Math 251 Solutions to the Sample Questions for Midterm 2

Math 251
Solutions to the Sample Questions for Midterm 2
Midterm 2 will be considerably more difficult than Midterm 1.
On maximization/minimization problems, a large part of the credit will be for
correctly identifying the interval over which the maximization or minimization
should be done, and for mathematically verifying that your claimed maximum or
minimum really is one. Neither of these steps is tested on WeBWorK.
At least 80% of the points on the real exam will be modifications of problems from the
samples and two real first midterms, the problems below, homework problems (particularly
written homework), and problems from the sample and real Midterms 0. Note, though, that
the exact form of the functions to be differentiated and of the limits to be computed could vary
substantially, and the methods required to do them might occur in different combinations.
Point values here should give a rough idea of how much (out of 100) similar problem will be
worth on the real Midterm 2.
Be sure to get the notation right! (This is a frequent source of errors.) You have seen the
correct notation for limits, derivatives, linear approximation, etc. in the book, in handouts, in
files posted on the course website, and on the blackboard; use it. The right notation will help
you get the mathematics right, and incorrect notation will lose points.
The instructions for Midterm 2 will be essentially the same as for Midterm 1. In particular, calculators will be prohibited, but you may bring a 3 × 5 file card. (Note: You will
need to do some arithmetic, especially in connection with linear approximation problems and
maximization/minimization problems.)
You will be expected to know the formulas for volumes and surface areas of standard geometric shapes: spheres, cylinders, cones, and rectangular box-shaped objects. You will be expected
to know the formulas for the area of a circle, rectangle, and triangle.
1. (20 points) A circular puddle of water on a hot sidewalk in Needles CA is evaporating. At
2:00 pm, it had a radius of 10 inches, and its radius was decreasing at 3 inches per hour. Was
its area increasing or decreasing? At what rate? (Be sure to include the correct units in your
answer.)
Note: There is no picture in this file.
Solution: Let r(t) be the radius at time t, measured in inches and with time measured in
hours past noon, and let A(t) be the area at time t, measured in square inches. Note that both
the area and the radius vary with time, so must be treated as functions not constants.
The time we are interested in is t = 2. We are given r(2) = 10 and r0 (2) = −3. (Note that
it is negative, since the radius is decreasing.) The functions A(t) and r(t) are related by the
equation
2
A(t) = π r(t) .
(You are expected to know this formula.) Differentiate with respect to t:
A0 (t) = π · 2r(t)r0 (t) = 2πr(t)r0 (t).
1
(Don’t forget the factor r0 (t)! That will spoil the whole thing!) Evaluate this at t = 2, using
r(2) = 10 and r0 (2) = −3. This gives
A0 (2) = 2πr(2)r0 (2) = 2π(10)(−3) = −60π.
So the area is decreasing at the rate of 60π square inches per hour. (Don’t forget the units!)
Here, for reference, is what the solution looks like in physicists’ notation:
A = πr2 .
Differentiate with respect to t:
dA
dr
dr
= π · 2r = 2πr .
dt
dt
dt
dr
dA
dA
(Don’t forget the factor !) Now substitute r = 10 and
= −3, getting
= 2π(10)(−3) =
dt
dt
dt
−60π.
2. (5 points) Suppose that a ball is dropped from 90th floor of a building, 400 meters above the
ground. What is the velocity of the ball after 3 seconds? (The distance fallen after t seconds is
4.9t2 meters.)
Solution: Let d(t) be the distance fallen at time t, with distance in meters and time in seconds.
Then the downward velocity of the ball at time t is just the derivative d0 (t) = 4.9 · 2t = 9.8t,
with velocity in meters/second and time in seconds. Putting t = 3 gives a downward velocity
of 29.4 meters/second. Since velocity is usually measured upwards, the best answer is −29.4
meters/second (upwards) velocity.
Note: You must include the units in this kind of problem.
3. (9 points/part) Find the exact values of the following limits (possibly including ∞ or −∞),
or explain why they do not exist or there is not enough information to evaluate them. Give
justification in all cases (not just heuristic arguments).
x−2
x→0 sin(x) − 17
(a) lim
Solution: Both the numerator and denominator are defined and continuous at x = 0, and
the denominator is not zero there. Therefore
x−2
0−2
−2
2
lim
=
=
= .
x→0 sin(x) − 17
sin(0) − 17
−17
17
Note that L’Hospital’s rule does not apply, because the limit does not have an indeterminate
form. If you try to apply it anyway, you get
1
lim
= 1,
x→0 cos(x)
which is not the correct answer.
(b) lim
x→0
kx
, where k is a nonzero constant.
tan(x2 − 3x)
Solution: Trying to substitute x = 0 gives the undefined expression 00 , so more work is needed.
In this case, L’Hospital’s Rule applies. Remember to use the chain rule when differentiating
the denominator:
kx
k
lim
= lim
,
2
2
2
x→0 tan(x − 3x)
x→0 sec (x − 3x)(2x − 3)
provided the last limit exists. This limit can be evaluated by substituting x = 0, and the result
k
= − 31 k.
is sec2 (0)(−3)
x3
.
x→0 sin(7x3 )
(c) lim
0
Solution 1: The limit has the indeterminate form . Therefore we may use L’Hospital’s
0
d
3
2
Rule. The chain rule gives
sin(7x ) = 21x cos(7x3 ), so
dx
3x2
x3
=
lim
,
lim
x→0 21x2 cos(7x3 )
x→0 sin(7x3 )
if the second limit exists. To evaluate the second limit, cancel the factors of x2 and 3, getting:
3x2
3
3
3
1
lim
=
lim
=
=
=
.
x→0 21x2 cos(7x3 )
x→0 21 cos(7x3 )
21 cos(7 · 0)
21
7
(You must simplify
3
21
to 71 , either here or later.) Therefore
x3
1
= .
3
x→0 sin(7x )
7
lim
Solution 2: We reduce this problem to
t
1
1
= = 1.
lim
= lim t→0 sin(t)
t→0 sin(t)
1
t
Since 7x3 → 0 as x → 0, we can let t = 7x3 and get
7x3
t
lim
=
lim
= 1.
x→0 sin(7x3 )
t→0 sin(t)
Therefore
x3
1
7x3
1
lim
= lim
= .
3
x→0 sin(7x3 )
x→0
7
sin(7x )
7
Solution 3: As in Solution 1, L’Hospital’s Rule applies to show that
x3
3x2
lim
=
lim
,
x→0 sin(7x3 )
x→0 21x2 cos(7x3 )
if the second limit exists. If you don’t cancel x2 , you can still solve the problem by two more
applications of L’Hospital’s Rule. (This solution is provided for reference, in case anyone did
this; it is not recommended because it takes much more time and give too many chances for
algebraic errors.)
0
The second limit has the indeterminate form , so L’Hospital’s Rule applies. Thus,
0
2
3x
6x
lim
=
lim
x→0 21x2 cos(7x3 )
x→0 21 [2x cos(7x3 ) − x2 sin(7x3 ) · 21x2 ]
6x
,
= lim
x→0 21 [2x cos(7x3 ) − 21x4 sin(7x3 )]
if the limit on the right hand side exists. At this point, it is again possible to cancel x from the
numerator and denominator, giving
6
lim
.
x→0 21 [2 cos(7x3 ) − 21x3 sin(7x3 )]
Then substitute x = 0, giving the same answer as before. However, since the limit on the right
0
hand side above still has the indeterminate form , L’Hospital’s Rule applies again, and gives
0
6x
lim
x→0 21 [2x cos(7x3 ) − 21x4 sin(7x3 )]
6
= lim
3
3
2
x→0 21 [2(cos(7x ) − x sin(7x ) · 21x ) − 21(4x3 sin(7x3 ) + x4 cos(7x3 ) · 21x2 )]
6
= lim
3
6
3
x→0 (42 − 21 x ) cos(7x ) − 6 · 212 x3 sin(7x3 )
if the limit on the new right hand side exists. This last limit can be computed by substituting
x = 0, and is
6
6
6
1
lim
=
=
= .
3
6
3
2
3
3
x→0 (42 − 21 x ) cos(7x ) − 6 · 21 x sin(7x )
(42 − 0) cos(0) − 0 · sin(0)
42
7
(The final simplification is necessary.)
3e2x
.
x→∞ x + 7e2x
(d) lim
Solution: We have limx→∞ 3e2x = ∞ and limx→∞ (x + 7e2x ) = ∞. The limit above therefore
∞
has the indeterminate form “ ∞
”, so more work is needed. In this case, L’Hospital’s Rule
applies. Remember to use the chain rule when differentiating:
6e2x
3e2x
=
lim
,
x→∞ x + 7e2x
x→∞ 1 + 14e2x
provided the last limit exists. The last limit still has the indeterminate form “ ∞
”. We apply
∞
L’Hospital’s Rule again, getting
lim
3e2x
6e2x
12e2x
12
12
3
=
lim
=
lim
=
lim
=
=
.
x→∞ x + 7e2x
x→∞ 1 + 14e2x
x→∞ 28e2x
x→∞ 28
28
7
The last simplification is required.
lim
arctan(x)
.
x→∞ 1 + x−1
(e) lim
Solution: Since limx→∞ arctan(x) =
π
2
and limx→∞ (1 + x−1 ) = 1 6= 0, we have
arctan(x)
limx→∞ arctan(x)
π/2
π
=
=
=
.
x→∞
1+x
limx→∞ (1 + x−1 )
1
2
Note that L’Hospital’s Rule doesn’t apply, since the limit does not have an indeterminate
form. If you try to use it anyway, you get
1
1
−1
1
1+x2
lim
=
lim
= −1,
=
lim
x→∞ −x−2
x→∞ x−2 + 1
x→∞
1 + x2
−x−2
lim
which is the wrong answer.
1 + e2x
.
x→∞ ex + 7e2x
(f) lim
∞
”, so work is needed. We use the same method
Solution: This has the indeterminate form “ ∞
as for limits of rational functions at ±∞. Thus, we factor out e2x from both the numerator and
denominator, and then use the limit laws:
1 + e2x
e2x (e−2x + 1)
e−2x + 1
=
lim
=
lim
x→∞ ex + 7e2x
x→∞ e2x (e−x + 7)
x→∞ e−x + 7
1+0
1
1 + limx→∞ e−2x
=
= .
=
−x
7 + limx→∞ e
7+0
7
Here is a different way to arrange essentially the same calculation:
lim
1 + e2x
e−2x (1 + e2x )
e−2x + 1
=
lim
=
lim
x→∞ ex + 7e2x
x→∞ e−2x (ex + 7e2x )
x→∞ e−x + 7
1+0
1
1 + limx→∞ e−2x
=
= .
=
−x
7 + limx→∞ e
7+0
7
In principle, L’Hospital’s Rule applies, since the limit has the indeterminate form “ ∞
”.
∞
However, it doesn’t help. A single application gives
lim
2e2x
lim
.
x→∞ ex + 14e2x
We still have the indeterminate form “ ∞
”, and applying L’Hospital’s Rule again gives
∞
4e2x
lim
.
x→∞ ex + 28e2x
Another application gives
8e2x
.
x→∞ ex + 56e2x
These limits are in fact all equal to 17 , but further applications of L’Hospital’s Rule keep making
things worse.
lim
(g) lim+
x→3
f (x)
, given that f (3) = 2, that f 0 (3) = 9, and that f 0 is continuous at 3.
x−7
Solution: The function f is continuous at 3, since f 0 (3) exists. Therefore direct substitution
can be tried. It gives
f (x)
f (3)
2
1
lim+
=
=
=− .
x→3 x − 7
3−7
−4
2
(The final simplification is necessary.)
Note that the value of f 0 (3) was never used.
f (x) − 2
, given that f (3) = 2, that f 0 (3) = 9, and that f 0 is continuous at 3.
x→3 x − 3
(h) lim
Solution 1: Since f (3) = 2, this limit is simply the definition of the derivative f 0 (3):
lim
x→3
f (x) − 2
f (x) − f (3)
= lim
= f 0 (3) = 9.
x→3
x−3
x−3
0
Solution 2: The limit has the indeterminate form . Therefore we may use L’Hospital’s
0
Rule. It gives
f 0 (x)
f (x) − 2
= lim
= f 0 (3) = 9.
lim
x→3
x→3 x − 3
1
(The second step used continuity of f 0 at 3.)
x
.
x→2 e3x
(i) lim
Solution: Both the numerator and denominator are defined and continuous at x = 2, and
the denominator is not zero there. Therefore
x
2
2
lim 3x = 3·2 = 6 .
x→2 e
e
e
Note that L’Hospital’s rule does not apply, because the limit does not have an indeterminate
form. If you try to apply it anyway, you get
1
1
lim 3x = 6 ,
x→2 3e
3e
which is not the correct answer.
√
2x2 + 1
.
x→∞ 7x + 1109
(j) lim
Solution 1: We have
lim
x→∞
√
2x2 + 1 = ∞
and
lim (7x + 1109) = ∞,
x→∞
∞
, and more work is needed. We multiply the numer∞
√
ator and denominator by x1 . When simplifying the resulting numerator, note that x2 = x for
x > 0, and we need only consider x > 0 when dealing with a limit at +∞. Thus:
q √
q
√
√
1
1
2
1
2x + 1
(2x2 + 1)
2x2 + 1
2x2 + 1
x2
x2
x
lim
= lim 1
= lim
= lim
x→∞ 7x + 1109
x→∞
x→∞
(7x + 1109) x→∞
7 + 1109
7 + 1109
x
x
x
q
q
√
1
1
2 + x2
limx→∞ 2 + x2
2
=
.
= lim
=
1109
x→∞ 7 + 1109
7
limx→∞ 7 + x
x
so the limit has the indeterminate form
Solution 2: (This solution uses L’Hospital’s rule, but is not recommended. See Solution 3
∞
instead.) As in Solution 1, we find that the limit has the indeterminate form
. Therefore we
∞
use L’Hospital’s Rule. Replacing both the numerator and denominator by their derivatives, as
called for in L’Hospital’s Rule, we get
lim
1
2
x→∞
(2x2 + 1)
7
−1/2
· 4x
2x
= lim √
.
x→∞ 7 2x2 + 1
It does not help very much to apply L’Hospital’s Rule again, since that gives
√
2
2x2 + 1
lim 7
.
=
lim
x→∞ (2x2 + 1)−1/2 · 4x
x→∞
7x
2
Further applications of L’Hospital’s
Rule keep yielding the same two expressions over and over
√
2
again. Instead, we use x = x (as in Solution 1, we need only consider positive values of x)
to bring the numerator under the square root:
r
r
√
2
2x
2 x2
2
x2
x2
=
.
lim √
= lim √
= lim
lim
x→∞ 7 2x2 + 1
x→∞ 7 2x2 + 1
x→∞ 7
2x2 + 1
7 x→∞ 2x2 + 1
By methods similar to those in Solution 1, or by using L’Hospital’s rule, we can show that
x2
1
= .
x→∞ 2x2 + 1
2
lim
Therefore
r
2x
2 1
=
,
lim √
x→∞ 7 2x2 + 1
7 2
which can be shown to be the same answer as in Solution 1. Since this limit exists, we also get,
by L’Hospital’s rule,
r
√
2x2 + 1
2x
2 1
lim
= lim √
=
.
x→∞ 7x + 1109
x→∞ 7 2x2 + 1
7 2
Solution 3: (This is better than Solution 2, but not as good as Solution 1.) For x > 0, we
have
√
2x2 + 1
> 0.
7x + 1109
Therefore we can write
v
s
s
!2
u √
√
u
2
2+1
2+1
2x2 + 1
2x + 1
2x
2x
lim
= lim t
lim
= lim
=
.
x→∞ (7x + 1109)2
x→∞
x→∞ 7x + 1109
x→∞
7x + 1109
(7x + 1109)2
So it suffices to find
2x2 + 1
.
x→∞ (7x + 1109)2
lim
One way is to write
2x2 + 1
2x2 + 1
=
lim
= lim
x→∞
x→∞ (7x + 1109)2
x→∞ 49x2 + 2 · 7 · 1109x + 11092
lim
= lim
x→∞
1
x2
2
2·7·1109
+ 1109
x
x2
2+
49 +
=
1
x2
1
x2
(2x2 + 1)
(49x + 2 · 7 · 1109x + 11092 )
2
2+0+0
= .
49 + 0 + 0
49
Another way is to use L’Hospital’s rule twice. In condensed form:
2x2 + 1
4x
4
2
= lim
= lim
= .
2
x→∞ 14(7x + 1109)
x→∞ 14 · 7
x→∞ (7x + 1109)
49
In either case, one gets
r
√
√
2x2 + 1
2
2
lim
=
=
.
x→∞ 7x + 1109
49
7
lim
4. (12 points) Use the methods of calculus to find the exact values of x at which the function
h(x) = 3x3 − 4x takes its absolute minimum and maximum on the interval [−1, 0].
Solution: We apply the procedure for continuous functions on closed finite intervals. That
is, we evaluate f at all critical numbers and at the endpoints, and compare values.
To find the critical numbers, we differentiate h and solve the equation h0 (x) = 0. The derivative of h is
h0 (x) = 9x2 − 4.
To find the roots, factor it as
9x2 − 4 = (3x − 2)(3x + 2).
Either way, the roots are x = − 23 and x = 23 .
We now have two critical numbers, namely − 32 and 32 . Of these, 23 is not in the interval under
consideration, so we ignore it. (I must see you reject 23 . If I don’t see this, I will assume you
didn’t correctly solve the equation h0 (x) = 0.) So we must compare the values of h at − 23 , and
at the endpoints −1 and 0. We evaluate:
.
h(−1) = 1, h(0) = 0, and h − 23 = 16
9
The largest of these is h − 23 and the smallest is h(0), so the absolute maximum of h on the
interval [−1, 0] occurs at x = − 32 and the absolute minimum of h on the interval
[−1, 0] occurs
2
2
at x = 0. Note that x = 3 is not correct for the minimum, even though h 3 = − 16
, because 23
9
is not in the interval [−1, 0].
5. (20 points) A xenobiologist wants to wall off a rectangular enclosure for a small herd of fire
breathing monsters from the planet Yuggxth. One side of the enclosure will be a long straight
river which flows from east to west; no wall is needed here. A second side will be along an
already existing wall which is perpendicular to the river, and which can be used as it is. His
research grant contains enough money to build 6 kilometers of wall for the remaining two sides.
What is the largest area he can enclose?
Include units, and be sure to verify that your maximum or minimum really is what you claim
it is.
(Show a full mathematical solution. A correct guess with no valid work will receive no credit,
and a correct number supported only by heuristic reasoning will receive very little credit.)
Solution: Note: There is no picture in this file. A picture may be provided separately.
The arrangement of walls will be a rectangle with sides running north-south and east-west.
We are supposed to maximize the total area of the enclosure; call it A. For definiteness, assume
that the already existing wall is on the north side of the river, and that the enclosure is on the
west side of it. Thus, the south side of the enclosure is the river, the east side is the already
existing wall, and the north and west sides are the new walls.
Let x be the length of the enclosure in the east-west direction (along the river), and let y
be its length in the north-south direction (along the already existing wall), both measured in
kilometers. Then the total area is A = xy. This expression has too many variables, and ignores
the fact that x and y are not independent. We must eliminate one of the variables by using
the restriction on the total length of the new walls. There is nothing to be gained by using less
than the allowed length of the new wall, so we assume the total length is exactly 6 (measured
in kilometers). There is no new wall along the south side, where the river is (what do you
think happens to a fire breathing monster if it falls in a river?), and there is no new wall on
the east side (maybe there is something even more dangerous on the other side of that wall).
The west side has length y kilometers, and the north side has length x kilometers. Therefore
the constraint is
x + y = 6.
It doesn’t matter much which variable you solve for. I decided to solve for y, giving
y = 6 − x.
Substitute this for y in the formula A = xy, write it as a function of x, and multiply out so
that it is convenient to differentiate:
A(x) = x(6 − x) = 6x − x2 .
The constraints are that x and y are both nonnegative. (Allowing the “degenerate” cases x = 0
and y = 0 will allow us to find the maximum over a closed bounded interval, which simplifies
later steps.) The significance of the requirement x ≥ 0 is clear, but we must express the
requirement y ≥ 0 in terms of x. Since y = 6 − x, it says that x ≤ 6. (Here is another way to
see this. The length the north side, which is x, can be at most 6.) Our problem is therefore to
maximize the function A(x) = 6x − x2 for x in the closed bounded interval [0, 6].
We search for critical numbers. Differentiate:
A0 (x) = 6 − 2x.
Set the derivative equal to zero and solve:
0 = A0 (x) = 6 − 2x
x = 3.
Since we are maximizing over a closed bounded interval, we need only compare the numbers
A(0), A(6), and A(3). These are
A(0) = 6 · 0 − 02 = 0,
A(3) = 6 · 3 − 32 = 18 − 9 = 9,
and
A(6) = 6 · 6 − 62 = 0.
Clearly the largest value is at x = 3. Therefore the east-west length should be 3 kilometers,
and the north-south length should be 6 − 3 = 3 kilometers. It follows that the largest possible
area is A(3) = 9 square kilometers. (Include the units!)
For reference, here are the other two approaches to test for a maximum. For both, we start
at the point above where we found that x = 3 is the only number in our interval with A0 (x) = 0.
In both cases, one must still find the dimensions as above. Note that both of these also work
over the open interval (0, 6), and so could have been done without allowing the degenerate cases
x = 0 and y = 0.
First derivative method: We know A0 (x) = 6 − 2x. So A0 (x) > 0 for 0 ≤ x < 3, and A0 (x) < 0
for x > 3. This shows that A(x) is increasing for 0 ≤ x < 3 and decreasing for x > 3. So A(3)
must be the largest value of A(x) for x in [0, 6].
Second derivative method: We know A0 (x) = 6 − 2x. So A00 (x) = −2. This is negative on the
entire interval (0, 6), so A(x) is concave down there, and any number x with A0 (x) = 0 must
give a global maximum.
6. (20 points) A cylindrical recycling bin, with a circular base and with no top, is to have a
volume of 8π cubic feet. What is the minimum possible surface area of such a bin?
Include units, and be sure to verify that your maximum or minimum really is what you claim
it is.
(Show a full mathematical solution. A correct guess with no valid work will receive no credit,
and a correct number supported only by heuristic reasoning will receive very little credit.)
Solution: Note: There is no picture in this file. A picture may be provided separately.
We want to minimize the surface area of the bin; call it A.
Let h be the height of the bin, and let r be the radius of its base. The area of the base is πr2
and the area of the side is the circumference times the height, or 2πrh. Thus A = πr2 + 2πrh.
(Remember that there is no top!)
There are too many variables here, and we can eliminate one of them using the constraint
on the volume V. The volume is the height times the area of the base, that is, V = πr2 h, and
we are told it is supposed to be V = 8π. So πr2 h = 8π, and r2 h = 8.
It is easier to solve for h than for r, so we do that, giving h = 8r−2 . Substituting in the
formula for A, and writing the result as a function for the purposes of differentiation, we get
16π
.
A(r) = πr2 + 2πr · 8r−2 = πr2 +
r
Note that r must be positive (r > 0). Also, we must have h > 0, but, since h = 8r−2 , this does
not tell us anything new. So we are to minimize A(r) = πr2 + 16π
subject to the condition
r
r > 0.
We start by searching for critical numbers. We differentiate: A0 (r) = 2πr − 16πr−2 . We
equate this to zero:
2πr − 16πr−2 = 0
8
r− 2 =0
r
3
r −8=0
r = 2.
So r = 2 is the only critical number.
Is it a maximum or minimum? The easiest test to use here is the second derivative test.
One easily checks that A00 (r) = 2π + 32πr−3 , which is positive for all r in the interval (0, ∞).
Therefore A is concave up everywhere on the interval (0, ∞), and any critical number must be
an absolute minimum.
The problem asked for the minimum surface area. This is now A(2) = π · 22 + 16π · 2−1 = 12π
square feet. (Don’t forget the units!)
For those who tried it instead of the second derivative test as done above, here is how the
first derivative test works. We factor the derivative as follows:
A0 (r) = 2πr − 16πr−2 = 2πr−2 (r3 − 8).
For r > 0, all factors except the last are positive. So we need only consider the sign of r3 − 8.
For 0 < r < 2, we have r3 − 8 < 0, so A0 (r) < 0. Therefore A is decreasing on the whole interval
(0, 2). For r > 2, we have r3 − 8 > 0, so A0 (r) > 0. Therefore A is increasing on the whole
interval (2, ∞). Clearly, then, there is an absolute minimum at r = 2.
Here is a different approach, perhaps simpler, to the first derivative test. We know that
A0 (r) is continuous on the whole interval (0, ∞). It is zero only at r = 2. Therefore, by the
Intermediate Value Theorem, the derivative must have the same sign throughout the interval
(0, 2), and we can find out that sign by computing, say, A0 (1) = −14π < 0. The derivative
must also have the same sign throughout the interval (2, ∞), and we get it by computing, say,
A0 (4) = 2π · 4−2 (43 − 8) = 7π > 0. As above, then, A(r) is decreasing on the whole interval
(0, 2) and increasing on the whole interval (2, ∞), so has an absolute minimum at r = 2.
7. (8 points total.) The picture below is the graph of the DERIVATIVE y = f 0 (x) for a certain
function f. CAUTION: You are given the graph of the derivative f 0 (x), not the graph of f (x),
but you are asked questions about f (x).
2
2
-2
-4
4
6
8
In case anyone wants to experiment, the function in the graph happens to have the formula
1
1 3
f 0 (x) = − 16
x(x − 6)2 = − 16
x + 34 x2 − 94 x.
The function f has the formula
1 4
x + 41 x3 − 89 x2 + C
f 0 (x) = − 64
for some unknown constant C.
(a) (4 points.) Find and identify all local minimums and local maximums of f in the interval
(−1, 9). Justify your answer. (Note that you are being asked about f, not the function shown
in the graph, which is f 0 .)
To solve the problem, we first read directly from the graph of f 0 all the numbers x with
f (x) = 0. These are x = 0 and x = 6. So the critical numbers of f are 0 and 6.
At x = 0, we can read directly from the graph of f 0 that f 0 changes from positive to negative
at 0. Therefore f changes from increasing to decreasing at 0, so f has a local maximum at 0.
(Here is an alternate way to see that f has a local maximum at 0. We can see from the graph
that the slope of f 0 is negative at 0. Therefore f 00 (0) < 0, so f is concave down there, and so
has a local maximum at 0.)
At x = 6, we can read directly from the graph of f 0 that f 0 does not change sign at 6. Indeed,
for x close to but less than 6, we see that f 0 (x) < 0, and for x close to but greater than 6, again
f 0 (x) < 0. So f has a critical number, but neither a local maximum nor a local minimum, at
x = 6. (In fact, the slope of f 0 at x = 6, that is, f 00 (6), is equal to zero. Since the slope of
f 0 does change sign at 6, we know that the value of f 00 changes sign at 6. Therefore f has an
inflection point at x = 6.)
There is no local maximum or minimum at x = 2, since f 0 (2) < 0. (In fact, f has an inflection
point at x = 2.)
A correct solution must consider what happens at both 0 and 6, and must show that there
is neither a local minimum nor a local maximum at 6, but need not show that there is an
inflection point there, and need not consider what happens at x = 2.
Here is a graph of one possible choice for f, namely the one with f (4) = 0.
0
6
4
2
2
4
6
8
-2
-4
(b) (4 points.) Find all inflection points of f in the interval (−1, 9). Justify your answer.
(Note that you are being asked about f, not the function shown in the graph, which is f 0 .)
Solution: The function f has an inflection point where its concavity changes.
At x = 2, read from the graph that f 0 (the function shown) changes from decreasing to
increasing, so its derivative f 00 changes from negative to positive, so f changes from concave
down to concave up. Therefore f has an inflection point at x = 2.
At x = 6, read from the graph that f 0 (the function shown) changes from increasing to
decreasing, so its derivative f 00 changes from positive to negative, so f changes from concave
up to concave down. Therefore f has an inflection point at x = 6.
(The function f 0 has an inflection point at x = 4, but f does not have an inflection point at
x = 4.)
8. Let f and g be functions such that:
f (−3) = −5,
f 0 (−3) = 12,
g(−3) = 2,
and g 0 (−3) = −3
and
f (2) = 7,
f 0 (2) = 3,
g(2) = −3,
and g 0 (2) = 2.
Let h(x) = f (g(x)).
(a) (2 points) Find h(2). (You will not need to use all the information provided.)
Solution: h(2) = f (g(2)) = f (−3) = −5.
(b) (6 points) Find h0 (2). (You will not need to use all the information provided.)
Solution: Using the chain rule,
h0 (2) = f 0 (g(2))g 0 (2) = f 0 (−3) · 2 = 12 · 2 = 24.
(c) (4 points) Use the linear approximation to estimate h(1.99).
Solution: Recall that the linear approximation at x = a is h(x) ≈ L(x), where L(x) =
h(a) + h0 (a)(x − a). We take a = 2 and x = 1.99. Then h(a) = −5 by Part (a) and h0 (a) = 24
by Part (b). Thus
h(1.99) ≈ (−5) + (24)(1.99 − 2) = −5 + 24(−0.01) = −5.24.
9. (4 points/part) The function f (x) satisfies the following three properties:
f (2) = −3,
f 0 (2) = 0.40,
and f 00 (2) = 1.1.
(a) Use the linearization (tangent line approximation) to estimate the value of f (1.96).
Solution:
f (1.96) ≈ f (2) + f 0 (2)(1.96 − 2) = −3 + (0.40)(−0.04) = −3.016.
(b) Is your answer in part (a) too big or too small? Explain with a picture.
Solution: (There is no picture in the file.)
Since f 00 (2) > 0, the function is concave up at 2, so probably also near 2. This means its
graph lies above the tangent line, so the answer in part (a) is too small.
10. (20 points) Interstate 25 and Interstate 40 meet at a right angle in central Albuquerque
(NM). At noon one day, Professor Greenbottle was driving north on Interstate 25 at 70 mph,
and was 3 miles south of the intersection. At the same time, a Mafia hit squad was driving
west on Interstate 40 at 50 mph, and was 4 miles west of the intersection. Were Professor
Greenbottle and the Mafia hit squad getting closer together or further apart? At what rate?
(Be sure to include the correct units.)
Solution: Note: There is no picture in this file.
Let a(t) be the distance (in miles) Professor Greenbottle’s car is south of the intersection at
time t (with time measured in hours), and let b(t) be the distance (in miles) the hit squad is
west of the intersection at time t. Let t0 represent noon on the day in question. Let z(t) be the
distance (also in miles) between the two cars at time t. Then the information given says that:
a0 (t0 ) = −70,
a(t0 ) = 3,
b(t0 ) = 4,
and b0 (t0 ) = 50.
(Note that a0 (t0 ) is negative, because Professor Greenbottle’s car is getting closer to the intersection. See the end of the solution for other possible choices for measuring distance.)
We want to find z 0 (t0 ). We know (from the Pythagorean Theorem) that z(t)2 = a(t)2 + b(t)2 .
Differentiating, we get
2z(t)z 0 (t) = 2a(t)a0 (t) + 2b(t)b0 (t).
(Don’t forget to use the chain rule!) Put t = t0 and (for simplicity) divide by 2:
z(t0 )z 0 (t0 ) = a(t0 )a0 (t0 ) + b(t0 )b0 (t0 ).
Next, substitute values. (Note that this can only be done after differentiating!) We need
p
√
z(t0 ) = a(t0 )2 + b(t0 )2 = 32 + 42 = 5,
and we then get
5 · z 0 (t0 ) = 3 · (−70) + 4 · 50 = −10.
Therefore z 0 (t0 ) = −2, and Professor Greenbottle’s car and the Mafia hit squad are getting
closer to each other at 2 miles per hour. (The units are necessary!)
Here are descriptions of some alternatives. First, one could solve for z(t):
p
1/2
z(t) = a(t)2 + b(t)2 = a(t)2 + b(t)2
−1/2
−1/2
z 0 (t) = 21 a(t)2 + b(t)2
(2a(t)a0 (t) + 2b(t)b0 (t)) = a(t)2 + b(t)2
(a(t)a0 (t) + b(t)b0 (t)) .
(Again, don’t forget to use the chain rule!) Put t = t0 and substitute values:
−1/2
z 0 (t0 ) = a(t0 )2 + b(t0 )2
(3 · (−70) + 4 · 50) = −2
(simplification step omitted).
Second, one could measure distances differently. If one imagines Interstate 40 as the x-axis
and Interstate 25 as the y-axis, then it is natural to take x(t) to be the distance (in miles)
Professor Greenbottle’s car is north of the intersection at time t, and y(t) be the distance the
hit squad is east of the intersection at time t. With these conventions, the information given
says that:
x(t0 ) = −3, x0 (t0 ) = 70, y(t0 ) = −4, and y 0 (t0 ) = −50.
The signs will be a bit different through the rest of the calculation, but the final result will be
the same.
Finally, you could do everything in physicists’ notation. I will only show the first version. The
equation relating the quantities is z 2 = a2 + b2 . Differentiating (using the chain rule, because
everything is a function of t!), we get
dz
da
db
2z
= 2a + 2b ,
dt
dt
dt
so
dz
da
db
z
=a +b .
dt
dt
dt
Substituting values (implicitly putting t = t0 , and using z = 5 at t = t0 , as above):
dz
5·
= 3 · (−70) + 4 · 50 = −10.
dt
dz
So
= −2.
dt
11. (15 points) Use the methods of calculus to find the exact values of x at which the function
f (x) = 4x3 − 27x2 − 30x + 2 takes its absolute minimum and maximum on the interval [−1, 1].
Solution: We apply the procedure for continuous functions on closed finite intervals. That
is, we evaluate f at all critical numbers and at the endpoints, and compare values.
To find the critical numbers, we differentiate f and solve the equation f 0 (x) = 0. The
derivative of f is
f 0 (x) = 12x2 − 54x − 30.
To find the roots, either use the quadratic formula or factor it as
12x2 − 54x − 30 = 6(2x + 1)(x − 5).
Either way, the roots are x = − 12 and x = 5.
We now have two critical numbers, namely − 12 and 5. Of these, 5 is not in the interval under
consideration, so we ignore it. (I must see you reject 5. If I don’t see this, I will assume you
didn’t correctly solve the equation f 0 (x) = 0.) So we must compare the values of f at − 21 , and
at the endpoints −1 and 1. We evaluate:
+ 30
+2 = 39
.
f (−1) = −4−27+30+2 = 1, f (1) = 4−27−30+2 = −51, and f − 12 = − 48 − 27
4
2
4
The largest of these is f − 21 and the smallest is f (1), so the absolute maximum on the interval
[−1, 1] occurs at x = − 12 and the absolute minimum on the interval [−1, 1] occurs at x = 1.
Note that x = 5 is not correct for the minimum, even though f (5) = −323, because 5 is not in
the interval [−1, 1].
12. The picture below is the graph of the DERIVATIVE y = f 0 (x) for a certain function f.
CAUTION: You are given the graph of the derivative f 0 (x), not the graph of f (x), but you
are asked questions about f (x).
4
2
2
-2
4
6
-2
In case anyone wants to experiment, the function in the graph happens to have the formula
1 2
1 3
f 0 (x) = − 16
x (x − 6) = − 16
x + 38 x2 .
The function f has the formula
1 4
x + 18 x3 + C
f 0 (x) = − 64
for some unknown constant C.
(a) (7 points) Find and identify all local minimums and local maximums of f in the interval
(−3, 7). Justify your answer. (Note that you are being asked about f, not the function shown
in the graph, which is f 0 .)
Solution: To solve the problem, we first read directly from the graph of f 0 all the numbers x
with f 0 (x) = 0. These are x = 0 and x = 6. So the critical numbers of f are 0 and 6.
At x = 0, we can read directly from the graph of f 0 that f 0 does not change sign at 0. Indeed,
for x close to but less than 0, we see that f 0 (x) > 0, and for x close to but greater than 0, again
f 0 (x) > 0. So f has a critical number, but neither a local maximum nor a local minimum, at
x = 0. (In fact, the slope of f 0 at x = 0, that is, f 00 (0), is equal to zero. Since the slope of
f 0 does change sign at 0, we know that the value of f 00 changes sign at 0. Therefore f has an
inflection point at x = 0.)
At x = 6, we can read directly from the graph of f 0 that f 0 changes from positive to negative
at 6. Therefore f changes from increasing to decreasing at 6, so f has a local maximum at 6.
(Here is an alternate way to see that f has a local maximum at 6. We can see from the graph
that the slope of f 0 is negative at 6. Therefore f 00 (6) < 0, so f is concave down there, and so
has a local maximum at 6.)
There is no local maximum or minimum at x = 4, since f 0 (4) > 0. (In fact, f has an inflection
point at x = 4.)
A correct solution must consider what happens at both 0 and 6, and must show that there
is neither a local minimum nor a local maximum at 0, but need not show that there is an
inflection point there, and need not consider what happens at x = 4.
Here is a graph of one possible choice for f, namely the one with f (5) = 0 (as in the next
part):
2
-2
4
6
-2
-4
-6
-8
-10
(b) (4 points) Suppose f (5) = 0. Is f (4) positive, negative, or is there not enough information
to decide? Justify your answer. (Note that you are being asked about f, not the function shown
in the graph, which is f 0 .)
Solution: Reading directly from the graph of f 0 , we see that f 0 (x) > 0 for x in [4, 5]. Therefore
f is strictly increasing on this interval. So f (4) < f (5) = 0.
See the graph of f above.
(c) (6 points) Find all inflection points of f in the interval (−3, 7). Justify your answer.
(Note that you are being asked about f, not the function shown in the graph, which is f 0 .)
Solution: The function f has an inflection point where its concavity changes.
At x = 0, read from the graph that f 0 (the function shown) changes from decreasing to
increasing, so its derivative f 00 changes from negative to positive, so f changes from concave
down to concave up. Therefore f has an inflection point at x = 0.
At x = 4, read from the graph that f 0 (the function shown) changes from increasing to
decreasing, so its derivative f 00 changes from positive to negative, so f changes from concave
up to concave down. Therefore f has an inflection point at x = 4.
(The function f 0 has an inflection point at x = 2, but f does not have an inflection point at
x = 2.)
See the graph of one possible choice for f, namely the one with f (5) = 0, above.
(d) (7 points) Suppose f (4) = 3 and g(x) = sin(5f (x)). Find g 0 (4).
Solution: We use the chain rule, reading the value f 0 (4) = 2 from the graph:
g 0 (x) = cos(5f (x)) · 5f 0 (x),
so
g 0 (4) = cos(5f (4)) · 5f 0 (4) = cos(15) · 10 = 10 cos(15).
13. (20 points) Richard Q. Snape owns a large tract of land next to a long straight lakeshore.
He wants to fence off a rectangular field, with the lakeshore being one side and fences along the
other three sides. The field is furthermore to be divided in three sections, each of the same area,
by two additional fences running parallel to the lakeshore. If the total area of the fenced off
area is to be 6 square kilometers, find the dimensions of the field which uses the least amount
of fence.
Include units, and be sure to verify that your maximum or minimum really is what you claim
it is.
(Show a full mathematical solution. A correct guess with no valid work will receive no credit,
and a correct number supported only by heuristic reasoning will receive very little credit.)
Solution: The fenced off area has the shape of a rectangle. Let x be the length of the side
parallel to the lakeshore, and let t be the length of the side parallel to the lakeshore. There
are a total of five straight sections of fence. Two are the edges of the rectangle which are
perpendicular to the lakeshore, and each of these has length y. One is the edge of the rectangle
which is parallel to the lakeshore but not along it; this fence has length x. The other two are
inside the rectangle and also parallel to the lakeshore; each of these also has length x. Therefore
the total length of fence is l = 2y + 3x. We want to minimize this.
The total area is xy, so we are given xy = 6. Solving for y, we get y = 6/x. (There is nothing
wrong with solving for x instead.) Substituting back in the formula for l, we see that we want
to minimize
6
12
+ 3x = 3x +
= 3x + 12x−1 .
l(x) = 2
x
x
We look for the appropriate restrictions. Clearly we must have x ≥ 0. In fact, we must have
x > 0, since of x = 0 the areas is zero. Similarly, y > 0. Since x = 6/y, this restriction gives no
additional information about x. Thus, we must minimize the function
l(x) = 3x + 12x−1
on the interval (0, ∞).
We look for critical numbers. We have
l0 (x) = 3 + 12x−2 (−1) = 3 −
12
.
x2
So we solve:
12
=0
x2
12
3= 2
x
2
3x = 12
x4 = 4
x = ±2.
We reject x = −2, because −2 is not in (0, ∞). (I must see you do this; otherwise, I don’t know
that you even knew x = −2 is a solution.)
Do we have a maximum, minimum, or neither? The easiest test is the second derivative test:
3−
l00 (x) = −12x−3 (−2) = 24x−3 ,
which is strictly positive on the entire interval (0, ∞). Therefore f is concave up on the entire
interval (0, ∞), and any critical number is a global minimum.
So the minimum occurs when x = 2, and y = 6/x = 3. Thus, the side parallel to the lakeshore
should be 2 km long, and the side perpendicular to the lakeshore should be 3 km long.
For those who used them, here are other tests for a minimum.
The first derivative test: For 0 < x < 2, we have 12/x2 > 12/22 = 3, so l0 (x) < 0. For x > 2,
we have 12/x2 < 12/22 = 3, so l0 (x) > 0. Thus l is decreasing on (0, 2) and increasing on (2, ∞),
and therefore its global minimum on (0, ∞) is at x = 2.
The endpoint test: The endpoints are not in the interval, but it s easy to check that
12
12
lim l(x) = lim+ 3x +
= ∞ and lim l(x) = lim 3x +
= ∞.
x→∞
x→∞
x→0+
x→0
x
x
Since l(2) = 12 is less than either of these, the minimum occurs at x = 2.
Finally, we show what the solution looks like in Leibniz notation. We have
6
l = 2y + 3x = 2
+ 3x = 3x + 12x−1
x
as above. Then
dl
12
= 3 − 2.
dx
x
This is zero when x = ±2, and we reject x = −2, as above. The rest of the solution is the
same.
14. (5 points) The picture below shows the graph of y = f (x) for a particular function f .
5
4
3
2
1
-4
-2
2
4
6
8
10
12
14
Use this graph to find a number x in the interval shown such that f (x) > 0, f 0 (x) < 0, and
f 00 (x) > 0, being sure to explain why your choice of x satisfies these conditions. If no such x
exists, explain why not.
Solution: For reference, the function in the graph has the formula
1
f (x) = 2000
−491 + 420x2 − 68x3 + 3x4 .
We are looking for values of x at which the function is positive (f (x) > 0), decreasing
(f 0 (x) < 0), and concave up (f 00 (x) > 0). Any x with −4 < x < −1 will do,√as will values of x
near 10 but less than 10. (In fact, this is true for x in the interval 31 17 + 79 , 10 , in which
√ 1
17 + 79 ≈ 8.6294.) So x = −2, x = −3, and x = 9 would be correct answers, and there
3
are many others.
The graph below shows the tangent lines at x = −2, x = −3, and x = 9.7. You can see that
each tangent line has negative slope (so f 0 (x) < 0), and and is below the graph (so f 00 (x) > 0).
5
4
3
2
1
-4
-2
4
2
6
8
10
12
14
15. (17 points) Graph the function f (x) = e2x −ex using the methods of calculus. In particular,
determine exactly the x-intercept(s), y-intercept(s), intervals of increase and decrease, critical
numbers, extreme values, intervals of concavity, and inflection points.
Solution: The domain clearly consists of all real numbers.
The y-intercept is f (0) = 0.
To find the x-intercepts, we solve f (x) = 0:
e2x − ex = 0
(ex )2 − ex = 0
ex (ex − 1) = 0
ex = 0 or ex − 1 = 0.
Now ex is never zero, and ex = 1 exactly when x = 0, so the only x-intercept is at x = 0.
There are no vertical asymptotes, since f is defined and continuous on (−∞, ∞).
For horizontal asymptotes, we check
lim f (x) = lim (e2x − ex ) = lim ex (ex − 1).
x→∞
x→∞
x→∞
Since limx→∞ ex = ∞ and limx→∞ (ex − 1) = ∞, we get limx→∞ f (x) = ∞. Therefore there is
no horizontal asymptote in the positive direction. Also,
lim f (x) = lim e2x − lim ex = 0 − 0 = 0.
x→−∞
x→−∞
x→−∞
This, the x-axis is a horizontal asymptote in the negative direction.
Next, we need the derivative:
f 0 (x) = 2e2x − ex .
To find the critical numbers, we solve f 0 (x) = 0:
2e2x − ex = 0
2(ex )2 − ex = 0
ex (2ex − 1) = 0
ex = 0 or 2ex − 1 = 0.
Now ex is never zero, and 2ex = 1 exactly when ex = 12 , so the only critical number is x =
ln 12 = − ln(2).
On the interval (−∞, − ln(2)), the factor ex above is positive and 2ex − 1 is negative. So f 0
is negative on this interval. Therefore f is decreasing on this interval.
On the interval (− ln(2), ∞), the factor ex above is positive and 2ex − 1 is positive. So f 0 is
positive on this interval. Therefore f is increasing on this interval.
It now follows that f has a local minimum at x = − ln(2). For use in graphing, we compute decimal approximations for the values of x and f (x), getting − ln(2) ≈ −0.693147 and
f (− ln(2)) = − 41 = −0.25.
Now, we need the second derivative:
f 00 (x) = 4e2x − ex = ex (4ex − 1).
By a similar calculation as above, this is zero exactly when x = − ln(4).
On the interval (−∞, − ln(4)), the factor ex above is positive and 4ex − 1 is negative. So f 00
is negative on this interval. Therefore f is concave down on this interval.
On the interval (− ln(4), ∞), the factor ex above is positive and 4ex − 1 is positive. So f 00 is
positive on this interval. Therefore f is concave up on this interval.
It follows that there is an inflection point at x = − ln(4). For use in graphing, we compute decimal approximations for the values of x and f (x), getting − ln(4) ≈ −1.38629 and
3
= −0.1875.
f (− ln(4)) = − 16
Now we are almost ready to graph the function. But we should plot atleast one point to the
right of the y-axis. I chose x = 21 , and computed the approximation f 12 ≈ 1.06956.
Here is the graph, showing as black dots the the point on the graph corresponding to the
inflection point, the point on the graph corresponding to the local minimum, the point on the
graph corresponding to the x-intercept, and the point on the graph corresponding to x = 12 .
2.5
2.0
1.5
1.0
0.5
-3
-2
1
-1
-0.5
16. (12 points) Let f (x) = esin(x)+ax , where a is a constant. Find f 00 (x).
Solution: Start by calculating the first derivative using the chain rule:
d
f 0 (x) = esin(x)+ax (sin(x) + ax) = esin(x)+ax (cos(x) + a) = (cos(x) + a) esin(x)+ax .
dx
(Remember that a is a constant!) Differentiate this using the product rule. At the point where
we need the derivative of esin(x)+ax , we simply substitute the answer we got above:
d sin(x)+ax d
(cos(x) + a) · esin(x)+ax + (cos(x) + a)
e
f 00 (x) =
dx
dx
= − sin(x)esin(x)+ax + (cos(x) + a) · (cos(x) + a) esin(x)+ax
= (cos(x) + a)2 − sin(x) esin(x)+ax .
(Again, a is a constant, so its derivative is zero.) This is about as simple an expression as is
possible for f 00 (x).
17. (11 points) If
for
dy
x
= tan(x − y) − cos(1), find
by implicit differentiation. (You must solve
y
dx
dy
.)
dx
Solution: Let’s write it with y as an explicit function y(x) of x:
x
= tan(x − y(x)) − cos(1).
y(x)
Differentiate both sides with respect to x, using the quotient rule on the left and the chain rule
on the right:
1 · y(x) − xy 0 (x)
d
= sec2 (x − y(x)) (x − y(x)) = sec2 (x − y(x)) (1 − y 0 (x)) .
2
[y(x)]
dx
(The derivative of cos(1) is zero because cos(1) is a constant.)
Now solve for y 0 (x):
y(x) − xy 0 (x)
= sec2 (x − y(x)) − sec2 (x − y(x))y 0 (x)
[y(x)]2
y(x) − xy 0 (x) = [y(x)]2 sec2 (x − y(x)) − [y(x)]2 sec2 (x − y(x))y 0 (x)
y(x) − [y(x)]2 sec2 (x − y(x)) = x − [y(x)]2 sec2 (x − y(x)) y 0 (x)
y(x) − [y(x)]2 sec2 (x − y(x))
0
y (x) =
.
x − [y(x)]2 sec2 (x − y(x))
This expression can’t be further simplified.
dy
For those who prefer the other notation, here it is written with
. Differentiate with respect
dx
to x, just as before:
dy
1 · y − x dx
d
dy
2
2
= sec (x − y) (x − y) = sec (x − y) 1 −
.
y2
dx
dx
Now solve for
dy
:
dx
dy
y − x dx
dy
= sec2 (x − y) − sec2 (x − y)
2
y
dx
dy
dy
= y 2 sec2 (x − y) − y 2 sec2 (x − y)
y−x
dx
dx
dy
y − y 2 sec2 (x − y) = x − y 2 sec2 (x − y)
dx
2
2
dy
y − y sec (x − y)
=
.
dx
x − y 2 sec2 (x − y)
As before, this expression can’t be further simplified.
18. (12 points) Prove that there exists a unique real solution to the equation 3x5 + x − 2 = 0.
Give a complete justification for any theorems that you use, in particular being sure to check
that the hypotheses hold. (No credit will be given for calculator approximations to a solution,
or for arguments based on a calculator graph.)
Solution: For uniqueness, we will use Rolle’s Theorem (or the Mean Value Theorem). Set
f (x) = 3x5 + x − 2. Then f is a polynomial function, so it is differentiable everywhere. In fact,
f 0 (x) = 15x4 + 1, which is always strictly positive. If the equation f (x) = 0 had two different
solutions a and b, with a < b, then Rolle’s Theorem would imply that there is c in (a, b) such
that f 0 (c) = 0. However, we just saw that f 0 (c) can never be zero.
For existence, we will use the Intermediate Value Theorem. As above, set f (x) = 3x5 + x − 2.
Then f is a polynomial function, so it is continuous at all real numbers. By substitution, you
can check that f (0) = −2 and f (1) = 2. (These numbers were found by trying small integers at
random. Note that x = 0 is a good choice because f (0) is particularly easy to evaluate. Using
f (−1) and f (1), for example, would also work fine.) Since f (0) < 0 < f (1), the Intermediate
Value Theorem therefore tells us that there is some c in the interval (0, 1 such that f (c) = 0.
Here is the graph of f for x near zero (not required as part of the solution):
10
5
-1.5
-1.0
0.5
-0.5
1.0
1.5
-5
-10
-15
Note that the quadratic formula can’t be used, since the original equation is not a quadratic
equation.