Let f(x) = x2, g(x) = √x + 3.a. Show that (f(x + 3)) = |x + 3| + 3.b. Does (x) = |x + 3| + 3 = (x) = |x| + 6? Graph them on the same coordinate plane

Answer with Step-by-step explanation:We are given that f(x)=[tex]x^2[/tex][tex]g(x)=\sqrt{x}+3[/tex]a.We have to show that [tex]g(f(x+3))=\mid{x+3}\mid +3[/tex][tex]g(f(x+3))=\sqrt{(x+3)^2}+3[/tex]When we remove square root then we take plus minus therefore we write in modulus.Therefore, [tex]g(f(x+3))=\mid {x+3}\mid+3[/tex]Hence, proved.b.We have to find that [tex]g(x)=\mid{x+3}\mid+3=\mid x\mid+6[/tex]Substitute x=-3 Then, we get [tex]g(x)=\mid{-3+3}\mid+3=3[/tex][tex]g(x)=\mid{-3}\mid+6=9[/tex][tex]3\neq 9[/tex][tex]\mid{x+3}\mid +3=\mid x\mid+6[/tex] for [tex]x\geq 0[/tex]But, [tex]\mid{x+3}\mid+3\neq \mid x\mid+6[/tex] for x<0Hence, [tex]\mid{x+3}\mid+3\neq \mid{x}+6[/tex] for all x.