/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 117 Suppose that the second derivati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the second derivative of the function \(y=f(x)\) is \begin{equation}y^{\prime \prime}=(x+1)(x-2).\end{equation} For what \(x\) -values does the graph of \(f\) have an inflection point?

Short Answer

Expert verified
The graph of \(f\) has inflection points at \(x = -1\) and \(x = 2\).

Step by step solution

01

Understand the Inflection Point Criterion

An inflection point occurs when the second derivative changes sign. We need to determine where \( y''=(x+1)(x-2) \) changes sign.
02

Find Critical Points of the Second Derivative

The critical points of \( y'' \) occur where \((x+1)(x-2) = 0\). Set each factor equal to zero and solve separately: \(x+1=0\) and \(x-2=0\). This gives \(x=-1\) and \(x=2\).
03

Test Intervals Between Critical Points

Choose test points from the intervals: \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\).- For \(x < -1\), choose \(x = -2\): \(y''=(-2+1)(-2-2)=(-1)(-4)=4 > 0\).- For \(-1 < x < 2\), choose \(x = 0\): \(y''=(0+1)(0-2)=1(-2)=-2 < 0\).- For \(x > 2\), choose \(x = 3\): \(y''=(3+1)(3-2)=4(1)=4 > 0\).
04

Identify Sign Change and Inflection Points

The sign of \(y''\) changes from positive to negative at \(x = -1\) and from negative to positive at \(x = 2\). Therefore, the function \(f(x)\) has inflection points at \(x = -1\) and \(x = 2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second Derivative
The second derivative of a function, denoted as \( y'' \), is a vital tool in determining the concavity of the function's graph. When you have \( y'' = (x+1)(x-2) \), you're looking at how the slope of the original function \( f(x) \) is changing. The second derivative tells us whether the graph of \( f(x) \) is curving upwards or downwards.
- When \( y'' > 0 \), the graph is concave up. - When \( y'' < 0 \), the graph is concave down.
An inflection point occurs where this concavity changes, meaning \( y'' \) changes its sign. Understanding the second derivative is key in calculus as it helps us identify points where the graph's shape alters.
Critical Points
Critical points related to the second derivative are where the function's concavity might switch. These are the \( x \)-values where \( y'' = 0 \) or doesn't exist. For our specific problem, the equation \( (x+1)(x-2) = 0 \) provides us with the critical points \( x = -1 \) and \( x = 2 \).
These values are candidates for inflection points. But it's necessary to test the intervals around these points to confirm if a sign change actually occurs. Finding these critical points allows us to focus our attention on where possible curvature changes might be happening on the graph.
Sign Change
The concept of a sign change in the second derivative is integral in identifying inflection points. Once you have the critical points, the next step is to investigate if \( y'' \) changes its sign across these points.
For example, by testing values in intervals determined by \(-1\) and \(2\), you will check the sign of \( y'' \) in each interval:
  • Choose \( x = -2 \) for the interval \((-\infty, -1)\); \( y'' > 0 \).
  • Choose \( x = 0 \) for the interval \((-1, 2)\); \( y'' < 0 \).
  • Choose \( x = 3 \) for \((2, \infty)\); \( y'' > 0 \).
In this analysis, \( y'' \) switches from positive to negative at \( x = -1 \) and negative to positive at \( x = 2 \). This confirms inflection points exist at these values.
Interval Testing
Interval testing is a method that uses selected values to determine the behavior of a function's second derivative across different sections of the real number line. This approach ensures that sign changes do occur at critical points, validating them as true inflection points.
By selecting test points in the intervals created by your critical values, you can examine the nature of \( y'' \) in each segment:
- In \((-\infty, -1]\), choose a test point like \( x = -2 \). Here, \( y'' \) is positive.- Between \(-1\) and \(2\), use \( x = 0 \). Here, \( y'' \) becomes negative.- Beyond \(2\), with \( x = 3 \), \( y'' \) returns to being positive.
These test results, showing an alternating pattern in the sign of \( y'' \), are what ultimately confirm the presence of inflection points at \( x = -1 \) and \( x = 2 \). Thus, interval testing is a strategic method in verifying where these crucial points occur.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Distance between two ships At noon, ship \(A\) was 12 nautical miles due north of ship \(B .\) Ship \(A\) was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2000 yd) and continued to do so all day. Ship \(B\) was sailing east at 8 knots and continued to do so all day. \begin{equation} \begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the distance }} \\ \quad{\text { } s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad{\text { noon? One hour later? }} \\\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad{\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal asymptote }} \\ \quad {\text { in the first quadrant. This in turn suggests that } d s / d t} \\ \quad {\text { approaches a limiting value as } t \rightarrow \infty . \text { What is this value? }} \\ \quad {\text { What is its relation to the ships' individual speeds? }} \end{array} \end{equation}

\(\begin{array}{l}{\text { Motion with constant acceleration } \text { The standard equation for }} \\ {\text { the position } s \text { of a body moving with a constant acceleration } a} \\ {\text { along a coordinate line is }}\end{array}\) $$ s=\frac{a}{2} t^{2}+v_{0} t+s_{0} $$ \(\begin{array}{l}{\text { where } v_{0} \text { and } s_{0} \text { are the body's velocity and position at time }} \\ {t=0 . \text { Derive this equation by solving the initial value problem }}\end{array}\) \(\begin{array}{l}{\text { Differential equation: } \frac{d^{2} s}{d t^{2}}=a} \\\ {\text { Initial conditions: } \quad \frac{d s}{d t}=v_{0} \text { and } s=s_{0} \text { when } t=0}\end{array}\)

For \(x>0,\) sketch a curve \(y=f(x)\) that has \(f(1)=0\) and \(f^{\prime}(x)=1 / x .\) Can anything be said about the concavity of such a curve? Give reasons for your answer.

The derivative \(d t / d x\) in Example 4 \begin{equation} \begin{array}{c}{\text { a. Show that }} \\\ {f(x)=\frac{x}{\sqrt{a^{2}+x^{2}}}} \\ {\text { is an increasing function of } x \text { . }}\\\\{\text { b. Show that }} \\\ {g(x)=\frac{d-x}{\sqrt{b^{2}+(d-x)^{2}}}} \\ {\text { is a decreasing function of } x .}\\\\{\text { c. Show that }} \\ {\frac{d t}{d x}=\frac{x}{c_{1} \sqrt{a^{2}+x^{2}}}-\frac{d-x}{c_{2} \sqrt{b^{2}+(d-x)^{2}}}} \\ {\text { is an increasing function of } x .}\end{array} \end{equation}

Tin pest When metallic tin is kept below \(13.2^{\circ} \mathrm{C},\) it slowly becomes britle and crumbles to a gray powder. Tin objects eventually crumble to this gray powder spontaneously if kept in a cold climate for years. The Europeans who saw tin organ pipes in their churches crumble away years ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gray powder is a catalyst for its own formation. A catalyst for a chemical reaction is a substance that controls the rate of reaction without undergoing any permanent change in itself. An autocatalytic reaction is one whose product is a catalyst for its own formation. Such a reaction may proceed slowly at first if the amount of catalyst present is small and slowly again at the end, when most of the original substance is used up. But in between, when both the substance and its catalyst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate \(v=d x / d t\) of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, \(v\) may be considered to be a function of \(x\) alone, and \begin{equation}v=k x(a-x)=k a x-k x^{2}\end{equation} where \begin{equation} \begin{aligned} x &=\text { the amount of product } \\ a &=\text { the amount of substance at the beginning } \\ k &=\text { a positive constant. } \end{aligned} \end{equation} At what value of \(x\) does the rate \(v\) have a maximum? What is the maximum value of \(v\) ?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.