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)#