Calculus 1: Sample Questions, Exam 2, Solutions
1. Compute the following derivatives, and show your work:
(a)
d
ln(t2 + 1)
dt
Solution: Use the Chain Rule to find
d
1
2t
ln(t2 + 1) = 2
· 2t = 2
.
dt
t +1
t +1
(b)
d tan−1 x
dx x
Solution: Use the Quotient Rule
x
d tan−1 x
=
dx x
1
x2 +1
− tan−1 x · 1
x2
=
x − (x2 + 1) tan−1 x
.
x2 (x2 + 1)
(c) (f −1 )0 (3), where f is a differentiable one-to-one function satisfying
f (1) = 3, f (3) = 2, f 0 (1) = −2, f 0 (3) = − 13 .
1
. Then, by taking f −1
Solution: Recall that (f −1 )0 (x) = f 0(f −1
(x))
of both sides of 3 = f (1), we find
f −1 (3) = f −1 (f (1)) = 1.
Now we may compute
(f −1 )0 (3) =
1
f 0 (f −1 (3))
=
1
f 0 (1)
=
1
1
=− .
−2
2
2. Identify the critical points and find the maximum value and the minimum value for f (x) = x3 − 3x + 2 on the interval [0, 2]. Show your
work.
Solution: Compute f 0 (x) = 3x2 − 3. f 0 (x) = 0 when 3x2 − 3 = 0,
x2 = 1. So x is 1 or −1. Only the critical point x = 1 is in our interval
[0, 2].
So we have 3 points to test: Endpoints x = 0, 2, and the critical point
x = 1. Now compute f (0) = 2, f (1) = 0, f (2) = 4. So the min is at
f (1) = 0, and the max is at f (2) = 4.
3. Sketch the graph of a function y = g(x) that has all the following
properties:
(a) The domain of g is the open interval (−3, 3).
(b) g is an odd function.
(c) lim− g(x) = +∞.
x→3
0
(d) g (x) < 0 for x in (−1, 1), while g 0 (x) > 0 for x in (−3, −1) and
for x in (1, 3).
(e) g 0 (−1) = g 0 (1) = 0.
(f) g 00 (x) < 0 for x in (−3, 0), while g 00 (x) > 0 for x in (0, 3).
(g) g 00 (0) = 0.
(h) g(1) = −2, g(−1) = 2.
4. Show that for a rectangle of given perimeter 4, the one with the maximum area is a square. (Recall that for a rectangle with side lengths a
and b, the perimeter is given by 2a + 2b, and the area is given by ab.)
Solution: The perimeter P = 4 = 2a + 2b, while the area A = ab. We
want to maximize A. To make A a function of a single variable, solve
for 4 = 2a + 2b to find a = 2 − b, and plug in to find
A = ab = (2 − b)b = 2b − b2 .
To find critical points, compute
dA
= 2 − 2b,
db
d2 A
= −2.
db2
So the only critical point is when 2 − 2b = 0, which is when b = 1. So
A00 (1) = −2 < 0 and the second derivative test shows b = 1 is a local
maximum. Since this is the only critical point, this must be the global
maximum. For b = 1, then we find a = 2 − b = 2 − 1 = 1. So for the
maximum area, the two sides b = 1 and a = 1, and so the rectangle is
a square.
5. Compute the following limits. Show your work if necessary. Each
answer should be a real number, ∞, −∞, or “does not exist.”
3x
.
x→1
−1
Solution: If we plug in at x = 1, we find
(a) lim+
x2
3·1
3
= ,
2
1 −1
0
which indicates an infinite limit. We must now determine its sign:
As x → 1+ , x > 1, which implies 3x > 3 · 1 > 0 is positive and
x2 > 12 = 1, so x2 − 1 > 0 is positive as well. So the signs of x23x−1
+
are of the form +
as x → 1+ , and so the fraction is positive, and
the answer is +∞.
1
(b) lim
.
x→−∞ 1 − ex
Solution: As x → −∞, ex → 0, and so
1
1
=
= 1.
x
x→−∞ 1 − e
1−0
lim
(c) limπ tan θ.
θ→ 2
Solution: One way is to recall the graph of tan θ, which has
a vertical asymptote at θ = π2 . But the one-sided limits differ,
as limθ→ π + tan θ = −∞, while limθ→ π − tan θ = +∞. Thus the
2
2
two-sided limit limθ→ π2 tan θ does not exist.
Another way is to recall that tan θ = sin θ/ cos θ. As θ → π2 ,
sin θ →
1, while cos θ = 0. Thus by plugging in, our limit would
sin π2
be cos π = 01 will be infinite. Near θ = π2 , sin θ is positive (since sin
2
is a continuous function and sin π2 = 1). In particular, as θ →
sin θ > 0. Here cos θ > 0 also, and so
lim
tan θ = lim
π−
π−
θ→ 2
θ→ 2
On the other hand, as θ →
cos θ < 0, and so
π+
,
2
sin θ
= +∞.
cos θ
sin θ > 0 as well. But in this case,
lim
tan θ = lim
π+
π+
θ→ 2
π−
,
2
θ→ 2
sin θ
= −∞.
cos θ
Since the two one-sided limits disagree, we see
limπ tan θ = limπ
θ→ 2
θ→ 2
sin θ
cos θ
does not exist.
6. Consider h(t) = 2t3 − 3t2 + 5.
(a) Compute the critical points of h(t). Show your work.
Solution: h0 (t) = 6t2 − 6t. h0 (t) always exists, so we just need to
find h0 (t) = 0.
6t2 − 6t = 0,
6t(t − 1) = 0.
So t = 0 and t = 1 are the only critical points of h(t).
(b) Compute the intervals where h(t) is increasing, and where h(t) is
decreasing. Show your work.
Solution: The critical points t = 0, 1 split the real line into 3
intervals (−∞, 0), (0, 1), (1, ∞). Pick representative points in
each interval.
For t = −1 in (−∞, 0), h0 (−1) = 6(−1)2 − 6(−1) = 12 > 0. So
h0 (t) > 0 on the interval (−∞, 0), and h is increasing there.
For t = 1/2 in (0, 1), h0 (1/2) = 6(1/2)2 − 6(1/2) = −6/4 < 0. So
h0 (t) < 0 on the interval (0, 1), and h is decreasing there.
For t = 2 in (1, ∞), h0 (2) = 6(2)2 − 6(2) = 12 > 0. So h0 (t) > 0
on the interval (0, ∞), and h is increasing there.
(c) Compute the intervals where h(t) is concave up, and where h(t)
is concave down. Show your work.
Solution: Compute h00 (t) = Dt (6t2 −6t) = 12t−6. h00 (t) = 0 only
if t = 1/2. This splits the real line into two intervals (−∞, 1/2)
and (1/2, ∞). Pick representative points in each interval.
For t = 0 in (−∞, 1/2), h00 (0) = 12(0) − 6 = −6 < 0. So h00 (t) < 0
on the interval (−∞, 1/2), and h is concave down there.
For t = 1 in (1/2, ∞), h00 (1) = 12(1) − 6 = 6 > 0. So h00 (t) > 0 on
the interval (1/2, ∞), and h is concave up there.
√
7. Use an appropriate linear approximation to approximate 3 1.03. Your
answer should be a decimal number. Show your work.
√
2
1
Solution: For f (x) = 3 x = x 3 , we have f 0 (x) = 31 x− 3 . Note that
x = 1 is a nice value for f near our value x = 1.03. Compute the
linearization of f at x = 1:
√
2
3
L(x) = f 0 (1)(x − 1) + f (1) = ( 31 · 1− 3 )(x − 1) + 1 = 13 (x − 1) + 1.
So the approximate value of
√
3
1.03 = f (1.03) is
L(1.03) = 31 (1.03 − 1) + 1 = 13 (0.03) + 1 = 0.01 + 1 = 1.01