If f(x) = the cube root of 2x - 3,
how do I find f^-1 (x)?
Hello,
$\displaystyle y=\sqrt[3]{2x-3}=(2x-3)^{1/3}$
Put everything to the power 3, so that it simplifies with 1/3 (recall $\displaystyle (a^b)^c=a^{bc}$)
so $\displaystyle y^3=2x-3$
and... $\displaystyle x=\tfrac{y^3+3}{2}=f^{-1}(y)$
does it look clear to you ?