MATH 110 TERM 2 ASSIGNMENT 4
Problem 1. Sketch the graph of the function
f (x) =
ex
.
1 + ex + e2x
Be sure to identify where the function f (x) is increasing and decreasing, and where it is concave up and concave down. Be sure
that you correctly identify the horizontal and vertical asymptotes of
the function. Also, make sure you correctly identify any x- and yintercepts.
For this problem, you will be marked based on both your work
and your final graph. Please explain how you came to arrive at each
aspect of the graph.
Hint: To solve an equation of the form ae2x + bex + c = 0, use
the substitution u = ex .
Note: There was an error in this problem, and students were instructed to use WolframAlpha to find the roots of the second derivative.
Solution
• f (0) = 1/3
• f (x) is never equal to 0.
• limx→−∞ f (x) = 0.
• limx→∞ f (x) = 0.
• No vertical asymptotes
• f ′ (x) =
(1+ex +e2x )ex −ex (ex +2e2 x)
(1+ex +e2x )2
=
ex −e3 x
.
(1+ex +e2x )2
• f ′ (x) = 0 when x = 0.
• f ′ (x) is negative when x > 0; positive when x < 0. So f (x) is
increasing for x < 0 and decreasing for x > 0.
• f ′′ (x) =
(1+ex +e2x )(ex −3e3x )−2(ex −e3x )(ex +2ex )
(1+ex +e2x )3
• f ′′ (x) is zero when x ≈ ±1.1131.
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• f ′′ (x) is positive for |x| > 1.1131 and negative for |x| < 1.1131.
• So f (x) is CCU for |x| > 1.1131 and CCD for |x| < 1.1131.
Marking Scheme
Total marks: 9
• 1 points for y-intercept and pointing out that f (x) 6= 0.
• 1 point for asymptotes.
• 1 point for f ′ (x).
• 1 point for intervals of increase/decrease.
• 1 point for f ′′ (x)
• 1 point for concavity (Students are permitted to use WolframAlpha for this).
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• 3 points for graph matching given description.
Problem 2. Sketch the graph of the function
f (x) = x3 − 3x2 − 5x + 14 .
Remember that the first and second derivatives of f can change sign
at places where f is not differentiable.
As in the previous problem, please show all of your work and
clearly explain how you decided where the graph was increasing/decreasing
and concave up/concave down. Be sure to note the x-and y-intercepts
of the graph and the horizontal and vertical asymptotes of the graph.
Solution
• No horizontal or vertical asymptotes.
• f (x) ≥ 0 for all x, with f (x) = 0 at
√
1± 29
2
and at x = 2.
• f (0) = 14.
(
√
√
1− 29
1+ 29
3
2
x
−
3x
−
5x
+
14
if
<
x
<
2
or
if
x
>
2
2
• f ′ (x) =
−x3 + 3x2 + 5x − 14 otherwise
(
√
√
1+ 29
1− 29
2
<
x
<
2
or
if
x
>
3x
−
6x
−
5
if
2
2
• f ′ (x) =
−3x2 + 6x + 5 otherwise
√
√
• f√′ (x) is negative for x √
< 1−2 29 , positive for 1− 292 < x < 1−
√
2 6
2 6
2 6
,
negative
for
1
−
<
x
<
2,
positive
for
2
<
x
<
1
+
,
3
3
√
√3
√
1+ 29
1+ 29
2 6
negative for 1 + 3 < x < 2 , and positive for x > 2 .
(
√
√
1+ 29
1− 29
<
x
<
2
or
if
x
>
6x
−
6
if
2
2
• f ′′ (x) =
−6x + 6 otherwise
• This is zero when x = 1.
√
√
• f ′′ (x) is positive if x < 1− 2 29 , negative if√ 1−2 29 < x < 1,
positive√ if 1 < x < 2, negative if 2 < x < 1+2 29 , and positive if
x > 1+2 29 .
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Marking Scheme
Total points: 9.
• 1 point for x-intercepts.
• 1 point for y-intercept.
• 1 point for f ′ (x).
• 1 point for intervals of increase/decrease.
• 1 point for f ′′ (x).
• 1 point for concavity.
• 3 points for correct drawing given calculations.
Problem 3. Suppose that you’re in the middle of a curve-sketching
problem, and you discover the following information about a function
f (x).
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• f (x) is defined on (−∞, ∞). Furthermore, f (x) is twice differentiable on (−∞, ∞) (That is, f and f ′ are differentiable
everywhere on (−∞, ∞)).
• f (x) has its only critical points at x = −1 and x = 3.
• f (−5) = 0.
• f (0) = −2.
• f (1) = 0.
• f (x) has its only inflection point at x = 7.
Explain how you know that you must have made a mistake. Change
one of the bulleted pieces of information about this function to make
the information consistent.
Solution
It’s clear from the fact that f ′ (x) = 0 at x = −1 and the fact
that f (x) increases on [0, 1] that f (x) has a minimum at x = −1.
However, f ′ (x) = 0 at x = 3, which is impossible because f (x) must
be concave up on (−∞, 7). We can fix the problem by placing an
inflection point somewhere between −1 and 3.
Marking Scheme
Total points: 4
• 1 point for concluding that f (x) has a minimum at x = −1.
• 1 point for concluding from this information that f (x) is concave up on (−∞, 7).
• 1 point for realizing that f ′ (x) can’t be zero at x = 3 because
of the concavity.
• 1 point for any correct fix.
Note: there may be more than one way to realize that there is a
contradiction here. Give 3 points to any student who identifies the
contradiction using some reasonable method, and 1 point for any fix
that resolves the contradiction.
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