0
$\begingroup$

I've run into a problem while trying to prove that $f(g(x))$ is even given $f$ & $g$ are even.

My ultimate goal is to find if the following is true: $(f \circ g)(x) = (f \circ g)(-x)$

Given $f$ is even --> $f(x) = f(-x)$

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

$(f \circ g)(-x) = f(g(-x)) = \mathbf{f(-g(-x))}$ // this line dosent seem to be true but I can't explain why.

Given $f$ & $g$ are even, is the bolded part equal to $f(g(x))$?

Really what i'm trying to ask is whether $-g(-x) = g(x)$, given we know $g$ is even.

If so, how?

  • 0
    $-g(x)$ is not equal to $g(-x)$ unless $g(x) = 0$, but $f$ of them has the same value if $f$ is even. – 2017-01-10
  • 0
    I just can't get my head wrapped around the idea. We know f(x) = f(-x), Therefore why dosent substituting g(x) in for x and after, evaluating fog(-x) dosent work?? – 2017-01-10

2 Answers 2

0

$f(g(-x)) = f(-g(-x))$ because $f$ is even, i.e. $f(t) = f(-t)$, with $t = g(-x)$.

But personally, known that $g$ is even, I'd have written $(f\circ g)(x) = f(g(x)) = f(g(-x)) = (f \circ g)(-x)$, and as you can see you don't need $f$ even.

  • 0
    Thank you, your answer helped the most!! – 2017-01-10
  • 0
    No problem. Check the answer, so people can immediately see how to deal with the question – 2017-01-10
2

All you need is that $g$ is even. For if $f$ is any function on $\mathbb R$ and $g$ is an even function,

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

  • 0
    see here i hope this will help you http://www.pace-monmouth.org/student/classes/precalculus/Functions%20Even%20Odd.pdf – 2017-01-10