Need Help with math problem :(

Question

Given that ƒ(x) = 2x + 1 and g(x) = 3 + 6/x

a) Find ƒ(g(x))

b) Find g^-1 (x)

c) Suppose that h(3+6/x) = 9 + 36/x^2

1. Find h(x)

Answer 1

f=2x+1

g=3 +6/x

f(g(x)) = 2(3 +6/x) + 1 = 19+12/x + 1 = 20+12/x

inverse of g is found by switching x and y, then solving for the new y

g(x) =y= 3+6/x

x=3 + 6/y

multiply by y

xy =3y + 6

xy-3y = 6

y= 6/(x-3)

g^-1(x) = 6(x-3)

g(g^-1(x)) = 3+ 6/(x-3/6)=x

g^-1(g(x) = 6(/(3 +6/x -3) = 6(x/6) = x

the inverse of the inverse = x

h(3+6/x) = 9 +36/x^2 = (3+6/x)^2 - 36/x

h(x0 = x^2 - .....