Question

.

Answer 1

This is a booger of a problem mate. Do you have any work done already?

Answer 2

To find (fog)(x), we substitute g(x) into h(x), which is\[\frac{ 1 }{ (x ^{2}+2) }-5\]

Answer 3

for g o f (x) you would have to use the FOIL method to distribute

(1/x - 5)^2

Go ahead and try it out.

Answer 4

Well, actually @harmonica1243 hinted you.

Answer 5

lets lay it out like this

(1/x - 5) * (1/x - 5)

First multiply the First items

(1/x)*(1/x)

then multiply the Outsides and add the Insides

(-5)*(1/x) + (1/x)*(-5)

then multiply the Last

(-5)*(-5)

add it all together and you get

\[1/x^2 - 10/x + 25\]

Answer 6

I think it might be easier to write it naturally.

Answer 7

Wait a second. For f(x), you mean 1/x-5 or 1/(x-5), @myusernameisfuntosay?

Answer 8

I see. Then you can find \[(gof)(x) = g(f(x)) = g(\frac{ 1 }{ x-5 }) = (\frac{ 1 }{ x-5 })^{2}+2\]Then substitute x for 6 to find (gof)(6). I think that's all.

Answer 9

Yes. All you need to do is replace x with 6 to get the answer.

Answer 10

It's 3, I think.

Answer 11

I got 3

Answer 12

I think you can easily see that we got x^2 in both numerator and denominator.

Answer 13

No since we got two x^2, you can get f(x) = x^2.

Answer 14

Then when you get g(x) = x/(x+4), you can substitute x with f(x) = x^2, which will give you F(x)= x^2/(x^2+4).