Even and odd composite functions.

• Nov 26th 2011, 04:09 PM
Fabio010
Even and odd composite functions.
Ok so
if f is even and g is even then fog is?

f(x) = f(-x)
g(x) = g(-x)

f(g(x)) = f(g(-x))
f(-g(x)) = f(g(x))
f(-g(x)) = f(-g(-x))

so f(-g(x)) = f(g(x) fog is even???
• Nov 26th 2011, 04:57 PM
Plato
Re: Even and odd composite functions.
Quote:

Originally Posted by fabio010
ok so
if f is even and g is even then fog is?
F(x) = f(-x) & g(x) = g(-x)
f(g(x)) = f(g(-x))
f(-g(x)) = f(g(x))
f(-g(x)) = f(-g(-x))
so f(-g(x)) = f(g(x) fog is even???

correct.
• Nov 26th 2011, 05:12 PM
Soroban
Re: Even and odd composite functions.
Hello, Fabio010!

This is simpler than you think . . .

Quote:

$\text{If }f\text{ is even and }g\text{ is even, then }f\!\circ\!g\text{ is?}$

We know that: . $\begin{Bmatrix}f(\text{-}x) &-& f(x) \\ g(\text{-}x) &=& g(x)\end{Bmatrix}$

The question is: . $\begin{Bmatrix}\text{Is }f\!\circ\!g \text{ even?} \\ \text{Does }f(g(\text{-}x)) \text{ equal }f(g(x))\,? \end{Bmatrix}$

Since $g(\text{-}x) = g(x)$, we have: . $f(g(\text{-}x)) \:=\:f(g(x))$ . . . . Yes!

• Nov 27th 2011, 04:30 AM
Fabio010
Re: Even and odd composite functions.
So if f is odd and g is odd then:

fog is odd because:

f(g(x)) not equal f(-g(x))
since g(-x) = -g(x)
and -f(g(x)) = f(-g(x))

right?
• Nov 27th 2011, 04:39 AM
Siron
Re: Even and odd composite functions.
If $f$ is odd then $f(-x)=-f(x)$, if $g$ is odd then $g(-x)=-g(x)$.

Show: $f(g(-x))=-f(g(x))$
$f(g(-x))=f(-g(x))=-f(g(x))$

Therefore $f\circ g$ is odd.
• Nov 27th 2011, 05:38 AM
Fabio010
Re: Even and odd composite functions.
Ok thanks for the help.