If h(x) = (fog) (x) and h(x) = 4 square root x+7, find g(x) if f(x) = 4 square root x+ 1

Answer:[tex]g(x)=x+6[/tex] is the answer given [tex]h(x)=4\sqrt{x+7}[/tex] and [tex]f(x)=4\sqrt{x+1}[/tex].Step-by-step explanation:[tex]h(x)=(f \circ g)(x)[/tex][tex]h(x)=f(g(x))[/tex]Inputting the given function for h(x)  into the above:[tex]4\sqrt{x+7}=f(g(x))[/tex]Now we are plugging in g(x) for x in the expression for f which is [tex]4\sqrt{x+1}[/tex] which gives us [tex]4\sqrt{g(x)+1}[/tex]:[tex]4\sqrt{x+7}=4\sqrt{g(x)+1}[/tex]We want to solve this for g(x).If you don't like the looks of g(x) (if you think it is too daunting to look at), replace it with u and solve for u.[tex]4\sqrt{x+7}=4\sqrt{u+1}[/tex]Divide both sides by 4:[tex]\sqrt{x+7}=\sqrt{u+1}[/tex]Square both sides:[tex]x+7=u+1[/tex]Subtract 1 on both sides:[tex]x+7-1=u[/tex]Simplify left hand side:[tex]x+6=u[/tex][tex]u=x+6[/tex]Remember u was g(x) so you just found g(x) so congratulations.[tex]g(x)=x+6[/tex].Let's check it:[tex](f \circ g)(x)[/tex][tex]f(g(x))[/tex][tex]f(x+6)[/tex] I replace g(x) with x+6 since g(x)=x+6.[tex]4\sqrt{(x+6)+1}[/tex] I replace x in f with (x+6).[tex]4\sqrt{x+6+1}[/tex][tex]4\sqrt{x+7}[/tex][tex]h(x)[/tex]The check is done. We have that [tex](f \circ g)(x)=h(x)[/tex].