/*! 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 96 Assume that \(|f(x)| \leq 1\) an... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 96

Assume that \(|f(x)| \leq 1\) and \(\left|f^{\prime \prime}(x)\right| \leq 1\) for all \(x\) on an interval of length at least \(2 .\) Show that \(\left|f^{\prime}(x)\right| \leq 2\) on the interval.

Short Answer

Expert verified
Given that the absolute value of a function and its second derivative are at most 1, we can use the Mean Value Theorem to prove that the absolute value of the first derivative is at most 2.

Step by step solution

01

Set up the equation

Let \(a\) be an arbitrary point in the interval. According to the Mean Value Theorem, there exists a \(c_1\) in the interval (a, a+1) such that \(f^{\prime}(c_1)=f(a+1)-f(a)\). Similarly, there exists \(c_2\) in the interval (a-1, a) such that \(f^{\prime}(c_2)=f(a)-f(a-1)\). Thus we derive two equations from the Mean Value Theorem.
02

Apply the given conditions

According to the given conditions, we know that \(|f(x)| \leq 1\), so \(|f(a+1)-f(a)| \leq 2\) and \(|f(a)-f(a-1)| \leq 2\). Since \(\left|f^{\prime \prime}(x)\right| \leq 1\), we can conclude that \(|f^{\prime}(c_1)-f^{\prime}(c_2)| \leq 2\).
03

Derive the result

We know that \(f^{\prime}(a)\) lies between \(f^{\prime}(c_1)\) and \(f^{\prime}(c_2)\). Thus, we know that \(|f^{\prime}(a)| \leq |f^{\prime}(c_1)-f^{\prime}(c_2)| \leq 2\). This proves the result.

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

Verifying a Formula In Exercises 45 and \(46,\) use a power series and the fact that \(i^{2}=-1\) to verify the formula. $$ g(x)=\frac{1}{2 i}\left(e^{i x}-e^{-i x}\right)=\sin x $$

the terms of a series \(\sum_{n=1}^{\infty} a_{n}\) are defined recursively. Determine the convergence or divergence of the series. Explain your reasoning. \(a_{1}=2, a_{n+1}=\frac{2 n+1}{5 n-4} a_{n}\)

Using a Binomial Series In Exercises \(17-26,\) use the binomial series to find the Maclaurin series for the function. $$ f(x)=\frac{1}{\sqrt{1-x}} $$

State the Ratio Test.

Finding a Taylor Series In Exercises \(1-12,\) use the definition of Taylor series to find the Taylor series, centered at \(c,\) for the function. $$ f(x)=e^{2 x}, \quad c=0 $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

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.