Answer
$a.$ $(f\circ g)(x)=x^{4}-6x^{2}+10$
$b.$ $(g\circ f)(x)=x^{4}+2x^{2}-2$
$c.$ $(f\circ g)(2)=2$
$d.$ $(g\circ f)(2)=22$
Work Step by Step
$f(x)=x^{2}+1$ $,$ $g(x)=x^{2}-3$
$a. (f\circ g)(x)$
To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify:
$(f\circ g)(x)=f(g(x))=(x^{2}-3)^{2}+1=...$
$...=x^{4}-6x^{2}+9+1=x^{4}-6x^{2}+10$
$b. (g\circ f)(x)$
To find $(g\circ f)(x)$, substitute $x$ by $f(x)$ in $g(x)$ and simplify:
$(g\circ f)(x)=g(f(x))=(x^{2}+1)^{2}-3=...$
$...=x^{4}+2x^{2}+1-3=x^{4}+2x^{2}-2$
$c. (f\circ g)(2)$
Substitute $x$ by $2$ in $(f\circ g)(x)$, which was found in part $a$, and evaluate:
$(f\circ g)(2)=f(g(2))=(2)^{4}-6(2)^{2}+10=...$
$...=16-24+10=2$
$d. (g\circ f)(2)$
Substitute $x$ by $2$ in $(g\circ f)(x)$, which was found in part $b$, and evaluate:
$(g\circ f)(2)=g(f(2))=(2)^{4}+2(2)^{2}-2=...$
$...=16+8-2=22$
