
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(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.
Chapter 1 Solutions
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