Math 2250 Exam #3 Practice Problem Solutions
1. Determine the absolute maximum and minimum values of the function
x
.
f (x) =
1 + x2
Answer: Notice that f is defined for all x. Also,
x
= 0,
x→±∞ 1 + x2
lim f (x) = lim
x→±∞
so f doesn’t go off to infinity.
Now, to find the critical points, compute
f 0 (x) =
1 − x2
(1 + x2 )(1) − x(2x)
=
,
2
2
(1 + x )
(1 + x2 )2
which equals zero precisely when x2 = 1, or x = ±1. Thus, we just need to evaluate f at the critical
points:
f (1) =
1
2
f (−1) = −
1
2
Since f limits to 0 in both directions, we see that these are the absolute maximum and absolute
minimum values of the function.
2. A specialty publisher has typically sold trade paperbacks for $15, averaging 300 sales per week. The
publisher has found that increasing the price by 50 cents reduces sales by 10 per week, so the demand
x
function is p(x) = − 20
+ 30. If the books cost $10 each to make, what price should the publisher
charge to maximize profit?
Answer: Since we know the demand function, we can figure out the publisher’s revenue:
R(x) = xp(x) = −
x2
+ 30x.
20
Also, the costs are
C(x) = 10x.
Therefore, the publisher’s profit is
P (x) = R(x) − C(x) = −
x2
x2
+ 30x − 10x = − + 20x.
20
20
To maximize this, find the critical points:
P 0 (x) = −
x
+ 20,
10
which equals zero when x = 200. Since P 00 (x) = −1/10 < 0, this critical point must be the maximum
(by the second derivative test). Therefore, the publisher should charge
p(200) = −
200
+ 30 = −10 + 30 = 20
20
in order to maximize profits.
1
3. Find the inflection points for the function
f (x) = 8x + 3 − 2 sin x,
0 < x < 3π.
Answer: Notice that
f 0 (x) = 8 − 2 cos x
and
f 00 (x) = 2 sin x.
Now, sin x changes from positive to negative at x = π and from negative to positive at x = 2π. Since
f (π) = 8π + 3 − 2 sin π = 8π + 3
f (2π) = 8(2π) = 3 − 2 sin 2π = 16π + 3
the inflection points for f between 0 and 3π are
(π, 8π + 3),
(2π, 16π + 3).
4. Evaluate the limit
lim x2 csc2 x.
x→0+
Answer: Re-write the limit as
x2
2 .
x→0 sin x
Since both numerator and denominator go to zero, we can use L’Hôpital’s Rule, so this limit equals
lim+
lim
x→0+
2x
.
2 sin x cos x
Again, both numerator and denominator go to zero, so apply L’Hôpital’s Rule again to get:
lim
x→0+
2 cos2
2
2
= = 1.
2
2
x − 2 sin x
5. Given that
f 0 (t) = 2t − 3 sin t,
f (0) = 5,
find f .
Answer: We know that
Z
f (t) =
f 0 (t)dt =
Z
(2t − 3 sin t)dt = t2 + 3 cos t + C.
Now, since
5 = f (0) = 02 + 3 cos 0 + C = 3 + C,
we see that C = 2, so
f (t) = t2 + 3 cos t + 2.
6. Find the absolute minimum value of the function
f (x) =
for x > 0.
2
ex
x
Answer: Notice that
lim+
x→0
ex
= +∞
x
and
ex
ex
= lim
=∞
x→∞ 1
x→∞ x
by L’Hôpital’s Rule. Therefore, we should expect the absolute minimum to occur at a critical point.
To find the critical points, take the derivative:
lim
f 0 (x) =
x−1
xex − ex
= ex 2 .
x2
x
This is zero only when x − 1 = 0, meaning that f has a single critical point at x = 1. Just to doublecheck that this is indeed the minimum, note that f changes sign from negative to positive at x = 1,
so, by the first derivative test, f has its minimum at x = 1. The minimum value of f is, thus,
f (1) =
e1
= e.
1
7. Evaluate the integral
Z
sec 3t tan 3tdt.
Answer: It’s easy to check that
sec 3t
3
is an antiderivative for sec 3t tan 3t, so
Z
sec 3t tan 3tdt =
3
sec 3t
+ C.
3