Question:

If $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$, then $f^{-1}(x)=$

Solution:

Given: A function $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$

$f(x)=6-(x-9)^{3}$

$\Rightarrow y=6-(x-9)^{3}$

$\Rightarrow(x-9)^{3}=6-y$

$\Rightarrow x-9=(6-y)^{\frac{1}{3}}$

$\Rightarrow x=9+(6-y)^{\frac{1}{3}}$

Thus, $f^{-1}(x)=9+(6-x)^{\frac{1}{3}}$

Hence, if $f: R \rightarrow R$ is defined by $f(x)=6-(x-9)^{3}$, then $f^{-1}(x)=9+(6-x)^{\frac{1}{3}}$.