Chapter 5.3, Problem 123AYU

(a)

To determine

To show:The function $f\left(x\right)=\cosh x$ is an even function.

Explanation of Solution

It is given that

  $\begin{array}{c}f\left(x\right)=\cosh x\\ =\frac{1}{2}\left(e^x+e^{-x}\right)\end{array}$

So,

  $\begin{array}{c}f\left(-x\right)=\frac{1}{2}\left(e^{-x}+e^x\right)\\ =\frac{1}{2}\left(e^x+e^{-x}\right)\\ =f\left(x\right)\end{array}$

Since $f\left(-x\right)=f\left(x\right)$ . Hence, $f\left(x\right)=\cosh x$ is an evenfunction.

(b)

To determine

To graph:The function $f\left(x\right)=\cosh x$

Explanation of Solution

The graph of $f\left(x\right)=\cosh x$ is shown in figure below.

  

Figure (1)

Therefore, the graph is shown in Figure (1).

(c)

To determine

To show :The equation ${\left(\cosh x\right)}^2-{\left(\sinh x\right)}^2=1$

Explanation of Solution

Consider the equation.

  $\begin{array}{c}{\left(\cosh x\right)}^2-{\left(\sinh x\right)}^2=1\\ {\left(\frac{e^x+e^{-x}}{2}\right)}^2-{\left(\frac{e^x-e^{-x}}{2}\right)}^2=1\\ \frac{1}{4}\left[\left(e^x+e^{-x}+e^x-e^{-x}\right)\left(e^x+e^{-x}-e^x+e^{-x}\right)\right]=1\end{array}$

Solve further,

  $\begin{array}{c}\frac{1}{4}\left[2e^x\cdot 2e^{-x}\right]=1\\ \frac{4}{4}=1\\ 1=1\end{array}$

Hence, proved.