Calculus 8th Edition

(a)$$f(x)=x^2+6$$ (b)$$g(x)=x^2+x-1$$
(a) To find $f$ such that $f \circ g=h$, we have$$f(g(x))=h(x) \quad \Rightarrow \quad f(2x+1)=4x^2+4x+7$$ $$\Rightarrow \quad f(2x+1)=(2x+1)^2+6$$ $$\Rightarrow \quad f(x)=x^2+6.$$ (b) To find $g$ such that $f \circ g=h$, we have$$f(g(x))=h(x) \quad \Rightarrow \quad 3g(x)+5=3x^2+3x+2$$ $$\Rightarrow \quad 3g(x)+5=3(x^2+x-1)+5$$ $$\Rightarrow \quad g(x)=x^2+x-1$$