Answer
$f(x)=x^2+6$
$g(x)=x^2+x-1$
Work Step by Step
Find $f$ such that $f\circ g=h$:
$(f\circ g)(x)=h(x)$
$f(g(x))=h(x)$
$f(2x+1)=4x^2+4x+7$
$f(2x+1)=4x^2+4x+1+6$
$f(2x+1)=(2x+1)^2+6$ (Replace $2x+1$ by $x$)
$f(x)=x^2+6$
Find $g$ such that $f\circ g=h$:
$(f\circ g)(x)=h(x)$
$f(g(x))=3x^2+3x+2$
$3g(x)+5=3x^2+3x+2$ (Subtract 5)
$3g(x)=3x^2+3x-3$ (Divide by 3)
$g(x)=x^2+x-1$
