breakout session 17 solutions · breakout session 17 solutions p1 outcome: determine how the graph...

Breakout Session 17 Solutions

P1 Outcome:Determine howthe graph of afunction looksbased on ananalytic de-scription of thefunction.

Problem 1

1. You are given that f ′′(x) > 0 for all x. Which of the following must betrue about f(x) on the region 0 ≤ x ≤ 2?

(a) There is a critical point between 0 and 2.

(b) An absolute maximum occurs at either x = 0 or x = 2.

(c) There is a local maximum, but not enough information is given todetermine where.

(d) f need not have a local maximum.

Solution: Only (b) and (d) must be true. The following picture pro-vides a counterexample for both (a) and (c):

2. You are told that f ′′(x) > 0 for all x. Which of the following must be trueabout the graph of y = f(x)?

(a) The graph is a straight line.

(b) The graph crosses the x-axis at most once.

(c) The graph is concave down.

(d) The graph crosses the y-axis more than once.

(e) The graph is concave up.

Solution: Only (e) must be true. For (a), the function f(x) = ex

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provides a counterexample:

For part (b), the function f(x) = x2 − 2 provides a counterexample:

Part (c) is clearly false since f ′′(x) > 0 means that f is concave up. Part(d) is false for any function.

P2 Outcome:Understandwhat infor-mation thederivativegives concern-ing when afunction isincreasing ordecreasing.

P2 Outcome:Understandwhat informa-tion the secondderivative givesconcerningconcavity of afunction.

P2 Outcome:Interpret limitsas giving infor-mation aboutfunctions.

Problem 2 Sketch (neatly!) the graph of a function f satisfying all of theconditions:

(a) f is continuous and odd, f(0) = 0,

(b) limx→∞

f(x) = −5,

(c) f ′(x) > 0 on (6,∞),

(d) f ′(x) < 0 on (0, 6),

(e) f ′′(x) > 0 on (0, 12), and

(f) f ′′(x) < 0 on (12,∞).

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Solution:

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P3 Outcome:Determine howthe graph of afunction lookswithout using acalculator.

P3 Outcome:Determine howthe graph of afunction looksbased on ananalytic de-scription of thefunction.

P3 Outcome:Find the in-tervals wherea function isincreasing ordecreasing.

P3 Outcome:Find the in-tervals wherea function isconcave up ordown.

Problem 3 Follow all the steps on the “Graph Sketching Summary Sheet” andthen graph the given function:

f(x) =x2 + x+ 1

x2

Solution:

• Domain

The function is a rational function, and so the domain of the function isall real numbers except where the denominator equals zero.

x2 = 0 =⇒ x = 0

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So the domain of f is (−∞, 0) ∪ (0,∞).

• x, y-interceptsTo find any x-intercept(s), set y = 0 and solve:

x2 + x+ 1

x2= 0 =⇒ x2 + x+ 1 = 0

=⇒ x =−1±

√1− 4(1)(1)

2(1)

which has no real solutions. Thus, f has no x-intercepts.

Since x = 0 is not in the domain of f , f has no y-intercepts as well.

• Symmetry

It is clear that f is not periodic. Note that f(1) = 3 and f(−1) = 1. So itcannot be for all values of x that either f(−x) = f(x) or f(−x) = −f(x).So f is neither even nor odd, and therefore f has no symmetry.

• AsymptotesVertical Asymptotes: Our only candidate is x = 0, and so we compute

P3 Outcome:Find anyasymptoticbehaviors afunction mayhave: vertical,horizontal, orslant.

the two one-sided limits:

limx→0−

x2 + x+ 1

x2=∞

limx→0+

x2 + x+ 1

x2=∞

Therefore, x = 0 is the only vertical asymptote of f .

Horizontal Asymptotes: We compute the following limits:

limx→∞

x2 + x+ 1

x2= 1

limx→−∞

x2 + x+ 1

x2= 1

and so the only horizontal asymptote of f is y = 1.

Slant Asymptote: Since our function is rational, we need to check forslant asymptotes. Since the degree of the numerator is not one greaterthan the degree of the denominator, we do not have a slant asymptote.

• Increasing/Decreasing

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f ′(x) =x2(2x+ 1)− (x2 + x+ 1)(2x)

x4

=2x3 + x2 − 2x3 − 2x2 − 2x

x4

=−x2 − 2x

x4

=−x− 2

x3

To find where f ′ is both positive and negative, we need to find wheref ′(x) = 0 and where f ′(x) does not exist. Clearly, f ′(x) does not existwhen x = 0. To find when f ′(x) = 0, we solve:

−x− 2

x3= 0 =⇒ −x− 2 = 0

=⇒ −x = 2

=⇒ x = −2

Since x = 0 is not in the domain of f , x = −2 is the only critical pointof f . To see where f is increasing and decreasing, consider the followingsign chart for f ′:

−2 0

f ′(−3) = −127

f ′(−1) = 1 f ′(1) = −3

(+) (-)(-)f ′

So we see that f is increasing on [−2, 0), and f is decreasing on (−∞,−2]∪(0,∞).

• Local Extremaf ′ changes from negative to positive at x = −2, so this is the location ofa local minimum. f ′ also changes from positive to negative at x = 0, butf is not defined at x = 0 and so this is not a local extreme value. f has a

local minimum at

(−2, 3

4

).

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• Concavity

f ′′(x) =x3(−1)− (−x− 2)(3x2)

x6

=−x3 + 3x3 + 6x2

x6

=2x3 + 6x2

x6

=2x+ 6

x4

=2(x+ 3)

x4

To find where f ′′ is both positive and negative, we need to find wheref ′′(x) = 0 and where f ′′(x) does not exist. Clearly, f ′′(x) does not existwhen x = 0. To find when f ′′(x) = 0, we solve:

2(x+ 3)

x4= 0 =⇒ 2(x+ 3) = 0

=⇒ x = −3

To see where f is concave up and concave down, consider the followingsign chart for f ′′:

−3 0

f ′′(−4) = −1128

f ′′(−1) = 4 f ′′(1) = 8

(+) (+)(-)f ′′

So we see that f is concave up on (−3, 0)∪ (0,∞), and f is concave downon (−∞,−3).

• Inflection Pointsf ′′(x) changes sign from negative to positive at x = −3, and f is contin-

uous at x = −3. So f has an inflection point at

(−3, 7

9

)

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• The graph of f

Extra problem for personal practice

Problem 4 Given that:

limx→−∞

f(x) = 0 , limx→−3−

f(x) =∞ , limx→−3+

f(x) = −∞ , limx→4−

f(x) =∞

limx→4+

f(x) = −∞ , f(1) = 1 , f(5) = −2 , f(−3) is undefined , f(4) is undefined

f(9) is undefined , f is continuous except at x = −3, 4, and 9

f ′(1) 6= 0 , f ′(7) = 0 , f ′(x) = 2 for x > 9 , f ′′(1) = 0

and the following sign chart for the first and second derivatives of f :

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find the following:

(a) Critical points.

(b) Intervals where f is increasing and decreasing.

(c) Local extrema.

(d) Inflection points.

(e) Intervals of concavity.

(f) Sketch the graph of f .

Solution: (a) Critical points.The critical points of f occur at points in the domain of f where eitherf ′(x) = 0 or where f ′(x) does not exist. We are given that f ′(7) = 0, andso x = 7 is a critical point of f . Even though f ′(−3), f ′(4), and f ′(9) donot exist, all three of those points are not in the domain of f . Therefore,x = 7 is the only critical point of f .

(b) Intervals where f is increasing and decreasing.f is increasing when f ′(x) > 0. From the sign chart and our criticalpoints, these are the intervals (−∞,−3), (−3, 4), (4, 7], and (9,∞). f(x)is decreasing when f ′(x) < 0. From the sign chart, this is on the interval[7, 9).

(c) Local extrema.Using the first derivative test, f(7) is a local maximum because the deriva-tive changes sign from positive to negative. Although the derivativechanges sign from negative to positive at f(9), this is not a local min-imum because the function is not defined at this point. Therefore, x = 7is the only local extremum of f , and it is a local maximum.

(d) Inflection points.Possible inflection points occur where f ′′(x) = 0 or where f ′′(x) does notexist. We are given that f ′′(1) = 0. In addition, f ′′(x) does not exist atx = −3, 4, 9. However, these are not inflection points because f is notdefined at these points. Since f ′′ changes sign at x = 1, this is in factan inflection point of f . We are given that f(1) = 1, and so the onlyinflection point of f is the point (1, 1).

(e) Intervals of concavity.f(x) is concave up when f ′′(x) > 0. From the sign chart, this is on theintervals (−∞,−3) and (1, 4). f(x) is concave down when f ′′(x) < 0.From the sign chart, this is on the interval (−3, 1) and (4, 9).

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(f) Sketch the graph of f .

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breakout session 17 solutions · breakout session 17 solutions p1 outcome: determine how the graph...

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