Given $f(x)=1/8x-3$ and $g(x)=x^3$ , how do you find $(fog)^-1$ ?

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Answer 1 · 2017-01-13T15:06:30

$root(3)(8 x + 24)$

First, let's evaluate $fog$ :

$Rightarrow fog = f (x^(3))$

$Rightarrow fog = (1) / (8) x^(3) - 3$

Then, let $y = fog$ :

$Rightarrow y = (1) / (8) x^(3) - 3$

Let's interchange the variables:

$Rightarrow x = (1) / (8) y^(3) - 3$

Now we must solve for $y$ :

$Rightarrow x + 3 = (1) / (8) y^(3)$

$Rightarrow 8 x + 24 = y^(3)$

$Rightarrow root(3)(8 x + 24) = y$

$therefore (fog)^(-1) = root(3)(8 x + 24)$

Given f(x)=1/8x-3f(x)=1/8x-3 and g(x)=x^3g(x)=x^3, how do you find (fog)^-1(fog)^-1?