Chapter 1, Problem 3PS

a .

To determine

To find:What can be said about the sum and difference of two even functions.

Answer to Problem 3PS

Even

Explanation of Solution

Concept Used:

A function $f\left(x\right)$ is said to be an even function if for all x in the domain of f , $f\left(-x\right)=f\left(x\right)$ .

A function $f\left(x\right)$ is said to be an odd function if for all x in the domain of f , $f\left(-x\right)=-f\left(x\right)$ .

Let f and g be two even functions.

Then,

  $f\left(-x\right)=f\left(x\right)$

And

  $g\left(-x\right)=g\left(x\right)$

Now consider the sum of these two functions $f+g$ ,

  $\begin{array}{c}\left(f+g\right)\left(-x\right)=f\left(-x\right)+g\left(-x\right)\\ =f\left(x\right)+g\left(x\right)\\ =\left(f+g\right)\left(x\right)\end{array}$

Thus, $f+g$ is an even function.

Now consider the difference of these two functions $f-g$ ,

  $\begin{array}{c}\left(f-g\right)\left(-x\right)=f\left(-x\right)-g\left(-x\right)\\ =f\left(x\right)-g\left(x\right)\\ =\left(f-g\right)\left(x\right)\end{array}$

Thus, $f-g$ is an even function.

b.

To determine

What can be said about the sum and difference of two odd functions.

Answer to Problem 3PS

Odd

Explanation of Solution

Concept Used:

A function $f\left(x\right)$ is said to be an even function if for all x in the domain of f , $f\left(-x\right)=f\left(x\right)$ .

A function $f\left(x\right)$ is said to be an odd function if for all x in the domain of f , $f\left(-x\right)=-f\left(x\right)$ .

Let f and g be two odd functions.

Then,

  $f\left(-x\right)=-f\left(x\right)$

And

  $g\left(-x\right)=-g\left(x\right)$

Now consider the sum of these two functions $f+g$ ,

  $\begin{array}{c}\left(f+g\right)\left(-x\right)=f\left(-x\right)+g\left(-x\right)\\ =-f\left(x\right)-g\left(x\right)\\ =-\left[f\left(x\right)+g\left(x\right)\right]\\ =-\left(f+g\right)\left(x\right)\end{array}$

Thus, $f+g$ is an odd function.

Now consider the difference of these two functions $f-g$ ,

  $\begin{array}{c}\left(f+g\right)\left(-x\right)=f\left(-x\right)-g\left(-x\right)\\ =-f\left(x\right)+g\left(x\right)\\ =-\left[f\left(x\right)-g\left(x\right)\right]\\ =-\left(f-g\right)\left(x\right)\end{array}$

Thus, $f-g$ is an odd function.

c .

To determine

What can be said about the sum and difference of an odd function and an even functions.

Answer to Problem 3PS

Even

Explanation of Solution

Concept Used:

A function $f\left(x\right)$ is said to be an even function if for all x in the domain of f , $f\left(-x\right)=f\left(x\right)$ .

A function $f\left(x\right)$ is said to be an odd function if for all x in the domain of f , $f\left(-x\right)=-f\left(x\right)$ .

Let f be an even function and g be an odd function.

Then,

  $f\left(-x\right)=f\left(x\right)$

And

  $g\left(-x\right)=-g\left(x\right)$

Now consider the sum of these two functions $f+g$ ,

  $\begin{array}{c}\left(f+g\right)\left(-x\right)=f\left(-x\right)+g\left(-x\right)\\ =f\left(x\right)+\left[-g\left(x\right)\right]\\ =f\left(x\right)-g\left(x\right)\\ =\left(f-g\right)\left(x\right)\end{array}$

Thus, $f+g$ is neither an even function nor an odd function.

Now consider the difference of these two functions $f-g$ ,

  $\begin{array}{c}\left(f-g\right)\left(-x\right)=f\left(-x\right)-g\left(-x\right)\\ =f\left(x\right)-\left[-g\left(x\right)\right]\\ =f\left(x\right)+g\left(x\right)\\ =\left(f+g\right)\left(x\right)\end{array}$

Thus, $f-g$ is neither an even function nor an odd function.