/*! 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 68 Assume that \(f\) is differentia... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 68

Assume that \(f\) is differentiable for all \(x\). The signs of \(f^{\prime}\) are as follows. \(f^{\prime}(x)>0\) on \((-\infty,-4)\) \(f^{\prime}(x)<0\) on \((-4,6)\) \(f^{\prime}(x)>0\) on \((6, \infty)\) Supply the appropriate inequality for the indicated value of \(c .\)

Short Answer

Expert verified
For \(c=-4\), we have \(f(-4) > f(x)\) for all \(x \neq -4\), and for \(c=6\), we have \(f(6) < f(x)\) for all \(x \neq 6\).

Step by step solution

01

Determine the behavior of the function at -4

The function is increasing before -4 and decreasing immediately after -4. According to the first derivative test, a function has a local maximum at a point when the derivative changes from positive to negative. Hence, for \(c=-4\), \(f(-4) > f(x)\) for all \(x \neq -4\).
02

Determine the behavior of the function at 6

As the function is decreasing before 6 and increasing immediately after 6, according to the first derivative test again, the function has a local minimum at a point where the derivative changes from negative to positive. Hence, for \(c=6\), \(f(6) < f(x)\) for all \(x \neq 6\).

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 91影视!

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

Use a graphing utility to graph the function. Use the graph to determine whether it is possible for the graph of a function to cross its horizontal asymptote. Do you think it is possible for the graph of a function to cross its vertical asymptote? Why or why not? $$ h(x)=\frac{\sin 2 x}{x} $$

The graph of \(f(x)=\frac{6 x}{\sqrt{x^{2}+2}}\) is shown. (a) Find \(L=\lim _{x \rightarrow \infty} f(x)\) and \(K=\lim _{x \rightarrow-\infty} f(x)\). (b) Determine \(x_{1}\) and \(x_{2}\) in terms of \(\varepsilon\). (c) Determine \(M\), where \(M>0\), such that \(|f(x)-L|<\varepsilon\) for \(x>M\) (d) Determine \(N\), where \(N<0\), such that \(|f(x)-K|<\varepsilon\) for \(x

Find the maximum value of \(f(x)=x^{3}-3 x\) on the set of all real numbers \(x\) satisfying \(x^{4}+36 \leq 13 x^{2}\). Explain your reasoning.

The average typing speed \(S\) (words per minute) of a typing student after \(t\) weeks of lessons is shown in the table. $$ \begin{array}{|l|c|c|c|c|c|c|} \hline t & 5 & 10 & 15 & 20 & 25 & 30 \\ \hline S & 38 & 56 & 79 & 90 & 93 & 94 \\ \hline \end{array} $$ $$ \text { A model for the data is } S=\frac{100 t^{2}}{65+t^{2}}, t>0 $$ (a) Use a graphing utility to plot the data and graph the model. (b) Use the second derivative to determine the concavity of \(S\). Compare the result with the graph in part (a). (c) What is the sign of the first derivative for \(t>0 ?\) By combining this information with the concavity of the model, what inferences can be made about the typing speed as \(t\) increases?

Find the minimum value of \(\frac{(x+1 / x)^{6}-\left(x^{6}+1 / x^{6}\right)-2}{(x+1 / x)^{3}+\left(x^{3}+1 / x^{3}\right)}\) for \(x>0\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation
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.