Use functions $f(x) =\frac{x}{8} + 1$ and $g(x) = x^3$ to find $(f o g)^-1(5)$

Question

Right answer to the question is $$ 3\sqrt[3]{2} $$ but i got $$ 2\sqrt[3]{4} $$ got it by finding inverse of $$ \frac{x^3}{8} + 1 $$ and then sustituting 5 into $$ 2\sqrt[3]{x-1} $$ who is right?

Answer 1

First of all recall that $$\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)={{\left( f\circ g \right)}^{-1}}\left( x \right)$$

it follows that you need to find the inverse functions for both

we have $f(x)=\frac{x}{8}+1$ then $8f(x)=x+8$ i.e $f^{-1}(x)=8(x-1)$

similarly for $g(x)=x^3$ then $g^{-1}(x)=\sqrt[3]{x}$

hence by above property we get

${{\left( f\circ g \right)}^{-1}}\left( x \right)=(g^{-1}(f^{-1}(x))=\sqrt[3]{8(x-1)}=2\sqrt[3]{x-1}$

or: you can find $$\left( {{f}}\circ {{g}} \right)\left( x \right)=f(g(x))=\frac{g(x)}{8}+1=\frac{x^3}{8}+1$$ then its easy to find its inverse function to get $${{\left( f\circ g \right)}^{-1}}\left( x \right)=\sqrt[3]{8(x-1)}=2\sqrt[3]{x-1}$$ Seems the error is not $x=5$ to get that result