Question

Let f : R → R, g : R → R be defined by f(x) = 2x - 3, g(x) = x³ + 5, then find (fog)⁻¹ (x).

Answer 1

Given that

f(x) = 2x - 3, g(x) = x³ + 5

We have to find (fog)⁻¹ (x).

First we are going to find (fog) x.

(fog) x = f [ g(x) ]

(fog) x = f [ x³ + 5 ]

(fog) x = 2( x³ + 5 ) - 3

(fog) x = 2 x³ + 10 - 3

          (fog) x = 2 x³ + 7

Now Let

y = (fog) x

y = 2 x³ + 7

y - 7 = 2 x³

2 x³ = y - 7

x³ = (y - 7)/2

Taking cube root on both sides

x = [(y - 7)/2]^1/3

$x=(\frac{y-7}{2} )^{\frac{1}{3}}\\ \\f^{-1}(y) =(\frac{y-7}{2} )^{\frac{1}{3}}\\ \\(fog)^{-1}(x)=(\frac{y-7}{2} )^{\frac{1}{3}}$