Chapter 13.3, Problem 30WE

a.

To determine

To show that $O(x)$ is odd and $E(x)$ is even.

Explanation of Solution

Given:

The two functions are defined as follows-

  $O(x)=\frac{f(x)-f(-x)}{2}$ and $E(x)=\frac{f(x)+f(-x)}{2}$

Calculation:

  $\begin{array}{l}O(-x)=\frac{f(-x)-f(x)}{2}\\ O(-x)=-\frac{f(x)-f(-x)}{2}\\ O(-x)=-O(x)\end{array}$

That mean $O(x)$ is odd function.

  $\begin{array}{l}E(-x)=\frac{f(-x)+f(x)}{2}\\ E(-x)=E(x)\end{array}$

This means $E(x)$ is even function.

b.

To determine

To show that $O(x)+E(x)=f(x)$ .

Explanation of Solution

Given:

The two functions are defined as follows-

  $O(x)=\frac{f(x)-f(-x)}{2}$ and $E(x)=\frac{f(x)+f(-x)}{2}$

Calculation:

  $\begin{array}{l}O(x)+E(x)=\frac{f(x)-f(-x)}{2}+\frac{f(x)+f(-x)}{2}\\ O(x)+E(x)=\frac{2f(x)}{2}\\ O(x)+E(x)=f(x)\end{array}$

Proved

c.

To determine

To find $O(x)$ and $E(x)$

Answer to Problem 30WE

  $O(x)=x^3+2x$ , $E(x)=-3x^2+1$

Explanation of Solution

Given:

The function $f(x)=x^3-3x^2+2x+1$

Calculation:

  $\begin{array}{l}O(x)=\frac{f(x)-f(-x)}{2}\\ =\frac{x^3-3x^2+2x+1+x^3+3x^2+2x-1}{2}\\ =\frac{2x^3+4x}{2}=x^3+2x\end{array}$

  $\begin{array}{l}E(x)=\frac{f(x)+f(-x)}{2}\\ =\frac{x^3-3x^2+2x+1-x^3-3x^2-2x+1}{2}\\ =\frac{-6x^2+2}{2}=-3x^2+1\end{array}$

d.

To determine

To find $O(x)$ and $E(x)$

Answer to Problem 30WE

  $O(x)=\frac{e^x-e^{-x}}{2}$ , $E(x)=\frac{e^x+e^{-x}}{2}$

Explanation of Solution

Given:

The function $f(x)=e^x$

Calculation:

  $\begin{array}{l}O(x)=\frac{f(x)-f(-x)}{2}\\ =\frac{e^x-e^{-x}}{2}\end{array}$

  $\begin{array}{l}E(x)=\frac{f(x)+f(-x)}{2}\\ =\frac{e^x+e^{-x}}{2}\end{array}$