Answer
$(f\circ g)(x)=-2x^{2}-x-1$
$(g\circ f)(x)=2x^{2}-17x+41$
$(f\circ g)(2)=-11$
$(g\circ f)(2)=15$
Work Step by Step
$f(x)=4-x$ $,$ $g(x)=2x^{2}+x+5$
$a.$ $(f\circ g)(x)$
To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify:
$(f\circ g)(x)=f(g(x))=4-(2x^{2}+x+5)=4-2x^{2}-x-5=...$
$...=-2x^{2}-x-1$
$b.$ $(g\circ f)(x)$
To find $(g\circ f)(x)$, substitute $x$ by $f(x)$ in $g(x)$ and simplify:
$(g\circ f)(x)=g(f(x))=2(4-x)^{2}+4-x+5=...$
$...=2(16-8x+x^{2})+4-x+5=...$
$...=32-16x+2x^{2}+4-x+5=2x^{2}-17x+41$
$c.$ $(f\circ g)(2)$
Substitute $x$ by $2$ in $(f\circ g)(x)$, which was found in part $a$, and evaluate:
$(f\circ g)(2)=f(g(2))=-2(2)^{2}-2-1=-2(4)-2-1=...$
$...=-8-2-1=-11$
$d.$ $(g\circ f)(2)$
Substitute $x$ by $2$ in $(g\circ f)(x)$, which was found in part $b$, and evaluate:
$(g\circ f)(2)=g(f(2))=2(2)^{2}-17(2)+41=2(4)-34+41=...$
$...=8-34+41=15$
