Question

Given that f(x) is an even function.
Prove that g(f(x)) is an even function regardless of the symmetry of g(x)

Answer 1

\[Even:~~f(-x)=f(x)\\Odd:~~~f(-x)=-f(x)\]

Answer 2

Hide the "\(g(f(x))\)" thing: call that \(F(x)\).

Now the question => what does it mean for \(F\) to be an even function?

\(F(-x) = F(x)\) right? Can you continue?

Answer 3

Yes. By definition, that's an even function.

Answer 4

You now have to write \(F(-x)\) using the \(g\) and \(f\) functions.

Answer 5

so you're saying
\[-x = -f(x)\]

Answer 6

This is not correct.

We have that \(F(x)=g(f(x))\). Using this expression, what is \(F(-x)\)? -> you must find the right place to put that “\(-\)” sign. To do so, observe \(g(f(x))\) and determine the place(s) where the \(x\) is located. Change any \(x\) into a \(-x\), but don't add any other “\(-\)” sign anywhere.

Answer 7

You're saying that F(x) = g(f(x)) by substitution.

And if F(-x), then g( -f(x) )

But -f(x) is impossible since f(x) is an even function because f(x) = f(-x) but never can it have an output of -f(x)

Answer 8

No, I `define` \(F\) as the function computed in this way: \(F(x) = g(f(x))\). There is only only one variable: it's \(x\).

Answer 9

Try this: change all \(x\) into \(y\)'s. What happened?

\(g(f(x))\) became \(g(f(y))\).

Answer 10

It is like a text editor: "replace 'x' with 'y' ".
Imagine \(y\) is "\(-x\)". What do you obtain?

Answer 11

if y = -x
inside f(y) and f is an even function then that's the same as
f(-x) = f(x)

You'd obtain f(x)

Answer 12

right!

\(F(\boxed{-x})=g(f(\boxed{-x}))\) and \(f(-x)=f(x)\),

So, what about \(F(-x) = g(f(-x)) = \dots\)

Answer 13

... = g(f(x)) = F(x)

Answer 14

Correct. You've proven the property.

(The thing was to see \(g(f(x))\) as a "black box", input -x instead of x and see what happens.)

Answer 15

kk thx

Answer 16

yw