Chapter 1.1, Problem 73E

To determine

Whether the function $f+g$ is even if f and g are even; whether the function $f+g$ is odd if f and g are odd; whether the function $f+g$ is even or odd if f  is even and g is odd.

Answer to Problem 73E

The function $f+g$ is an even function if f and g are even.

The function $f+g$ is an odd function if f and g are odd.

The function $f+g$ neither an even function nor an odd function if f is even and g is odd.

Explanation of Solution

Definition used:

If $f\left(x\right)=f\left(-x\right)$, then the function $f\left(x\right)$ is said to be an even function.

If $f\left(x\right)\ne f\left(-x\right)$, then the function $f\left(x\right)$ is not an even function.

If $f\left(-x\right)=-f\left(x\right)$, then the function $f\left(x\right)$ is said to be an odd function.

If $f\left(-x\right)\ne -f\left(x\right)$, then the function $f\left(x\right)$ is not an odd function.

Calculation:

Section (i)

If f and g are even functions, then by the definition, $f\left(x\right)=f\left(-x\right)$ and $g\left(x\right)=g\left(-x\right)$.

Recall the fact that, $\left(f+g\right)\left(-x\right)=f\left(-x\right)+g\left(-x\right)$ (1)

As f and g are even functions, substitute $f\left(-x\right)=f\left(x\right)$ and $g\left(-x\right)=g\left(x\right)$ in equation (1) as follows.

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

Therefore, $f+g$ is an even function as it satisfies the definition of even function, $\left(f+g\right)\left(-x\right)=\left(f+g\right)\left(x\right)$.

Section (ii)

If f and g are odd functions, then by definition, $f\left(-x\right)=-f\left(x\right)$ and $g\left(-x\right)=-g\left(x\right)$.

As f and g are odd functions, substitute $f\left(-x\right)=-f\left(x\right)$ and $g\left(-x\right)=-g\left(x\right)$ in equation (1) as follows.

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

Therefore, $f+g$ is an odd function as it satisfies the definition of odd function, $\left(f+g\right)\left(-x\right)=-\left(\left(f+g\right)\left(x\right)\right)$.

Section (iii)

Given that, f is an even function and g is an odd function.

Then by the definition, $f\left(x\right)=f\left(-x\right)$ and $g\left(-x\right)=-g\left(x\right)$.

Therefore, substitute $f\left(x\right)=f\left(-x\right)$ and $g\left(-x\right)=-g\left(x\right)$ in equation (1),

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

Observe that the function $f\left(x\right)-g\left(x\right)$ neither satisfies the condition of even function nor the condition of odd function.

Therefore, the function $f+g$ is neither an even function nor an odd function if f is even and g is odd.