If f : R → R be defined by

Solution:

Injectivity of $f$ :

Let $x$ and $y$ be two elements in domain $(R)$,

such that, $x^{3}-3=y^{3}-3$

$\Rightarrow x^{3}=y^{3}$

$\Rightarrow x=y$

So, $f$ is one-one.

Surjectivity of $f$ :

Let $y$ be in the co-domain (R) such that $f(x)=y$

$\Rightarrow x^{3}-3=y$

$\Rightarrow x^{3}=y+3$

$\Rightarrow x=\sqrt[3]{y+3} \in R$

$\Rightarrow f$ is onto.

So, $f$ is a bijection and, hence, it is invertible.

Finding $f^{-1}$.

Let $f^{-1}(x)=y$                        ...(1)

$\Rightarrow x=f(y)$

$\Rightarrow x=y^{3}-3$

$\Rightarrow x+3=y^{3}$

$\Rightarrow y=\sqrt[3]{x+3}=f^{-1}(x) \quad[$ from (1) $]$

So, $f^{-1}(x)=\sqrt[3]{x+3}$

Now, $f^{-1}(24)=\sqrt[3]{24+3}=\sqrt[3]{27}=\sqrt[3]{3^{3}}=3$

and $f^{-1}(5)=\sqrt[3]{5+3}=\sqrt[3]{8}=\sqrt[3]{2^{3}}=2$