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Expert-verified Found in: Page 341 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Hyperbolic Cosine Function The hyperbolic cosine function is defined by$\mathrm{sin}h\left(x\right)=\frac{{e}^{x}-{e}^{-x}}{2}$(a) Sketch the graph of this function using graphical addition as in Exercise 17.(b) Use the definition to show that $\mathrm{sin}h\left(-x\right)=-\mathrm{sin}h\left(x\right)$.

a. The graph of $y=\mathrm{sin}h\left(x\right)$ is: b. We have shown that $\mathrm{sin}h\left(-x\right)=-\mathrm{sin}h\left(x\right)$.

See the step by step solution

## Part a. Step 1. Given.

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

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

## 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 $y=\mathrm{sin}h\left(x\right)$.

## 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 $y=\mathrm{sin}h\left(x\right)$ we get: ## Part b. Step 1. Given.

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

## Part b. Step 2. To determine.

We have to show that $\mathrm{sin}h\left(-x\right)=-\mathrm{sin}h\left(x\right)$.

## Part b. Step 3. Calculation.

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

width="183" height="172" role="math" style="max-width: none; vertical-align: -87px;" localid="1648448499993" $\mathrm{sin}h\left(-x\right)=\frac{{e}^{-x}-{e}^{-\left(-x\right)}}{2}\phantom{\rule{0ex}{0ex}}\begin{array}{c}=\frac{{e}^{-x}-{e}^{x}}{2}\\ =-\frac{{e}^{x}-{e}^{-x}}{2}\\ =-\mathrm{sinh}\left(x\right)\end{array}$

Hence, we have shown that $\mathrm{sin}h\left(-x\right)=-\mathrm{sin}h\left(x\right)$. ### Want to see more solutions like these? 