Math 1102: Calculus I (Math/Sci majors)
MWF 3pm, Fulton Hall 230
Exam 2, v.2 Solutions
Please write neatly, and show all work. Caution: An answer with no work is wrong! Note that point
values for problems are indicated, and partial credit will be given for partial solutions.
1. (10 pts) State the Mean Value Theorem.
Solution 1. If f (x) is a function that is differentiable on the interval (a, b) and continuous on the interval
[a, b], then there exists a c in (a, b) so that
f (b) − f (a)
f 0 (c) =
.
b−a
2. (10 pts) Let f be a function on the interval [a, b]. Define the following:
(i) f is decreasing on [a, b].
(ii) f has a minimum at x = c.
(iii) f has a critical point at x = c.
Solution 2.
(1) If f (x1 ) ≥ f (x2 ) for any x1 ≤ x2 in [a, b], then f is decreasing on [a, b].
(2) If f (x) ≥ f (c) for any x in [a, b] then f has a minimum at x = c.
(3) If f 0 (c) = 0, then f has a critical point at x = c.
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3. (10 pts) Find the maxima and minima of 2x2 + 4x − 3 on the interval [−4, 1].
Solution 3. Let f (x) = 2x2 + 4x − 3. Since the derivative f 0 (x) is given by 4x + 4, the only critical
point of f (x) is at x = −1. We check the critical value, and the values at the endpoints:
f (−1) = 2 − 4 − 3 = −5
f (−4) = 32 − 16 − 3 = 13
f (1) = 2 + 4 − 3 = 3.
Thus the maximum of f on the interval [1, 5] is x = −4 and the minimum is x = −1.
4. (10 pts) Suppose f (x) is a differentiable function on an interval (a, b), and suppose that f 0 (x) = 0 for
all x in (a, b). Show that f is constant.
Solution 4. Suppose f were not constant on (a, b), so that there exist x1 and x2 in (a, b) such that
f (x1 ) 6= f (x2 ). By the Mean Value Theorem applied to the interval between x1 and x2 , there is a c so
that
f (x2 ) − f (x1 )
f 0 (c) =
.
x2 − x1
But f (x1 ) 6= f (x2 ) implies that f 0 (c) 6= 0, contradicting the assumption that f 0 (x) = 0 for x in (a, b).
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5. (15 pts) Use Taylor’s Theorem to estimate cos 21 , with n = 3. Estimate the error in your computation.
Solution 5. Let f (x) = cos x. We have f 0 (x) = − sin xx, f 00 (x) = − cos x, f (3) (x) = sin x, and
f (4) (x) = cos x. We compute the Taylor polynomial P3 , with x0 = 0:
c0 = f (0) = 1,
c1 = f 0 (0) = 0,
f 00 (0)
= −1, and
2
f (3) (0)
c3 =
= 0,
3!
c2 =
so that
x2
.
2
By Taylor’s Theroem, there is a c between 0 and 1/2 so that
4
1
f (4) (c) 1
1
f
= P3
+
2
2
4!
2
1 1 cos c
7 cos c
=1− · + 4 = + 4 ,
2 4
2 4!
8
2 4!
and we have
cos c 1
1
7
f
=
≤
.
−
2
8
24 4!
24 4!
P3 (x) = 1 −
This implies that cos 21 is within .003 of 7/8.
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6. (15 pts) Suppose a and b are two positive numbers whose sum is 1. Find the values of a and b that
minimize the sum
2 3
+ .
a b
Solution 6. Since both a and b are positive numbers that sum to 1, each is between 0 and 1. Let
b = 1 − a. We seek to minimize the function
3
2
.
f (a) = +
a 1−a
We find the critical points:
2
3
3
2
f 0 (a) = − 2 −
· (−1) =
− 2.
2
2
a
(1 − a)
(1 − a)
a
0
If f (a) = 0 then
2
3
=
a2
(1 − a)2
so that
2(1 − a)2 = 3a2 .
To solve for a we put everything on one side:
a2 + 4a − 2 = 0.
Thus the solutions are given by
√
√
−4 ± 24
= −2 ± 6.
a=
2
√
Of these solutions, only −2 + 6 is positive. We check the ‘values’ of f on the boundary:
2
3
lim− +
= +∞, and
a→1 a
1−a
2
3
lim+ +
= +∞.
a→0 a
1−a
√
Thus a = −2 + 6 is the minimum value
√ interval (0, 1), and the values of a and b that
√ of f (a) on the
minimize the sum above are a = −2 + 6 and b = 3 − 6.
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7. (15 pts) Consider the lines in the plane containing the origin (0, 0) and a point on the graph of
f (x) = log x. Which such line has maximum slope?
Solution 7. A point on the graph of f (x) has coordinates (x, log x) for x > 0. The line through the
origin and (x, log x) has slope given by
log x − 0
m(x) =
.
x−0
Thus we seek to maximize m(x) for x > 0. We compute the critical points:
1
x
· x − log x
1 − log x
=
.
2
x
x2
Thus the only critical point occurs when log x = 1, or x = e. It remains to check the ‘values’ of f as x
approaches the boundary.
0
m (x) =
Since
lim 1/x = +∞
x→0+
and
lim log x = −∞
x→0+
, we may use the product law for infinite limits to conclude that
1
log x
= lim+ · log x = −∞.
lim+
x→0 x
x→0
x
To compute
log x
lim
x→+∞ x
we use l’Hôpital’s Rule: Evidently
limx→+∞ log x
+∞
=
limx→+∞ x
+∞
is an indeterminate form, and
1
(log x)0
= lim
= 0.
lim
0
x→+∞ x
x→+∞ (x)
Thus
log x
lim
= 0.
x→+∞ x
Since
log e
1
m(e) =
=
e
e
is greater than 0 and −∞, the point x = e is the maximum of m(x) for x > 0.
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8. (15 pts) Show that ex + x5 + 1 has exactly one real root.
(Hint: Use the Intermediate Value Theorem to show there is at least one root, and Rolle’s Theorem to
find out there’s not two).
Solution 8. Let f (x) = ex + x5 + 1. Since
lim f (x) = +∞
x→+∞
and
lim f (x) = −∞
x→−∞
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(note that we are using the fact that (−x) = −x5 ), there are values x1 and x2 so that f (x1 ) > 0 and
f (x2 ) < 0. By the Intermediate Value Theorem, there is a c between x1 and x2 so that f (c) = 0, so that
f (x) has at least one real root.
If there were two roots, say c1 and c2 , then by Rolle’s Theorem there would be a c0 between c1 and c2 so
that f 0 (c0 ) = 0. But
f 0 (x) = ex + 5x4
satisfies f 0 (x) ≥ ex > 0 for all x, so no such c0 can exist. Thus f has at most one real root, and we are
done.
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