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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?
Because all your doing when you compose functions is replacing x values. You don't modify constants like +2. +2 was always in g(x). We're interested in replaces all x's in g(x) with f(x) so +2 remains as is.
It was - 2
And I'm still convinced f(-x) = x^2
Ok. What f(-x) is stating is "take my original self and make all my x's negative". f(x)=x^2 so f(-x) would equal -x^2 because we are now taking the negative of all x's values in that function. That is why it is notated as f(x).
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.
Exactly. That's what you're doing. You're multiplying x^2 by -1 to get -x^2. If it's just multiplying "x" by -1 then it's -x * -x which equals x^2.
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 :)