Question

Hyperbolic Cosine Function The hyperbolic cosine function is defined by

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

(a) Sketch the graph of this function using graphical addition as in Exercise 17.

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

Short answer

a. The graph of $y=\sin h\left(x\right)$is:

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

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

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

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$y=\sin h\left(x\right)$ we get:

Part b. Step 1. Given.

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

Part b. Step 2. To determine.

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

Part b. Step 3. Calculation.

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

width="183" height="172" role="math">sinh(−x)=e−x−e−(−x)2=e−x−ex2=−ex−e−x2=−sinh(x)

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