Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

9–12 ■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers \(c\) that satisfy the conclusion of the Mean Value Theorem.
9. \(f(x) = 2{x^2} - 3x + 1\), \((0,2)\)

The numbers \(c = 1\) satisfies the conclusion of Mean Value Theorem.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given

The function \(f(x) = 2{x^2} - 3x + 1\) is a polynomial.

Step 2: Concept of Mean Value Theorem

Assume \(f\) to be a function satisfying the following properties.

(1) \(f\)Continuous on the interval \((a,b)\).

(2) \(f\) Is differentiable on the interval \((a,b)\).

Then there is a number \(c\) in \((a,b)\) such that \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently, \(f(b) - f(a) = {f^\prime }(c)(b - a)\).

Step 3: Check the conditions of Mean Value Theorem

Since the function \(f(x) = 2{x^2} - 3x + 1\) is a polynomial function, it is continuous on\(R\).

Therefore, the function is continu\(f(x)\)ous on the closed interval \((0,2)\).

2. Since the function \(f(x) = 2{x^2} - 3x + 1\) is a polynomial function, it is differentiable everywhere.

Therefore, the function is differentiable on the open interval \((0,2)\).

Hence, the function \(f(x)\) satisfies the conditions of Mean Value Theorem.

Since the number c satisfies the conditions of Mean Value Theorem, it should lie on the open interval \((0,2)\).

Step 4: Differentiate the function and find the values

Find the derivative of .

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}\left( {2{x^2} - 3x + 1} \right)\\ &= 2(2x) - 3(1) + 0\\ &= 4x - 3\end{aligned}\)

Replace \({\rm{x }} \to c\) in \({f^\prime }(x)\).

\({f^\prime }(c) = 4c - 3\)

Obtain the functional values at the end points of the given interval.

Substitute \(0{\rm{ }} \to x\) in \(f(x)\).

\(\begin{aligned}{c}f(a) &= f(0)\\ &= 2{(0)^2} - 3(0) + 1\\ &= 1\end{aligned}\)

Substitute \(2{\rm{ }} \to x\) in \(f(x)\).

\(\begin{aligned}{c}f(b) &= f(2)\\ &= 2{(2)^2} - 3(2) + 1\\ &= 8 - 6 + 1\\ &= 3\end{aligned}\)

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Essential Calculus: Early Transcendentals

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Short Answer

9–12 ■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers \(c\) that satisfy the conclusion of the Mean Value Theorem.
9. \(f(x) = 2{x^2} - 3x + 1\), \((0,2)\)

Step by Step Solution

Step 1: Given

Step 2: Concept of Mean Value Theorem

Step 3: Check the conditions of Mean Value Theorem

Step 4: Differentiate the function and find the values

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

9–12 ■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers \(c\) that satisfy the conclusion of the Mean Value Theorem.9. \(f(x) = 2{x^2} - 3x + 1\), \((0,2)\)

Step by Step Solution

Step 1: Given

Step 2: Concept of Mean Value Theorem

Step 3: Check the conditions of Mean Value Theorem

Step 4: Differentiate the function and find the values

Most popular questions for Math Textbooks

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9–12 ■ Verify that the function satisfies the hypotheses of the Mean Value Theorem on the given interval. Then find all numbers \(c\) that satisfy the conclusion of the Mean Value Theorem.
9. \(f(x) = 2{x^2} - 3x + 1\), \((0,2)\)