Answer
(a.)$(f\circ g)(x) =x^4-6x^2+10$.
(b.)$(g\circ f)(x) =x^4+2x^2-2$.
(c.)$(f\circ g)(2) =2$.
Work Step by Step
The given functions are
$f(x)=x^2+1$ and $g(x)=x^2-3$.
(a) $(f\circ g)(x) = f(g(x))$
Replace $x$ with $g(x)$ in the function $f(x)$.
$f(g(x))=(g(x))^2+1$
Plug value of $g(x)$ in the right hand side.
$f(g(x))=(x^2-3)^2+1$
$f(g(x))=x^4+9-6x^2+1$
$f(g(x))=x^4-6x^2+10$.
Hence, $(f\circ g)(x) =x^4-6x^2+10$.
(b) $(g\circ f)(x) = g(f(x))$
Replace $x$ with $f(x)$ in the function $g(x)$.
$g(f(x))=(f(x))^2-3$
Plug value of $f(x)$ in the right hand side.
$g(f(x))=(x^2+1)^2-3$
$g(f(x))=x^4+1+2x^2-3$
$g(f(x))=x^4+2x^2-2$.
Hence, $(g\circ f)(x) =x^4+2x^2-2$.
(c) Replace $x$ with $2$ in the part (a) solution.
$(f\circ g)(2) =(2)^4-6(2)^2+10$
$(f\circ g)(2) =16-6(4)+10$
$(f\circ g)(2) =16-24+10$
$(f\circ g)(2) =26-24$
$(f\circ g)(2) =2$.
