Q. 122 - Precalculus Enhanced with Graphing Utilities

Question

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

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

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

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

Step-by-Step Solution

Answer

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

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

Part (a) Step 1. Given information.

Consider the given question,

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

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

Then we can say,

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

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

Part (b) Step 1. Graph the function.

Consider the given function,

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

Plot the function,