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Do the opposite for every instance of x in f(x). So all x's become negative.

@PhantomCrow So h(x) = 1/3(x^2) - 2 just like f(g(x) would be?

h(x) would be (1/3)(-x^2) +2

@PhantomCrow I don't think so. I think because f(x) = x^2, then g(f(-x)) = g(f(x))

if f(x)=x^2 then f(-x)=-x^2. Now replace f(-x) with every instance of x in g(x) to get g(f(x)).

or rather g(f(-x)). Sorry.

@PhantomCrow Why + 2?

It was - 2

And I'm still convinced f(-x) = x^2

And apologies for misinterpreting the 2. -2 remains the same, then.

I have to disagree. That's -f(x).

-f(x) would be multiplying the entire function by -1.

In this case, -f(x) happens to equal f(-x). That's just arbitrary, however, and not always the case.

No. It does not. You are wrong.

Actually, I am. I'm terribly sorry. I wasn't making the distinction between (-x)^2 and -(x^2).

f(-x) is indeed x^2.

No problem man. Glad you figured it out. And I'm glad I figured it out :)