# 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.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

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.

Expert Solution

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(x)=f(x), then the function f(x) is said to be an even function.

If f(x)f(x), then the function f(x) is not an even function.

If f(x)=f(x), then the function f(x) is said to be an odd function.

If f(x)f(x), then the function f(x) is not an odd function.

Calculation:

Section (i)

If f and g are even functions, then by the definition, f(x)=f(x) and g(x)=g(x).

Recall the fact that, (f+g)(x)=f(x)+g(x) (1)

As f and g are even functions, substitute f(x)=f(x) and g(x)=g(x) in equation (1) as follows.

(f+g)(x)=f(x)+g(x)=f(x)+g(x)=(f+g)(x)

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

Section (ii)

If f and g are odd functions, then by definition, f(x)=f(x) and g(x)=g(x).

As f and g are odd functions, substitute f(x)=f(x) and g(x)=g(x) in equation (1) as follows.

(f+g)(x)=f(x)g(x)=(f(x)+g(x))=((f+g)(x))

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

Section (iii)

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

Then by the definition, f(x)=f(x) and g(x)=g(x).

Therefore, substitute f(x)=f(x) and g(x)=g(x) in equation (1),

(f+g)(x)=f(x)+g(x)=f(x)g(x)

Observe that the function f(x)g(x) 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.

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