Answer:
(gp) (5)= sqrt(32)+4
(pog)(5) = 28+8sqrt(5)
Explanation:
We are given the following functions:
![p(x)=x^2+7,g(x)=√(x)+4](https://img.qammunity.org/2023/formulas/mathematics/college/v2qxst8s2b128p9ole3b1mn48xxrm8tdjh.png)
(gp) (5)=
(gp) means that the outside function is g and the inside is p:
So
![g\circ p(x)=g(x^2+7)=√(x^2+7)+4](https://img.qammunity.org/2023/formulas/mathematics/college/d2bkjtrekr5jzt96hzqiyrawt1kaytycgq.png)
At x = 5
![(g\circ p)(5)=√(5^2+7)+4=√(32)+4](https://img.qammunity.org/2023/formulas/mathematics/college/phkjr2ks82p5qr7p392tbi0qoyenbjogx5.png)
(gp) (5)= sqrt(32)+4
(pog)(5) =
Outside p, inside g. So
![(p\circ g)(x)=p(√(x)+4)=(√(x)+4)^2+7=(√(x))^2+8√(x)+16+7=x+8√(x)+23](https://img.qammunity.org/2023/formulas/mathematics/college/gh8t00smyk3wrqmjtwvw0kolcmvy7povfy.png)
At x = 5
![(p\circ g)(5)=5+8√(5)+23=28+8√(5)](https://img.qammunity.org/2023/formulas/mathematics/college/qvsvh43860b9pyed7qmniulosmgwexyviy.png)
(pog)(5) = 28+8sqrt(5)