Question

let
$f(x) = x^2 + 1$ and $g(x) = x^2 - 1$
then
$(f \circ f)(x) = \boxed{x^4 + 2x^2 + 2}$
$(f \circ g)(x) = \boxed{x^2 + 2x}$
$(g \circ f)(x) = \boxed{x^2 - 2x + 2}$

Explanation:

Step1: Define function composition

$(f \circ g)(x) = f(g(x))$

Step2: Substitute $g(x)$ into $f(x)$

$f(g(x)) = f(x^2 - 1) = (x^2 - 1)^2 + 1$

Step3: Expand and simplify

$(x^2 - 1)^2 + 1 = x^4 - 2x^2 + 1 + 1 = x^4 - 2x^2 + 2$

Step4: Define $(g \circ f)(x)$

$(g \circ f)(x) = g(f(x))$

Step5: Substitute $f(x)$ into $g(x)$

$g(f(x)) = g(x^2 + 1) = (x^2 + 1)^2 - 1$

Step6: Expand and simplify

$(x^2 + 1)^2 - 1 = x^4 + 2x^2 + 1 - 1 = x^4 + 2x^2$

Answer:

$(f \circ g)(x) = x^4 - 2x^2 + 2$
$(g \circ f)(x) = x^4 + 2x^2$