Question

If I'm given f(x) and g(x) and I need to find h(x) if h(x) = g(f(-x)) what do I do to f(x) for the negative x transformation?

f(x) = x^2 and g(x) = (1/3)x - 2

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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)).

Answer 6

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

Answer 7

@PhantomCrow Why + 2?

Answer 8

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.

Answer 9

It was - 2

Answer 10

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

Answer 11

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).

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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

Answer 17

No. It does not. You are wrong.

Answer 18

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

Answer 19

f(-x) is indeed x^2.

Answer 20

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