Show that if f(x) is an even function, then g(f(x)) is an even function.

Question

The question does not say whether g(x) is odd, even or neither.

Proving that g(f(x)) = g(-f(x)) (Proof that g(f(x)) is even)

I swapped f(x) with f(-x)

So that g(f(-x)) = g(-f(x)).

But from there, I don’t know how else to rearrange it to finish off the proof. Attempting to logic it out, I’m getting confused, because if g(x) was odd, then wouldn’t plugging in opposite numbers (f(-x) and -f(x)) keep it odd? I know this thought is wrong, because I’ve sampled different equations, it’s just that I’m struggling to understand it how it works and how to prove it from that final step.

Thanks, any help is appreciated. Sorry I’m not sure if what I said makes sense, I’m just really confused.

Answer 1

$f (-x)=f(x)$. Just apply $g$on both sides to get $g(f(-x))=g(f(x))$. So $g(f(x))$ is even.

Answer 2

From your comments, I infer that you try to prove that $g$ is an even function, with $g(-x)=g(x)$, or $g(-f(x))=g(f(x))$.

But the question is about $g\circ f$, which is a different function. The correct reasoning is

$$(g\circ f)(-x)=g(f(-x))=g(f(x))=(g\circ f)(x)$$

whatever $g$.

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Example:

• $f(x)=|x|$ is even.

• $g(x)=x+1$ is neither even nor odd.

• $(g\circ f)(x)=|x|+1$ is even.

(And don't confuse with $(f\circ g)(x)=|x+1|$.)