Calculus with Analytic Geometry I
Exam 4-Take Home Part.
Solutions
1. (Similar to §2.8#4) The picture shows the graph of a function f . Sketch the graph of f 0 . You may use
the grid on the last page, or you can trace the graph on your own paper, and then draw the graph of f 0
underneath it.
Solution. See the picture on the page before the last page (the penultimate page), showing both curves,
with the graph of the derivative in red.
2. (Probably not similar to any text exercise) Suppose f is an even function. Explain as best you can why f 0
will be odd.
Solution. The graph of an even function is symmetric about the y-axis; the part of the graph on the left
side of the y-axis is the reflection of the part on the right side with respect to that axis. That means that the
tangent line at −x is the reflection with respect to the y-axis of the tangent line at x. The only difference
between the line and its reflection is that if you compare rises and runs, the rises will be the same, while the
runs will be going in the opposite direction, thus the slope of the reflected line will be minus (the additive
inverse of) the slope of the original line. To be slightly more precise, if you have any figure, graph in the
plane, if a point of coordinates (x, y) is in the graph, then (−x, y) is in the reflection of the graph about the
y-axis. If the graph is a line of equation y = mx + b, the reflected line will have to be y = −mx + b. Reflecting
a line does not change the y-intercept, but changes the slope from m to −m. Returning to graphs of functions
and their tangent lines, it means that if f 0 (x) = m, then f 0 (−x) = −m; in other words, f 0 (−x) = −f 0 (x).
We can, of course, be more precise.
f (x + h) − f (x)
. Because f is even, f (−x) = f (x),
h→0
h
f (−x + h) = f (x − h), and if we consider that as h goes to 0, so does −h, we get
The derivative at x is defined as being f 0 (x) = lim
f 0 (−x)
f (−x + h) − f (−x)
f (x − h) − f (x)
f (x − h) − f (x)
= lim
= lim −
h→0
h→0
h
h
(−h)
f (x − h) − f (x)
= − lim
= −f 0 (x)
h→0
(−h)
=
lim
h→0
showing that f 0 is odd. While this is very precise, it is an approach lacking intuition.
dxn
Comment: Many of you answered by stating that the reason was that
= nxn−1 and if n is even, then
dx
n − 1 is odd. As it turns out, using mathematics more advanced than the one we have at our disposal right
now, one can use this fact to justify the result. But it shows no intuition whatsoever, no feeling for why the
derivative of an even function is odd.
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3. (Exercise #46 of Section 2.8.)The figure shows the graph of four functions. One is the position function of a
car, one is the velocity of the car, one is its acceleration, and one is its jerk. Identify each curve and explain
your choices.
Solution. Let us assume that the textbook author isn’t lying, and that the curves actually represent what
they are supposed to represent. One thing we might observe first (and this may or may not be the most
obvious thing; how to approach this problem can differ from person to person) is that every curve with one
exception develops a negative slope at some point. The derivative of any such curve will have to be negative
at those points. The curve without a negative slope is also always positive. The conclusion is that this
curve, curve d, cannot be the graph of the derivative of any of the other functions whose graph is given,
thus d is the graph of the position function. If we now look at the slope (or slopes of tangent lines) of d, we
see that d has horizontal slope at 0, the slope then increases little by little. It is never negative, and of the
remaining curves only c is never negative. I have a small problem with the picture. The slopes of tangent lines
of d seem to keep increasing, yet c begins to get smaller a little bit before it intersects d. I think the problem is
due to a bad reproduction of an original picture in which the curve d actually begins to slowly curve downward.
It keeps on increasing, but at a slower and slower rate. However, there being no other curve that does not
cross the x-axis, we have to declare that c represents the velocity. For the acceleration we need a curve that
is above the x-axis up to close to the point where curves c and d intersect, since c has positive slopes until
then, then gets slightly negative slopes. Curve b fits the bill, thus b is the graph of the acceleration function.
By elimination we can declare a represents the jerk. We should observe, however, that a does indeed behave
like the graph of the derivative of the function whose graph is b. For example, it crosses the x axis exactly at
the points where the graph of b has a horizontal tangent, etc.
4. (Similar to §2.8,#27) Find the derivative of the function
√
g(x) = 36 − x
using the definition of derivative. State the domain of the function and the domain of its derivative.
Solution.
√
√
√
36 − (x + h) − 36 − x
36 − x − h − 36 − x
lim
= lim
h→0
h→0
h
h
√
√
√
√
36 − x − h − 36 − x
36 − x − h + 36 − x
√
lim
·√
h→0
h
36 − x − h + 36 − x
(36 − x − h) − (36 − x)
−6h
√
√
√
lim √
= lim
h→0 h( 36 − x − h +
h→0
36 − x)
6 h( 36 − x − h + 36 − x)
−1
1
√
lim √
=− √
h→0
2 36 − x
36 − x − h + 36 − x
p
0
g (x)
=
=
=
=
3
1
The derivative is g 0 (x) = − √
.
2 36 − x
The domain of g is (−∞, 36], the domain of g 0 is (−∞, 36).
5. (Not too different from part of # 35 of §3.1) Find the equation of the tangent line to the curve y = x5 − 3ex
at the point where x = 0.
dy
= 5x4 − 3ex , evaluating at x = 0 we get the slope of the tangent line, namely
Solution. We see that
dx
m = −3. The value of y when x = 0 is y = −3. Thus the equation of the tangent line is y + 3 = −3x or
y = −3x − 3.
6. (Exercise #52, §3.1) For what value of x does the graph of f (x) = ex − 2x have a horizontal tangent line?
Solution. This exercise can be rephrased as: For what value of x is f 0 (x) = 0. Now f 0 (x) = ex − 2, setting
it to 0 we get ex = 2 thus x = ln 2.
7. (Similar to Exercise 75, §3.1) Let
f (x) =
x3
if x ≤ 2
mx2 + b if x¿2
Find the values of m and b that make f differentiable everywhere.
Solution. A somewhat unfortunate typo appears in the statement of the exercise. However, I think there
is only one interpretation, and comparing to exercise 75 of 3.1 should have cleared it up. The fact that
nobody e-mailed me to ask about it is evidence that the correct interpretation was clear. The
function should have appeared as
3
x
if x ≤ 2
f (x) =
mx2 + b if x > 2
The function f coincides with a polynomial to the right of 2 and also to the left of 2. So it is differentiable
everywhere except possibly at 2. For differentiability, it is necessary (though not at all sufficient) that the
function be continuous, so we must have limx→2− f (x) = limx→2+ f (x) = f (2). Now:
lim f (x) = lim x3 = 8 = f (2);
x→2−
x→2−
lim f (x) = lim f (x)4m + b,
x→2+
x→2+
so a first condition is that 4m + b = 8. The derivative of f , except at x = 2 is given by
3x2 if x < 2
0
f (x) =
2mx if x > 2
(Notice that I am not saying anything about the value at 2). Now, a derivative does not have to be continuous,
but it stands to reason that if we set things up so the function f 0 is continuous at 2, then f will be differentiable.
There is actually a result that escapes the level of this course that a derivative can be discontinuous, but
it cannot have jump discontinuities. Anyway, it stands to reason to equate limx→2− f 0 (x) = limx→2+ f 0 (x).
We have limx→2− f 0 (x) = limx→2− 3x2 = 12, and limx→2+ f 0 (x) = limx→2+ 2mx = 4m. We get 12 = 4m, or
m = 3. returning to the first equation 4m + b = 8, we now get b = −4. The answer is m = 3, b = −4.
The function f is thus defined by
f (x) =
x3
3x2 − 4
if x ≤ 2
if x > 2
I am attaching as last page a graph generated by Wolframalpha of the function. The function H I am
using there is the Heaviside function; H(x − 2) is the Heaviside function shifted two units to the right. The
p(x) + H(x − 2)q(x) will equal p(x) if x < 2, because H(x − 2) = 0 if x < 2. If x > 2 it equals p(x) + q(x),
because then H(x − 2) = 1. To get the function f (x) of the problem I used p(x) = x3 , q(x) = (3x2 − 4) − x3 .
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The graph shows a little hole at 2; that is probably due to the fact that Wolframalpha’s H(x) is not defined
for x = 0.
It is also possible to check now by the definition that f is differentiable at 2. We have
f (x) − f (2)
x−2
x→2
f (x) − f (2)
lim
x−2
x→2+
lim−
=
=
x3 − 8
(x − 2)(x2 + 2x + 4)
= lim−
= lim− (x2 + 2x + 4) = 12
x−2
x−2
x→2
x→2
x→2
3x2 − 4 − 8
3(x − 2)(x + 2)
lim
= lim
= lim (3x + 2) = 12
x−2
x−2
x→2+
x→2−
x→2−
lim−
Since
lim
x→2−
f (x) − f (2)
f (x) − f (2)
= lim
= 12
+
x−2
x−2
x→2
f (x) − f (2)
exists (and equals 12); the function is differentiable at 2.
x→2
x−2
we conclude that lim
NOTE: You’ll get full credit for this exercise if you figured out m = 3, b = −4. What I wrote here is only
for those who are interested in learning a bit more than what is absolutely necessary.
8. (Exercise 44, §3.2) Suppose that f (2) = −3, g(2) = 4, f 0 (2) = −2, and g 0 (2) = 7. Find h0 (2) if
(a) h(x) = 5f (x) − 4g(x).
Solution.
h0 (x) = 5f 0 (x) − 4g 0 (x);
h0 (2) = −38.
(b) h(x) = f (x)g(x).
Solution.
h0 (x) = f 0 (x)g(x) + f (x)g 0 (x);
h0 (2) = −29.
f (x)
.
g(x)
Solution.
(c) h(x) =
h0 (x) =
f 0 (x)g(x) − f (x)g 0 (x)
;
g(x)2
h0 (2) =
13
.
16
g(x)
.
1 + f (x)
Solution.
(d) h(x) =
h0 (x) =
g 0 (x)(1 + f (x)) − g(x)f 0 (x)
;
(1 + f (x))2
h0 (2) =
13
.
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9. (Exercise 48, §3.2) If f (2) = 10 and f 0 (x) = x2 f (x) for all x, find f 00 (2).
Solution.
We have
f 00 (x) =
d 2
(x f (x)) = 2xf 0 (x) + x2 f (x) = (2x3 + x2 )f (x).
dx
Evaluating for x = 2 we get f 00 (2) = 200.
10. (Partially Similar to Exercise #35, §3.3) A mass on a spring vibrates horizontally on a smooth level surface.
Its equation of motion is x(t) = 36 cos t, where t is in seconds and x is in centimeters. In other words, the
mass is moving back and forth on a straight horizontal line that is taken as the x-axis, and the formula gives
its position at all times.
(a) Find the velocity and acceleration at time t.
Solution. The velocity is v(t) = x0 (t) = −36 sin t, the acceleration, a(t) = v 0 (t) = −36 cos t.
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(b) What is the first value of positive t for which the velocity is 0?
Solution. −36 sin t = 0 for t = π. That is the first positive value at which the velocity is 0.
(c) Find the position, velocity, and acceleration of the mass at time t = 2π/3. In what direction is it moving
at that time? (Just in case: The directions are left and right.)
Solution. Setting t = 2π/3 in the formulas we are given for the position, and those we√obtained for
the velocity and acceleration, we see that the position is x = −18 cm, the velocity v = −18 3 cm/s, and
the acceleration a = 18 cm/s2 . Since the velocity is negative, it is moving to the left.
d35
(x sin x) by finding the first few derivatives and observing the pattern that
11. (Exercise # 50, §3.3) Find
dx35
occurs.
Solution.
d35
(x sin x) = −x cos x − 35 sin x.
dx35
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