Chapter 8.4, Problem 76E

(a)

To determine

A constant multiple of an even function being even.

Answer to Problem 76E

  $g\left(-x\right)=g\left(x\right)$

Explanation of Solution

Given information:

Any constant multiple of an even function is even.

Formula used:

  $g\left(-x\right)=af\left(-x\right)$

Calculation:

True

Let $f\left(x\right)$ be an even function. Let $g\left(x\right)=af\left(x\right)$ where $a\in ℝ$

  $g\left(-x\right)=af\left(-x\right)$

Since f is even function, we get

  $\begin{array}{l}g\left(-x\right)=af\left(x\right)\\ =g\left(x\right)\end{array}$

Thus, g is an even function.

Conclusion:

  $g\left(-x\right)=g\left(x\right)$

(b)

To determine

A constant multiple of an odd function being odd.

Answer to Problem 76E

  $g\left(-x\right)=-g\left(x\right)$

Explanation of Solution

Given information:

Any constant multiple of an odd function being odd.

Formula used:

  $g\left(-x\right)=af\left(-x\right)$

Calculation:

True

Let $f\left(x\right)$ be an even function. Let $g\left(x\right)=af\left(x\right)$ where $a\in ℝ$

  $g\left(-x\right)=af\left(-x\right)$

Since f is even function, we get

  $\begin{array}{l}g\left(-x\right)=af\left(-x\right)\\ =-g\left(x\right)\end{array}$

Thus, g is an even function.

Conclusion:

  $g\left(-x\right)=-g\left(x\right)$

(c)

To determine

The sum or difference of two even function is even.

Answer to Problem 76E

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

Explanation of Solution

Given information:

The sum or difference of two even function is even.

Formula used:

  $g\left(-x\right)=f\left(-x\right)+ah\left(-x\right)$

Calculation:

True

Let $f\left(x\right)$ be an even function. Let $g\left(x\right)=f\left(x\right)+ah\left(x\right)$ where $a\in \left\{-1,+1\right\}$

  $g\left(-x\right)=f\left(-x\right)+ah\left(-x\right)$

Since f, h is even function, we get

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

Thus, g is an even function.

Conclusion:

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

(d)

To determine

The sum or difference of two odd functions is odd.

Answer to Problem 76E

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

Explanation of Solution

Given information:

The sum or difference of two odd functions is odd.

Formula used:

  $g\left(-x\right)=f\left(-x\right)+ah\left(-x\right)$

Calculation:

True

Let $f,h$ be an even function. Let $g\left(x\right)=f\left(x\right)+h\left(x\right),a\in \left\{-1,+1\right\}$

  $g\left(-x\right)=f\left(-x\right)+ah\left(-x\right)$

Since f, h are odd function, we get

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

Thus, g is an odd function.

Conclusion:

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

(e)

To determine

The sum or difference of an even function and odd function is odd.

Answer to Problem 76E

g is not odd function.

Explanation of Solution

Given information:

The sum or difference of an even function and an odd function is odd.

Formula used:

  $\begin{array}{l}f\left(x\right)=x^2-f\text{is}\text{even}\text{function}\\ h\left(x\right)=x^2-h\text{is}\text{odd}\text{function}\\ \text{g}\left(x\right)=f\left(x\right)+h\left(x\right)\end{array}$

Calculation:

False

A counter-example is following:

  $\begin{array}{l}f\left(x\right)=x^2-f\text{is}\text{even}\text{function}\\ h\left(x\right)=x^2-h\text{is}\text{odd}\text{function}\\ \text{g}\left(x\right)=f\left(x\right)+h\left(x\right)\end{array}$

Note that

  $f\left(1\right)=h\left(1\right)=1,g\left(1\right)=2\ne -g\left(-1\right)=0$

Thus, g is not odd function.

Conclusion:

g is not odd function.