Question regarding Odd and Even function $g(x)=f(x)(f(x)+f(-x))$

Question

Let $f\colon\mathbb{R} \to \mathbb{R}$. Define $g: \mathbb{R}\to \mathbb{R}$ by

$g(x)=f(x)(f(x)+f(-x))$

Then which of following is/are correct?

A. $g$ is even for all $f$

B. $g$ is odd for all $f$

C. $g$ is even if $f$ is even

D. $g$ is even if $f$ is odd

Taking $f(x)=\sin x$ eliminated option B. What if I take $f$ to be odd function then my $g=0$. Is $0$ a even function or odd function?

Thanks

Write out $g(-x)$ and simplify. Then you will be able to solve the question using definition of even and odd — Shailesh, Dec 08 '15 at 05:31
Note that $f(x)+f(-x)$ is even. So, what is the product of an even with an even or an odd with an even? So, the only correct answer is $C$. — Mark Viola, Dec 08 '15 at 05:59

score 0 · Answer 1 · edited Jun 23 '17 at 00:26

0

$g(x)=f(x)(f(x)+f(-x))\tag{1}$

$g(-x)=f(-x)(f(-x)+f(x))\tag{2}$

adding (1) and (2) we get

$g(x)+g(-x)=(f(x)+f(-x))^2$,this equation tells us that $g$ is odd if $f$ is odd.But this is not among the options.

If $f$ is even then equation (1) becomes $g(x)=2(f(x))^2$.Here,$2$ being a constant is an even function in this case $g$ is product of two even functions therefore even itself.

if $f$ is odd, then $g=f(f−f)=0,$, rendering g to be both even and odd. Hence,$g$ is even if $f$ is both even and odd.Hence (C) & (D) are correct options

edited Jun 23 '17 at 00:26

answered May 27 '17 at 13:36

Styles

3,539

(D) is also correct – Hagen von Eitzen May 27 '17 at 13:43
@HagenvonEitzen:$g(x)+g(-x)=(f(x)+f(-x))^2 \implies g$ is odd if $f$ is odd. – Styles May 27 '17 at 18:13
@HagenvonEitzen:Have i done something wrong? – Styles May 27 '17 at 18:14
@PKStyles He's referring to the original assumptions in the problem. Note that if $f $ is odd, then $g = f(f-f) = 0$, rendering g to be both even and odd. – bloomers May 27 '17 at 21:14
@bloomers:hope this works now. – Styles May 28 '17 at 06:23

Question regarding Odd and Even function $g(x)=f(x)(f(x)+f(-x))$

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