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