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Q18.

Expert-verifiedFound in: Page 341

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

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

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

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

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:

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

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

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" $\mathrm{sin}h(-x)=\frac{{e}^{-x}-{e}^{-(-x)}}{2}\phantom{\rule{0ex}{0ex}}\begin{array}{c}=\frac{{e}^{-x}-{e}^{x}}{2}\\ =-\frac{{e}^{x}-{e}^{-x}}{2}\\ =-\mathrm{sinh}\left(x\right)\end{array}$

Hence, we have shown that $\mathrm{sin}h\left(-x\right)=-\mathrm{sin}h\left(x\right)$.

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