To determine (1) If f and g are both even functions, whether f+g is even? (2) If f and g are both odd functions, whether f+g is odd? (3) If f is even and g is odd function, what will f+g?

Question

To determine    (1) If f and g are both even functions, whether f+g is even?    (2) If f and g are both odd functions, whether f+g is odd?    (3) If f is even and g is odd function, what will f+g?

sweererlirumeX · Accepted Answer

A function is even when f(x)=f(−x) for all x. A functon is odd when −f(x)=f(−x) for all x. Consider (f+g)(−x), (f+g)(−x)=f(−x)+g(−x) =f(x)+g(x) (f and g are even functions) =(f+g)(x) So, f+g is also even function. f is odd so, f(−x)=−f(x), for all x. g is odd so, g(−x)=−g(x), for all x. Consder (f+g)(−x), (f+g)(−x)=f(−x)+g(−x) =−f(x)+g(−x) (−g(x)) (f and g are odd function) =−f(x)−g(x) =−(f(x)+g(x)) =−(f+g)(x) So, f+g is also odd fuction. f is an even function so, f(−x)=f(x), for all x. g is an odd function so, g(−x)=−g(x), for all x. Let f(x)=x2 Since, f(−x)=(−x)2=x2 So, f(x)=x2 is an even function. Let g(x)=x3 Since, g(−x)=(−x)3=−x3 So, g(x)=x3 is an odd function. Consider (f+g)(−x), (f+g)(−x)=f(−x)+g(−x) =(−x)2+(−x)3 =x2−x3 and (f+g)(x)=f(x)+g(x) =x2+x3 We observe that (f+g)(−x)≠(f+g)(x) and (f+g)(−x)≠−(f+g)(x). So, the sum of an even and an odd function is neither even non odd.

To determine (1) If f and g are both even functions,

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