Quiz 4 Solutions
Sean Fitzpatrick
November 17, 2013
1. Section 4.3, Problem 46: For the function f (x) =
x2 − 4
,
x2 + 4
(a) Find any vertical or horizontal asymptotes for the graph of f .
Solution: Since the denominator is never negative (x2 + 4 ≥ 4 > 0 for all x),
there are no vertical asymptotes. Since
x2 − 4
= 1,
x→±∞ x2 + 4
lim
we see that y = 1 is a horizontal asymptote.
(b) Find the intervals of increase or decrease.
Solution: The derivative of f is given by
f 0 (x) =
2x(x2 + 4) − 2x(x2 − 4)
16x
= 2
,
2
2
(x + 4)
(x + 4)2
so we have f 0 (x) > 0 for x > 0 and f 0 (x) < 0 for x < 0. Thus, f is decreasing on
(−∞, 0) and increasing on (0, ∞).
(c) Find the local maximum and minimum values.
Solution: The only critical point of f is x = 0, and from part (b) and the first
derivative test, we see that this is a local minimum.
(d) Find the intervals of concavity and the inflection points.
Solution: The second derivative of f is given by
√
√
16(x2 + 4)2 − 2(x2 + 4)(2x)(16x)
64 − 48x2
16(2 − 3x)(16 + 3x)
= 2
=
.
f (x) =
(x2 + 4)4
(x + 4)3
(x2 + 4)3
√
√
√
Thus,
f
is
concave
up
on
(−2/
3.2/
3),
concave
down
on
(−∞,
−2/
3) ∪
√
√
(2/ 3, ∞), and has inflection points (±2/ 3, −1/2).
00
1
(e) Use the information from parts (a)–(d) to sketch the graph of f .
Solution: The graph of f is given as below:
2. Section 4.3, Problem 76: Suppose f and g are both concave upward on (−∞, ∞).
Under what condition on f will the composite function h(x) = f (g(x)) be concave
upward?
Solution: Let h(x) = f (g(x)). Then h0 (x) = f 0 (g(x))g 0 (x) by the chain rule, so using
the product and chain rules, we have
h00 (x) = (f 00 (g(x))g 0 (x))g 0 (x) + f 0 (g(x))g 00 (x) = f 00 (g(x))(g 0 (x))2 + f 0 (g(x))g 00 (x).
Since f and g are concave up, we have f 00 (g(x)) > 0 and g 00 (x) > 0 for all x, and of
course (g 0 (x))2 ≥ 0. Thus, we are guaranteed that h00 (0) > 0 if f 0 (g(x)) > 0 for all x.
Thus, if f is increasing on (−∞, ∞), then h will be concave up.
3. Section 4.4, Problem 42: Evaluate the limit
√
lim xe−x/2 .
x→∞
Solution: Since e−x/2 goes to zero as x → ∞, this is a limit of indeterminate type
0 · ∞. We rewrite it as
√
x
lim x/2 ,
x→∞ e
∞
which is of type
, so l’Hospital’s rule may be applied, giving us
∞
√
1/2 x
1
lim x/2
= lim √ x/2 = 0.
x→∞ e
x→∞
/2
xe
2
4. Section 4.4, Problem 44: Evaluate the limit
lim sin x ln x.
x→0+
Solution: Since sin(0) = 0 and ln x → −∞ as x → 0+ , the limit is of indeterminate
type 0 · ∞. We rewrite it as
ln x
,
lim+
x→0 csc x
∞
which is of type
, so l’Hospital’s rule may be applied, giving us
∞
− sin2 x
sin x
1/x
= lim+
= lim+
tan x = 0,
lim
x→0
x→0
x→0+ − csc x cot x
x cos x
x
since tan(0) = 0 and
sin x
→ 1 as x → 0.
x
5. Section 4.4, Problem 66: Evaluate the limit
2x+1
2x − 3
lim
.
x→∞
2x + 5
Solution: Since the numerator and denominator are both of degree one, we have
2
2x − 3
= = 1,
x→∞ 2x + 5
2
∞
so the limit is of indeterminate type 1 . We rewrite it as
2x+1
2x − 3
lim
= lim e(2x+1) ln((2x−3)/(2x+5)) = elimx→∞ (2x+1) ln((2x−3)/(2x+5)) ,
x→∞
x→∞
2x + 5
lim
since the exponential function is continuous. The resulting limit is
2x − 3
,
lim (2x + 1) ln
x→∞
2x + 5
which is of indeterminate form 0 · ∞, since ln(1) = 0. We rewrite this as
lim
x→∞
ln((2x − 3)/(2x + 5))
,
(2x + 1)−1
0
, so l’Hospital’s rule may be applied, giving us
0
−8(2x + 1)2
2/(2x − 3) − 2/(2x + 5)
lim
=
lim
= −8.
x→∞
x→∞ (2x − 3)(2x + 5)
−2(2x + 1)−2
which is of type
Thus, we have
lim
x→∞
2x − 3
2x + 5
3
2x+1
= e−8 .
6. Section 4.7, Problem 56: What is the smallest possible area of the triangle that is cut
off by the first quadrant and whose hypotenuse is tangent to the parabola y = 4 − x2
at some point?
Solution: Let (a, 4 − a2 ) be a point on the parabola, with a ∈ [0, 2] (so that (a, 4 − a2 )
is in the first quadrant). Since the derivative of f (x) = 4 − x2 is f 0 (x) = −2x, the
slope of the tangent line to the curve at (a, 4 − a2 ) is m = −2a, and thus the equation
of the tangent line is
y − (4 − a2 ) = −2a(x − a),
or
y + 2ax = a2 + 4.
The desired triangle is formed by the coordinate axes and this line, so its base length is
given by the x-intercept of the tangent line, and its height is given by the y-intercept.
These are given by x0 = (a2 + 4)/(2a) and y0 = a2 + 4, respectively, so the area of the
triangle is
(a2 + 4)2
1
.
A(a) = x0 y0 =
2
2a
We note that A(a) → ∞ as a → 0 and A(2) = 16. The derivative of A(a) is
√
2(a2 + 4)(2a)(2a) − 2(a2 + 4)2
2(a2 + 4) 2
2(a2 + 4) √
=
(3a
−4)
=
( 3a−2)( 3a+2).
2
2
2
(2a)
(2a)
(2a)
√
We thus have a critical point a = 2/ 3 in [0, 2] where A0 (a) = 0, and we can check
that A0 (a) changes from negative to positive at this point, so it is a local minimum.
Thus, the smallest possible area is given by
√
√
64
(2/ 3)2 + 4)2
√
= √ .
A(2/ 3) =
2(2/ 3)
3 3
√
(Note that the simplification of A(2/ 3) given in the last step is not necessary. However, it does let us see that the result
√ is less than the endpoint value A(2),
√ in case we
weren’t already sure of that, since 3 3 is a little more than 5, so 64/(3 3) is less than
64/5, which is less than 65/5 = 13, which is less than 16 = A(2).)
A0 (a) =
7. Section 4.7, Problem 70: A steel pipe is being carried down a hallway 9 feet wide. At
the end of the hall there is a right-angled turn into a narrower hallway 6 feet wide.
What is the length of the longest pipe that can be carried horizontally around the
corner?
Solution: From the given diagram, we note that there are two other angles equal to
the given angle θ, as below:
4
From this we can conclude that the total length of the pipe, in terms of θ, is given by
l(θ) = 9 csc θ + 6 sec θ.
We note that l(θ) → ∞ as θ → 0 (when the pipe is vertical) and l(θ) → ∞ as θ → π/2
(when the pipe is horizontal). Now for each θ ∈ (0, π/2), l(θ) describes the length of
a pipe that does not fit into the second hallway (since it gets stuck on the corner). In
order to find the longest pipe that will fit, we need to find the minimum length among
all pipes that do not fit, since anything shorter will be able to fit. Since l(θ) → ∞ at
the end points of the interval, the minimum must occur at a critical point. We have
l0 (θ) = −9 csc θ cot θ + 6 sec θ tan θ =
6 sinθ −9 cos3 θ
−9 cos θ 6 sin θ
=
+
.
cos2 θ
sin2 θ
sin2 θ cos2 θ
p
sin3 θ
9
3
Thus l (θ) = 0 when
=
,
which
gives
us
θ
=
arctan(
3/2, and therefore the
6
3
cos θ
desired length is
p
p
l = 9 csc(arctan( 3 3/2) + 6 sec(arctan( 3 3/2).
0
Note: leaving the answer as above is fine. If you really, really want to simplify, you
can proceed as follows: we have tan3 x = 9/6, so tan x = (9/6)1/3 . Since sec x =
√
1 + tan2 x (positive root in the first quadrant) we get
√
62/3 + 92/3
.
sec x =
61/3
Similarly, using cot3 x = 6/9 and csc2 x = cot2 x + 1, we get
√
62/3 + 92/3
csc x =
.
91/3
5
Putting this together, we get
l=
p
62/3 + 92/3 (9/61/3 + 6/91/3 ).
8. Section 4.7, Problem 72: A rain gutter is to be constructed from a metal sheet of width
30 cm by bending up one-third of the sheet on each side through an angle of θ. How
should θ be chosen so that the gutter will carry the maximum amount of water?
Solution: We note that as per the given diagram, the angle θ corresponds to the
angles in the top corners of the gutter, as shown below:
We can maximize the amount of water carried by the gutter by maximizing the crosssectional area of the gutter. This area is give by the area of the two triangles, which
each have height 10 sin θ and base 10 sin θ, and the rectangle, which has height 10 sin θ
and width 10. The total area is therefore
A(θ) = 100 sin θ cos θ + 100 sin θ = 50 sin 2θ + 100 sin θ.
We must have θ ∈ [0, π/2], since clearly we have less area if we bend the two sides
inwards, beyond 90 degrees. If θ = 0 then A = 0, and if θ = π/2 then A = 100. The
derivative of A with respect to θ is
A0 (θ) = 100 cos 2θ + 100 cos θ = 100(2 cos2 θ + cos θ − 1) = 100(2 cos θ − 1)(cos θ + 1).
Since cos θ = −1 does not give a value in [0, π/2], we have the critical value θ = π/3
such that 2 cos θ − 1 = 0. Since cos θ > 1/2 for θ < π/3 and cos θ < 1/2 for θ > π/3,
we see that A0 (θ) changes from positive to negative at π/3, so that θ = π/3 is a local
maximum by the first derivative test.
(This is enough to conclude that we also obtain an absolute maximum at π/3,√since
it’s the only critical point in [0, π/2], but we can also check that A(π/3) = 75 3 >
100 = A(π/2).)
6
√
9. Section 4.9, Problem 42: Find f , given that f 00 (t) = 3/ t, with f (4) = 20 and
f 0 (4) = 7.
Solution: If f 00 (t) = 3t−1/2 then√f 0 (t) = 6t1/2 + C for some C ∈ R. Since we must
have f 0 (4) = 7, we get f 0 (4) = 6 4 + C = 12 + C = 7, which gives C = −5. Thus,
f 0 (t) = 6t1/2 − 5. Taking the antiderivative of this gives us f (t) = 4t3/2 − 5t + D for
some D ∈ R, so f (4) = 4(8) − 20 + D = 20, which gives D = 8. Therefore, we have
f (t) = 4t3/2 − 5t + 8.
10. Section 4.9, Problem 50: Find a function f such that f 0 (x) = x3 and the line x + y = 0
is tangent to the graph of f .
Solution: If f 0 (x) = x3 then we must have f (x) = x4 /4 + C for some C ∈ R. The
line x + y = 0 has slope −1, so if the curve y = f (x) is to be tangent to this line, we
must have f 0 (a) = a3 = −1 at the point of intersection. This is possible if and only if
a = −1. If this curve is to intersect the line x + y = 0 when x = −1, then we must
have f (−1) = (−1)4 /4 + C = 1, since y = −x gives y = 1 when x = −1. This implies
that C = 1 − 1/4 = 3/4, so desired function is therefore f (x) = x4 /4 + 3/4.
11. Section 5.1, Problem 20: Use the definition
A = lim Rn = lim [f (x1 )∆x + f (x2 )∆x + · · · + f (xn )∆x]
n→∞
n→∞
√
to find an expression for the area under the graph of f (x) = x2 + 1 + 2x, for 4 ≤ x ≤ 7,
as a limit. Do not evaluate the limit.
3i
. Therefore
n
r
2 s
3i
3i
24i 9i2
6i
f (xi ) = 4 +
+ 1+2 4+
= 16 +
+ 2 + 9+ ,
n
n
n
n
n
Solution: We have ∆x = (7 − 4)/n = 3/n and xi = 4 + i∆x = 4 +
and so the area is given by
A = lim
n→∞
n
X
i=1
24i 9i2
16 +
+ 2 +
n
n
r
6i
9+
n
! 3
.
n
12. Section 5.1, Problem 22: Determine a region whose area is given by the limit
10
n
X
2
2i
lim
5+
.
n→∞
n
n
i=1
Do not evaluate the limit.
7
Solution: Comparing with
n
X
f (xi )∆x we have ∆x = 2/n with f (x) = x10 and xi =
i=1
5 + i(2/n) = 5 + i∆x, which tells us that a = 5 and b = 7, since ∆x = (b − a)/n = 2/n.
The region is therefore the area above the x-axis but below the graph y = x10 , for
5 ≤ x ≤ 7.
8