sqrt(5)). No, we don't expect you to notice this: perhaps it would be best to just leave the answer alone, as it was initially written! The angle theta can be found in various ways. For example, the tangent of theta is (5+sqrt(60))/(12+sqrt(60)), and therefore theta itself is approximately .573 radians.
Problem 3
a) The functions ln and exp are inverses. Also, 3 ln
(sqrt(x))= ln (x^{3/2}), so exp(3 ln(sqrt(x)) ) = exp (ln
(x^{3/2}))= x^{3/2}.
b) The standard properties of logs will be used. ln(B^{2}
sqrt(A)) = 2 ln(B) + (1/2) ln A = 2 (5.3) + (1/2) (1.2)= 11.2. For the
second part of the problem, we recall (look at the graph of ln!) that
ln(Frog) < ln(Toad) exactly when Frog < Toad. Therefore we can
compare A^{9} and B^{2} by comparing their lns. But
ln(A^{9}) = 9 ln(A)= 9 (1.2) =10.8 and ln(B^{2})=2
ln(B) = 2 (5.3) = 10.6, so that the first number is larger than the
second.
Problem 4 a) The equation of the circle is (x-2)^{2} +
y^{2} = 9.
b)The equation of the line is y=-2x+6.
c) Maybe the easiest way is to substitute "y" from b) into the
equation in a). This yields a quadratic in x which can be solved using
the quadratic formula (important moral lesson: very few quadratics,
even with small integer coefficients, can be "solved" by factoring!).
The second coordinates can be gotten by using the equation in b). The
two points gotten are ((14+sqrt(41))/5, (2-2sqrt(41))/5) and
((14-sqrt(41))/5, (2+2sqrt(41))/5). These points are approximately
(1.52,-2.16) and (4.08,2.96). The picture may help some people
understand the answers.
Problem 5 a) The graph of y=Z(x) in the accompanying figure is
the solid broken line.
b) Since 2 is in the interval [0,3], Z(2) should be evaluated using
the first expression so Z(2) = 2*2 = 4. Since 4 belongs to the interval
[3,infinity), we evaluate Z(4) using the second expression: Z(4) = 4+3=7.
c) Z is strictly increasing so it has an inverse N. For x in [0,3],
Z(x)=2x, so the values of Z are in [0,6]. That means N(x) = (1/2)x for
x in [0,6] (just solve y=2x for x!). Otherwise (for x > 6) N(x)= x-3
(gotten by solving y=x+3 for x). To find N(2) we use the first
formula, so N(2)=(1/2)2=1, and N(8) uses the second formula, so
N(8)=5.
d) The graph of y=N(x) in the accompanying figure is
the dashed broken line.
Problem 6 The inverse function is given by n(p) = ((p-100)/3)^{2} and its domain coincides with the range of p(n) which is [103,130]. n(p) represents the number of boxes of detergent which can be produced for p dollars in this model.
Problem 7
a) The limit of the top is 0, and the limit of the
bottom is 8 which is not 0. Therefore the limit of the
expression is 0.
b) The bottom is x^{2} -4 = (x-2)(x+2), so if x is not equal
to 2, the whole expression simplifies to 1/(x+2). The limit as x
approaches 2 of this is just 1/4.
c) The bottom has a non-zero limit as x approaches 4, so we can get
the limit just by evaluating the top and bottom at 4. The result is 0.
d) Multiply the top and bottom by sqrt(x)+2. Then the bottom becomes
(sqrt(x)-2)(sqrt(x)+2) = x-4 which cancels the factor of (x-4) in the
top and we're left with just sqrt(x)+2 on top. This has limit 4 as x
approaches 4.
Problem 8 The derivative of f is f'(x) = 6x and f'(1)=6. So 6 is the slope of the line tangent to this curve when x=1. The tangent line has to pass through the point (1,f(1))=(1,7), so one equation for this line is y-7 = 6(x-1).
Problem 9 In this problem, dx will denote the expression "delta x". Then f(x+dx)-f(x) = (3/(x+dx))-(3/x)= (3(x)-3(x+dx))/((x+dx)x)= (-3dx)/((x+dx)x). Therefore (f(x+dx)-f(x))/dx = -3/((x+dx)x). The limit of this last expression as dx approaches 0 is just -3/(x^{2}).
Problem 10
a) (x^{3}-3x+17)' = 3x^{2}-3
b) (exp(x) sin(x))'= exp(x)' sin(x) + exp(x) (sin(x))' = exp(x) sin(x)
+ exp(x)cos(x).
c) ((2x+3)/(x^{2}+1))' is a complicated fraction. For
typographical purposes, think of it as Top/Bottom where Top=
2(x^{2} +1) -(2x+3)(2x) and Bottom =
(x^{2}+1)^{2}. Please note: you are not asked
to "simplify" the answer by any additional algebra.
d) (3exp(x) + (x/(2+5cos(x)) )' = 3exp(x) + another Top/Bottom where
here Top= 1(2+5cos(x))-(0-5sin(x))x and Bottom=
(2+5cos(x))^{2}. Again, no additional "simplification" is
needed.
Problem 11 By the chain rule the derivative of the left-hand side f'(g(x)) g'(x) which when x=2 is f'(g(2))g'(2)= f'(4)g'(2)= 2g'(2). The derivative of the right-hand side is 16x which when x=2 is 32. So we know that 2g'(2) = 32, and we get g'(2)=16.
Problem 12
a) tan(?) = 10/15 = 2/3, so ? = arctan(2/3) which is
approximately 0.588 radians.
b) Here tan(alpha) = 10/A, so alpha = arctan(10/A).
Problem 13
a) If f is continuous at 0, the limit from the left
should be equal to f(0). But the left-hand limit of f at 0 is equal to
the left-hand limit of x^{2}+2 at 0, and this limit is just
2. That means f(0) should be 2. We also know that f(0)=0^{2}
+A = A, and we know that the right-hand limit of f at 0 is the
right-hand limit of the formula x^{2}+A at 0, which is also
A. Now the two limits and the value of f at 0 should agree to make f
continuous at 0. This will occur when A=2.
b) The left-hand limit of f at 2 uses the formula x^{2} + 2
Here we substitute 2 for A, since 2 was the value found in the
previous part of the problem. Therefore the left-hand limit of f at 2
is 2^{2} +2 =6. The right-hand limit at 2 uses the formula
x^{2}+1, which has limit 5 at x=2. Since 5 is not equal to 6,
f is not continuous at x=2.
Problem 14
The share price has a jump in the middle of
Wednesday (the left-hand and right-hand limits are certainly not the
same on either side of the vertical dashed line). The price seems to
be continuous at all other points. The graph seems to be made up of
some line segments pieced together. At each "corner" the graph is not
differentiable. The limit defining the derivative would have different
values approaching each such point from the left and from the right.
The events suggested may be associated with certain stock price
"movements".
Problem 15 Vertical asymptotes are x=0 (the y-axis) and x=3. The accompanying graph satisfies the conditions given in the question. Note the "empty hole" at (5,1) since the function's domain does not include 5 (tricky!).
Problem 16
a) Consider the interval [2.5,3]. Enzyme
concentration is increasing, since the slope of the tangent line is
positive. But the rate of increase is "slowing" as time increases
(moving from left to right along the curve) because the tangent line's
slope is decreasing (such a curve is concave down).
b) Now consider the interval [1,2], for example. Here the enzyme
concentration is also increasing because the slope of the tangent line
is always positive. But in this interval as time increases the slope
of the tangent line is increasing -- so the rate of increase is
increasing here. This part of the curve is concave up. In both parts
of this problem, the slope of the tangent line must be recognized as
the rate of change of enzyme concentration. Also, there are other
correct answers to this problem than the ones given here.
Problem 17
F(1) = f(1)g(1) = 2(5) = 10 and the product rule for
derivatives gives the following: F'(1) = f'(1)g(1) + f(1)g'(1) = -1.
G(1) = f(1)/(g(1)+1^{2}) = 2/(5+1)= 2/6. The quotient rule
shows that G'(x) =
(f'(x)(g(x)+x^{2})-f(x)(g'(x)+2x))/((g(x)+x^{2})^{2})
and evaluating this when x=1 yields G'(1)=-1.
Problem 18
a)(exp(3cos(2x)))'= exp(3cos(2x)) (3 cos(2x))' =
exp(3cos(2x)) 3 (cos(2x))' = exp(3cos(2x))(3) (-sin(2x)) (2x)'=
-6exp(3cos(2x))sin(2x)
b)(ln(x^{3}+5x+9))'=
(1/(x^{3}+5x+9))(x^{3}+5x+9)'= (1/(x^{3}+5x+9)
(3x^{2}+5)
c) (sin((x+1)/(x^{2}+2)))' = cos((x+1)/(x^{2}+2))
((x+1)/(x^{2}+2))' = cos((x+1)/(x^{2}+2)) (Top/Bottom)
where here Top = (1)(x^{2}+2) - 2x (x+1) and Bottom =
(x^{2}+2)^{2}.