Mathematic 108, Fall 2015: Solutions to assignment #7
Problem #1. Suppose f is a function with f 00 continuous on the open interval I and so that f has a local
maximum at both x = a and x = b for a, b ∈ I with a < b. Explain why there must be value c ∈ (a, b) so
f 00 (c) ≥ 0. (Hint: If f 00 (x) < 0 for all x ∈ (a, b), then any critical number of f in (a, b) is a local maximum).
Solution #1. First, note that, since f has local maxima at x = a and x = b, it follows that there is at
least one x in (a, b) where f (x) ≤ f (a) and at least one x in (a, b) where f (x) ≤ f (b). We know that, if we
restrict f (x) to the interval [a, b], that there is a global minimum of f restricted to that region, and that,
from before, there can’t be an exclusive global minimum at an endpoint. Therefore, the function must have
a local minimum on the interior (i.e., in the interval (a, b)). At this point, since f is twice differentiable,
f 00 (x) ≥ 0 (since if it was less than 0, the point would be a local maximum instead of a local minimum).
Problem #2. Give an example of a continuous function with domain [−1, 1] with a local maximum, but
no local minimum.
Solution #2. In order for this to be true, since the function is continuous on a closed bounded set, the
function must attain its absolute minimum on the endpoint. One such function is f (x) = −x2 (restricted to
the appropriate domain), which has a local maximum at 0.
Problem #3. Give an example of a function f with continuous second derivative for which f 00 is zero at
some point and whose graph does not have an inflection point.
Solution #3. x4 is probably the simplest such function (note that in order for this to be true at a point x,
f 000 (x) must either equal 0 or not exist; otherwise, f 00 (x) will be positive on one side of the zero and negative
on the other side). Alternatively, f (x) = C, where C is constant, gives a trivial example.
Problem #4. Determine whether the following functions have an absolute maximum value and absolute
minimum value on the given domain. If it does determine the value.
a) f (x) =
1
1+e−x2
−1
on D = (−∞, ∞).
1
(x) + 1+x
2 on D = (−∞, ∞).
√
c) f (x) = x − x2 + 3 on D = [0, ∞).
b) f (x) = tan
Solution #4.
a) We use the quotient and chain rules to evaluate f 0 (x):
2
2xe−x
f (x) =
(1 + e−x2 )2
0
2
Since the denominator is always positive, as is e−x , it holds that f 0 (x) is positive when the numerator
(and therefore when x) is positive, and negative when x is negative. Therefore, there is a minimum at
x = 0.
b) By the chain and quotient rules again, we see that
1
2x
−
1 + x2
(1 + x2 )2
x2 − 2x + 1
=
(x2 + 1)2
(x − 1)2
= 2
(x + 1)2
f 0 (x) =
This is equal to 0 when x = 1 and positive otherwise. Therefore, the only critical point, x = 1, is
neither a local minimum nor a local maximum. Since this is the only candidate, there are no global
minima or maxima.
√
2
c) The derivative
√ of this
√ function is 1√− (x/ x + 3). This is always positive, since for x > 0, it follows
2
2
2
that x = x < x + 3 (so x/ x + 3 is strictly less than 1). Therefore, the function is always
increasing, so the only candidate for a minimum or maximum is at the endpoint x = 0. Since the
function increases to the right of 0, f (x) has a minimum at 0.
Problem #5. Let f (x) =
x2 sin(1/x) x 6= 0
and g(x) = sin(x).
0
x=0
a) Use the limit laws to show that limx→0
b) Determine limx→0
f 0 (x)
g 0 (x)
f (x)
g(x)
= 0 (Hint: Consider limx→0
f (x)/x
g(x)/x ).
how do you reconcile this with a) and L’Hospital’s Rule.
Solution #5.
a) As in previous homework assignments, we use the squeeze theorem and the inequality −|x| ≤ x sin(1/x) ≤
|x| to show that limx→0 f (x)/x = 0. Additionally, limx→0 sin(x)/x = 1 is an established limit. There(x)
(x)/x
fore, limx→0 fg(x)
= limx→0 fg(x)/x
= 0.
b) f 0 (x), by the product and chain rules, is equal to 2x sin(1/x)−cos(1/x) whenever x 6= 0. g 0 (x) = cos(x).
0
(x)
The numerator doesn’t have a limit as x goes to 0, so this limit doesn’t exist. This implies that fg0 (x)
isn’t a continuous function (formally, this breaks the requirement that limx→0
f 0 (x)
g 0 (x)
exists).
Problem #6. Use L’Hospital’s Rule to evaluate the following limits
a) limx→0
arcsin(2x)
.
x
b) limx→0+
1
x
−
1
arctan(x)
c) limx→0+ (1 + sin(2x))
.
1/x
.
Solution #6.
a) This is a 0/0 type limit, so we can use L’Hospital’s Rule directly. By the chain rule, the derivative
of the numerator is √ 2 2 and the derivative of the denominator is 1. This quotient, evaluated at
1−(2x)
x = 0 is equal to 2.
b) This isn’t a 0/0 limit as written, so we need to use the equivalent expression limx→0+ arctan(x)−x
x arctan(x) .
0
This is 0 , so we can use L’Hospital’s Rule. If f (x) = arctan(x) − x and g(x) = x arctan(x), then
x
1
0
f 0 (x) = 1+x
2 − 1 and g (x) = arctan(x) + x2 +1 . Both of these are 0 at 0, so we can apply L’Hospital’s
Rule again. f 00 (x) =
−2x
(x2 +1)2
and g 00 (x) =
1
1+x2
+
1
1+x2
−
f 00 (0)
2x2
(1+x2 )2 . g 00 (0)
= 0, so this limit is 0.
∞
c) This is a 1 type limit. Therefore, we can take the natural log and evaluate the limit of that, then
convert it back to our original limit by exponentiating. We know ln((1 + sin(2x))1/x ) = ln(1+sin(2x))
.
x
This is a proper 0/0 limit, so we can now apply L’Hospital’s rule. Given f (x) = ln(1 + sin(2x)) and
2 cos(2x)
and g 0 (x) = 1. Evaluating this directly,
g(x) = x, it follows from the chain rule that f 0 (x) = 1+sin(2x)
we see that at 0 the quotient of these two (and therefore the log of the original limit) is equal to 2.
Therefore, the value of the original limit is e2 .
Problem #7. Suppose f is differentiable, f (3) = 1 and f 0 (3) = −2. Evaluate limx→0
f (3+x)−f (3−4x)
.
x
Solution #7. This is a 0/0 limit, so we can use L’Hospital’s rule. By the chain rule:
f 0 (3 + 0) + 4f 0 (3 + 0)
f (3 + x) − f (3 − 4x)
=
x→0
x
1
= −10
lim
Problem #8. Suppose g is differentiable, g(2) = 0 and g 0 (2) = 3. Evaluate limx→1
g(1+x)+g(4−2x)
.
x
Solution #8. For this problem, we can evaluate directly (because all functions involved are continuous
and the denominator is nonzero). Plugging in x = 1, this expression evaluates to 0.
Problem #9. Use the methods of Section 4.5 to sketch the following curves
a) y =
(x−2)2
x2 +1
b) y = x − sin(2x)
√
c) y = 1 + x2 − x
Solution #9. Sketches are left to the reader’s imagination, or alternatively, graphing software.
a) Domain: Since x2 + 1 is positive, the domain is all real numbers.
Intercepts: y(0) = 4, y = 0 at x = 2 (since the numerator is 0 then).
Symmetry: This function has none.
2
,
Asymptotes: Since the function is continuous, there are no vertical asymptotes. Since y = x x−4x+4
2 +1
and the limit of this as x goes to ±∞ is 1 either way, there is a horizontal asymptote at y = 1.
(x2 +1)(2x−4)−(x2 −4x+4)(2x)
, which can be simplified to
(x2 +1)2
4x2 −6x−4
subsequently to (x2 +1)2 . The numerator factors into (x−2)(4x+2).
Intervals of Increase or Decrease: y 0 =
(2x3 −4x2 +2x−4)−(2x3 −8x2 +8x)
, and
(x2 +1)2
1
This is positive for x < − 2 and x
> 2, and negative between the two roots.
Local Maximum and Minimum Values: From the previous calculations, there is a local max at − 12 and
a min at 2.
Concavity and Points of Inflection: The second derivative of this doesn’t have roots that are easy
to deal with, so looking at convexity and concavity wouldn’t be particularly enlightening beyond the
information we can derive from the increase/decrease behavior and asymptotes.
b) Domain: The domain of this function is all real numbers.
Intercepts: y(0) = 0. Other than that, other intercepts are too difficult to find.
Symmetry: This is an odd function.
Asymptotes: This function has no asymptotes, due to the oscillatory nature of sine.
Intervals of Increase or Decrease: y 0 = 1 − 2 cos(2x). This is positive whenever cos(2x)
< 12 and
1
1
1
0
negative when cos(2x) > 2 . Therefore, x must be in the
interval
k − 6 π, k + 60 π in order for y
1
5
to be negative, where k is any integer, and in k + 6 π, k + 6 π in order for y to be positive.
Local Maximum and Minimum Values: As before, the derivative
is periodic, so there are infinitely
many points where y 0 = 0. By the previous part, k + 16 π is a local min, where k ranges over the
integers, and k − 61 π is a local max.
00
Concavity and Points of Inflection:
y = 4 sin(2x). This is positive (and therefore y 1is concave up)
1
when k is between kπ, k + 2 π and negative (so
y
is
concave
down)
between
k
+
π,
(k
+
1)π
.
2
1
Therefore, inflection points are at kπ and k + 2 π.
c) Domain: The domain of this function is all real numbers.
√
Intercepts: As 1 + x2 > |x|, y is always going to be positive.
Symmetry: This has no symmetry.
Asymptotes: This has a horizontal asymptote of x = 0 as x goes√to ∞ (to see this, use rational
1
conjugation to rewrite y as √x2 +1+x
). Since we have that limx→∞ x2 + 1 − x = 0, it follows that
√
limx→−∞ x2 + 1 − |x| =√0, since this function is even and behaves like
√ our original function for x > 0,
and therefore limx→−∞ x2 + 1 + x = 0 (i.e., this shows us that √
x2 + 1 has an oblique, or slant,
asymptote of y = −x as x goes to −∞. This tells us that limx→−∞ ( x2 + 1 − x) − (−2x) = 0, so y
has a slant asymptote of −2x.
Intervals of Increase or Decrease: y 0 =
function is always decreasing.
√ x
1+x2
− 1, which is necessarily less than 0. Therefore, the
Local Maximum and Minimum Values: There are no local minima or maxima.
x(2x)
1
− 12 (1+x
Concavity and Points of Inflection: y 00 = √1+x
2 )3/2 , or equivalently,
2
always positive, so the function is always concave up.
1
.
(1+x2 )3/2
The latter is
Problem #10. Consider the family of polynomials Pc (x) = x3 + 3cx2 + 3x.
a) Determine the values of c so that Pc has both a local maximum and a local minimum.
b) Sketch the graph of y = Pc (x) for a value c for which c has both a local maximum and minimum and
sketch the graph for a value c for which it does not.
Solution #10.
a) Pc0 (x) = 3x2 + 6cx + 3 = 3(x2 + 2cx + 1). In order to find the critical points, we√apply the quadratic
2
formula to x2 + 2cx + 1 (or the original 3x2 + 6x + 3) in order to find roots at −2c± 2 4c −4 . The number
2
of roots therefore depends on the behavior of the discriminant 4c − 4: if it’s positive, there are two
distinct roots, if it’s 0, there’s only one root, and if it’s negative, there are no real roots. Therefore,
in order to have a local minimum and a local maximum, there must be two distinct critical points, so
c > 1 or c < −1.
When this happens, by the properties of quadratic equations, Pc0 (x) is positive to the left of the first
(leftmost) root, negative between the
√ two roots, and positive to the right of√the second root. Therefore,
there is a local maximum at −c − c2 − 1 and a local minimum at −c + c2 − 1.
b) You can graph these yourselves, A case for which a local maximum and minimum exist is c = 2 and a
case for which they don’t is c = 0. It might also be interesting to graph the ”border” cases c = ±1,
in which f has a critical point which is neither a minimum nor maximum (in fact, each of these cases
are translations of the graph of y = x3 , which has the same properties).
Book Problems.
a) Section 4.4: #4, #32, #44, #76, #88
b) Section 4.5: #2, #12, #50, #72
c) Section 4.6: #28
Solution #4.4.4.
Only (b) and (e) fail to be indeterminate forms, with limits 0 and ∞ respectively.
Solution #4.4.32.
This is ∞/∞. The derivative of the numerator is 2 ln(x)/x and the derivative of
the denominator is 1, so the quotient, 2 ln(x)/x, once again requires L’Hospital’s Rule. The derivative of
the numerator is 2/x and the derivative of the denominator is 1. The quotient here goes to 0 as x goes to
infinity.
Solution #4.4.44.
This is a ∞ · 0 case, so we can rewrite it as
us √xe1x/2 , which goes to 0.
√
x
.
ex/2
Applying L’Hospital’s Rule gives
x tan x
tan x
Solution #4.4.76.
The first application of L’Hospital’s rule gives us sec
sec2 (x) , or sec x . The second
application brings us back to where we started, and further applications keep alternating between the two.
If we write the quotient as the equivalent expression sin1 x , the limit is easily shown to be 1.
Solution #4.4.88.
We start off by writing this as
sin(2x) + ax3 + bx
lim
x→0
x3
This is 0/0, so we use L’Hospital’s Rule to get
2 cos(2x) + 3ax2 + b
lim
x→0
3x2
In order for this limit to have a chance of existing, since the denominator is 0, the numerator must be 0
as well. Therefore, b = −2. Talking L’Hospital’s Rule, we next get
lim
x→0
and, once more,
−4 sin(2x) + 6ax
6x
lim
x→0
−8 cos(2x) + 6a
6
Therefore, in order for the limit to be 0, a = 4/3.
Solution #4.5.2, 4.5.12, 4.5.50.
These use techniques similar to 9, so for the sake of getting the
solutions finished before the midterm, I won’t go through all of the details, but notable features include a
vertical asymptote at x = 0 for problem 12, and a restricted domain of (−1, ∞) (and a corresponding vertical
asymptote at the left endpoint) for problem 50.
√
=
Solution #4.5.72. The y = x+2 slant asymptote is equivalent to saying that limx→∞ x2 + 4x−(x+2)
√
0. Note that we chose positive infinity here from cursory examination of the behavior of x + 2 and x2 + 4x,
which both go to infinity as x goes to infinity;
if x goes to −∞, the two functions blow up in different
√
2
−4
directions. We can multiply by the conjugate √xx2 +4x+(x+2)
to get √x2 +4x+(x+2)
, which goes to 0, so x + 2
+4x+(x+2)
is a slant asymptote. The proof of the other slant asymptote proceeds similarly, except to −∞. The graph
can be drawn from this behavior, as well as first and second derivative behavior.
Solution #4.6.28.
This is similar to Problem 10 from the worksheet, where there are maxima and
minima when c < 0, and none otherwise (but a critical point when c = 0).