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Found in: Page 216

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Suppose $$f$$ is an odd function and is differentiable everywhere. Prove that for every positive number b, there exists a number c in $$( - b\;,\;b)$$ such that $${f^\prime }(c) = \frac{{f(b)}}{b}$$.

It is proved that for every positive number $$b$$, there exists a number $$c$$ in $$( - b,b)$$ such that $${f^\prime }(c) = \frac{{f(b)}}{b}$$.

See the step by step solution

## Step 1: Given data

The given function $$f$$ is an odd function and differentiable everywhere.

## Step 2: Concept of Mean value theorem

“Let $$f$$ be a function that satisfies the following hypothesis:

1. $$f$$ is continuous on the closed interval $$\left( {a\;,{\rm{ }}b} \right)$$.

2. $$f$$ is differentiable on the open interval $$\left( {a,\;b} \right)$$.

Then, there is a number $$c$$ in $$\left( {a\;,{\rm{ }}b} \right)$$ 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: Calculation to prove that for every positive number $$b$$, there exists a number $$c$$ in $$( - b\;,\;b)$$

It is given that $$f$$ is differentiable everywhere.

Thus, $$f$$ is differentiable on the interval, $$( - b,b)$$.

Since $$f$$ is differentiable everywhere, $$f$$ must be continuous everywhere.

Therefore, $$f$$ is continuous on $$( - b,b)$$ and differentiable on, $$( - b,b)$$.

Then by Mean value theorem there exist a number $$c$$, in $$( - b,b)$$ such that:

$${f^\prime }(c) = \frac{{f(b) - f( - b)}}{{b - ( - b)}}$$

$${f^\prime }(c) = \frac{{f(b) - f( - b)}}{{2b}}$$ …… (1)

Since $$f$$ is an odd function, $$f( - b) = - f(b)$$.

Substitute $$f( - b) = - f(b)$$ in equation (1).

\begin{aligned}{c}{f^\prime }(c) &= \frac{{f(b) - ( - f(b))}}{{2b}}\\{f^\prime }(c) &= \frac{{2f(b)}}{{2b}}\\{f^\prime }(c) &= \frac{{f(b)}}{b}\end{aligned}

Hence, it is proved that for every positive number $$b$$, there exists a number $$c$$ in $$( - b,b)$$ such that $${f^\prime }(c) = \frac{{f(b)}}{b}$$.