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Q18.

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Precalculus Mathematics for Calculus

Found in: Page 341

Book edition 7th Edition

Author(s) James Stewart, Lothar Redlin, Saleem Watson

Pages 948 pages

ISBN 9781337067508

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Short Answer

Hyperbolic Cosine Function The hyperbolic cosine function is defined by
$\sin h (x) = \frac{e^{x} - e^{- x}}{2}$
(a) Sketch the graph of this function using graphical addition as in Exercise 17.
(b) Use the definition to show that $\sin h (- x) = - \sin h (x)$ .

a. The graph of $y = \sin h (x)$ is:

b. We have shown that $\sin h (- x) = - \sin h (x)$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part a. Step 1. Given.

The given functions are $y = \frac{1}{2} e^{x}$ and $y = \frac{1}{2} e^{- x}$ .

And, $\sin h (x) = \frac{e^{x} - e^{- x}}{2}$ .

Part a. Step 2. To determine.

We have to graph the given functions on the same axes. Then we have to use graphical addition to sketch the graph of $y = \sin h (x)$ .

Part a. Step 3. Calculation.

We’ll use a graphing calculator to graph $y = \frac{1}{2} e^{x}$ and $y = \frac{1}{2} e^{- x}$ .

Using the graphical subtraction method to sketch the graph of $y = \sin h (x)$ we get:

Part b. Step 1. Given.

Given: $\sin h (x) = \frac{e^{x} - e^{- x}}{2}$ .

Part b. Step 2. To determine.

We have to show that $\sin h (- x) = - \sin h (x)$ .

Part b. Step 3. Calculation.

We’ll plug $- x$ in place of x in the formula: $\sin h (x) = \frac{e^{x} - e^{- x}}{2}$ .

width="183" height="172" role="math" style="max-width: none; vertical-align: -87px;" localid="1648448499993" $\sin h (- x) = \frac{e^{- x} - e^{- (- x)}}{2} \begin{matrix} = \frac{e^{- x} - e^{x}}{2} \\ = - \frac{e^{x} - e^{- x}}{2} \\ = - \sinh (x) \end{matrix}$

Hence, we have shown that $\sin h (- x) = - \sin h (x)$ .

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