A function of a function?

Question

Let f(x) = x² − 1. Evaluate the following:

(a) f(x^2 + 3x)

(b) f(sqrt{x^2 - 9}/(x-3)

Answer 1

f(x) = x^2 - 1
f(x^2 - 3x) = (x^2 -1)^2 - 3(x^2 -1) =
x^4 - 2 x^2 + 1 - 3x^2 + 3 =
x^4 - 5x^2 + 4
That factors to (x^2 -4)(x^2 -1) = (x+2)(x-2)(x+1)(x-1) if you need to factor it.

f(sqrt{x^2 -9}/(x-3)) =
{sqrt(x^2-1)^2 - 9} / {(x^2-1)-3} =
sqrt(x^4 - 2x^2 -8) / (x^2 -4)

Answer 2

so f(x) = x^2 - 1 is your original function
(a) f(x^2 + 3x) means plug (x^2 + 3x) into the equation everytime you see x, so:
f(x^2 + 3x) = (x^2 + 3x)^2 - 1
= x^4 + 6x^3 + 3x^2 - 1
(b) f(sqrt{x^2 - 9}/(x-3) = (sqrt{x^2 - 9}/(x-3)^2 - 1
= (sqrt(x+3)(x-3)/(x-3))^2 -1
= [(x+3)(x-3) / (x-3)^2] - 1
= (x+3) / (x-3) - 1