Question

Hyperbolic Cosine Function The hyperbolic cosine function is defined by

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

(a) Sketch the graphs of the functions$y=\frac{1}{2}{e}^x$and$y=\frac{1}{2}{e}^{-x}$on the same axes, and use graphical addition (see Section 2.7) to sketch the graph of$y=\cos h\left(x\right)$

(b) Use the definition to show that$\cos h\left(-x\right)=\cos h\left(x\right)$.

Short answer

a. Graphs of $y=\frac{1}{2}{e}^x$and $y=\frac{1}{2}{e}^{-x}$is:

The graph of$y=\cos h\left(x\right)$ is:

b. We have shown that $\cos h\left(-x\right)=\cos h\left(x\right)$.

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

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 $y=\cos h\left(x\right)$.

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 addition method to sketch the graph of$y=\cos h\left(x\right)$ we get:

Part b. Step 1. Given.

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

Part b. Step 2. To determine.

We have to show that $\cos h\left(-x\right)=\cos h\left(x\right)$.

Part b. Step 3. Calculation.

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

width="188" height="172" role="math">cosh(−x)=e−x+e−(−x)2=e−x+ex2=ex+e−x2=cosh(x)

Hence, we have shown that $\cos h\left(-x\right)=\cos h\left(x\right)$.