Q. 123 - Precalculus Enhanced with Graphing Utilities

Question

The hyperbolic cosine function, designated by $\cos h\left(x\right)$, is defined as:

$\cos h\left(x\right)=\frac{1}{2}\left(e^x+e^{-x}\right)$

Part (a): Show that $f\left(x\right)=\cos h\left(x\right)$ is an even function.

Part (b): Graph $f\left(x\right)=\cos h\left(x\right)$ using a graphing utility.

Part (c): Refer to Problem 122. Show that, for every x.

Step-by-Step Solution

Answer

Part (a): To prove $f\left(x\right)=\cos h\left(x\right)$ is an even function, substitute $f\left(-x\right)$ and solve the equation.

Part (b): The graph of the function $f\left(x\right)=\cos h\left(x\right)$ is given below,

Part (c): To show for every x, first assume $g\left(x\right)={\left(\cos h x\right)}^2-{\left(\sin h x\right)}^2$ and solve the equation.

Part (a) Step 1. Given information.

Consider the given question,

$\cos h\left(x\right)=\frac{1}{2}\left(e^x+e^{-x}\right)$

Assume $f\left(x\right)=\cos h\left(x\right)$.

Then we can say,

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

Hence, $f\left(x\right)$ is an even function.

Part (b) Step 1. Graph the function.

Consider the given function,

$f\left(x\right)=\cos h\left(x\right)$

Plot the function,

Part (c) Step 1: Showing for every value of x.

Assume $g\left(x\right)={\left(\cos h x\right)}^2-{\left(\sin h x\right)}^2$.

Then,

  $$\begin{gathered} g\left(x\right)={\left(\frac{1}{2}\left(e^x+e^{-x}\right)\right)}^2-{\left(\frac{1}{2}\left(e^x-e^{-x}\right)\right)}^2 \\ g\left(x\right)=\frac{1}{4}\left(e^{2x}+e^{-2x}+2-\left(e^{2x}+e^{-2x}-2\right)\right) \\ g\left(x\right)=\frac{1}{4}\left(4\right) \\ g\left(x\right)=1 \end{gathered}$$