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If $f: R \rightarrow R$ be mapping defined by $f(x)=x^{3}+5$, then $f^{-1}(x)$ is equal to
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1925 Upvotes
Verified Answer
The correct answer is:
$(x-5)^{1 / 3}$
Let
$$
y=f(x)=x^{3}+5
$$
$$
\begin{array}{lr}
\Rightarrow \quad x=(y-5)^{1 / 3} \\
\therefore \quad f^{-1}(x)=(x-5)^{1 / 3} \\
\text { Given, } & f(x)=\frac{a x+b}{c x+d}
\end{array}
$$
$$
y=f(x)=x^{3}+5
$$
$$
\begin{array}{lr}
\Rightarrow \quad x=(y-5)^{1 / 3} \\
\therefore \quad f^{-1}(x)=(x-5)^{1 / 3} \\
\text { Given, } & f(x)=\frac{a x+b}{c x+d}
\end{array}
$$
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