Math 113 Exam #3 Solutions
1. Evaluate the limit
lim x tan
x→+∞
1
.
x
Answer: Notice that, as x → +∞, x1 goes to zero. Since tan(0) = 0, we see that the above limit takes
the form of ∞ · 0. Therefore, I can convert it into a standard form for applying L’Hôpital’s Rule as
follows:
tan x1
1
.
= lim
lim x tan
1
x→+∞
x→+∞
x
x
Now, both numerator and denominator go to zero, so L’Hôpital’s Rule says that the above limit is
equal to
sec2 x1 · −1
1
2
x2
lim
=
lim
sec
.
−1
x→+∞
x→+∞
x
x2
1
cos θ
In turn, since sec θ =
for any θ, the above limit is equal to
lim
x→+∞
1
cos2
2. For what value of c does the function f (x) = x +
1
x
=
c
x
1
= 1.
cos 0
have a local minimum at x = 3?
Answer: If f has a local minimum at x = 3, then it must be the case that f has a critical point at
x = 3, meaning that f 0 (3) = 0. Now,
c
f 0 (x) = 1 − 2 ,
x
so f 0 (3) = 0 implies that
c
1 − 2 = 0,
3
or, equivalently,
c
= 1.
9
Hence, f has a critical point at x = 3 only if c = 9. To double-check that f really has a local minimum
here, let c = 9 and use the second derivative test. Since
f 00 (x) =
we see that f 00 (3) =
minimum at x = 3.
18
27
=
2
3
2·9
,
x3
> 0, so the second derivative test says that f does indeed have a local
3. Draw the graph of the function g(x) = 4x3 − x4 . Label any local maxima or minima, inflection points,
and asymptotes, and indicate where the graph is concave up and where it is concave down.
Answer: Notice that
lim
x→±∞
3
4x − x4 = −∞,
so the graph of g(x) has no horizontal asymptotes. Moreover, g(x) is defined for all real numbers, so
its graph has no vertical asymptotes.
Now,
g 0 (x) = 12x2 − 4x3 ,
so g has a critical point when
0 = 12x2 − 4x3 = 4x2 (3 − x).
1
Thus, the critical points of g occur at x = 0 and x = 3. Take the second derivative:
g 00 (x) = 24x − 12x2 = 12x(2 − x).
Hence, g 00 (x) = 0 when x = 0 or x = 2. Notice that g 00 (x) < 0 when x < 0, that g 00 (x) > 0 when
0 < x < 2, and g 00 (x) < 0 when x > 2. Hence g has inflection points at x = 0 and x = 2, and the
second derivative test tells us that g has a local maximum when x = 3 (since g 0 (3) = 0 and g 00 (3) < 0).
In fact, this local maximum is an absolute maximum, since g(x) goes to −∞ when x → ±∞. Putting
this all together yields the following graph
Absolute max: (3,27)
24
16
Inflection point:
(2,16)
8
Inflection point: (0,0)
-5
-4
-3
-2
-1
0
1
2
-8
-16
-24
4. Suppose that
h0 (u) =
u2 + 1
u2
h0 (u) =
u2 + 1
u2
1
= 2 + 2 = 1 + u−2 .
2
u
u
u
and that h(1) = 3.
What is h(2)?
Answer: Notice that
Hence,
Z
h0 (u)du = u − u−1 + C = u −
1
+ C,
u
so
1
+C
u
for some real number C. We can solve for C by plugging in u = 1:
h(u) = u −
3 = h(1) = 1 −
so we see that h(x) = u −
1
u
1
+ C = 0 + C = C,
1
+ 3. Therefore,
h(2) = 2 −
2
1
9
+3= .
2
2
3
4
5
√
5. A rectangle is bounded by the x-axis and the graph of the function f (x) = 25 − x2 as shown in the
figure below. What length and width should the rectangle be so that its area is maximized?
√
Answer: If the top-right corner of the rectangle is at the point (x, y), then y = 25 − x2 and the four
corners of the rectangle will be at the points
p
p
(−x, 0), (−x, 25 − x2 ), (x, 25 − x2 ), (x, 0).
√
Hence, the length of the rectangle is 2x and the width is 25 − x2 . If A(x) is the area of the rectangle,
then
p
A(x) = 2x 25 − x2 .
Notice that x can be no smaller than 0 and no bigger than 5, so we want to maximize the function
A(x) on the interval [0, 5]. First, find the critical points of A. The derivative of A is given by
A0 (x) = 2
p
p
1
2x2
25 − x2 + 2x (25 − x2 )−1/2 (−2x) = 2 25 − x2 − √
,
2
25 − x2
which, by finding a common denominator, can be simplified to
2(25 − x2 )
2x2
50 − 4x2
A0 (x) = √
−√
=√
.
25 − x2
25 − x2
25 − x2
Therefore, A0 (x) = 0 when
0 = 50 − 4x2
25
√5
or, equivalently, when x2 = 50
4 = 2 . Therefore, the critical points of A occur when x = ± 2 . Only
the positive one of these is in the interval [0, 5], so we have three points to check: the two endpoints
x = 0 and x = 5 and the critical point x = √52 .
p
A(0) = 2 · 0 25 − 02 = 0
r
√
5
25
A(5/ 2) = 2 · √
25 −
= 25
2
2
p
A(5) = 2 · 5 25 − 52 = 0.
3
Hence, the absolute maximum of the area of the rectangle occurs when x =
of length
√
5
2x = 2 · √ = 5 2
2
and width
y=
r
p
25 − x2 =
4
25 −
25
5
=√ .
2
2
√5 .
2
This gives a rectangle