Explain why a composite function in which g(x) is any function, is applied to the output of any even function h(x) , f(x) is always even.

Question

This is what I have so far, but I don't feel like its a full explanation

The even function gets rid of the negation

Since $ h(x) $ is even then $ h(x)=h(-x) for \ all \ x \ $

$ f(x) = g(h(x)) $

$ f(-x) = g(h(-x)) = g(h(x)) \ or \ g(-(-h(x))) = g(h(x)) $

therefore $ f(x) $ is even

Thanks in advance!!

Answer 1

You want to prove that $f$, defined by $f(x) = g(h(x))$ is even. That means proving that for any $x$, you have $$ f(x) = f(-x) $$

The stuff you've written out in the middle is almost a proof of that. To give a real proof, start by writing out the definition of $$ f(-x) $$ and simplifying. When you've done a step or two, you should be able to say "...which is just $f(x)$".