A function f : R → R is defined as f(x)

Solution:

Injectivity of $f$.

Let $x$ and $y$ be two elements of domain $(R)$, such that

$f(x)=f(y)$

$\Rightarrow x^{3}+4=y^{3}+4$

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

$\Rightarrow x=y$

So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (R), such that f(x) = y.

$\Rightarrow x^{3}+4=y$

$\Rightarrow x^{3}=y-4$

$\Rightarrow x=\sqrt[3]{y-4} \in R$ (domain)

$\Rightarrow f$ is onto.

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

Finding $f^{-1}$ :

Let $f^{-1}(x)=y \quad \ldots(1)$

$\Rightarrow x=f(y)$

$\Rightarrow x=y^{3}+4$

$\Rightarrow x-4=y^{3}$

$\Rightarrow y=\sqrt[3]{x-4}$

So, $f^{-1}(x)=\sqrt[3]{x-4} \quad[$ from $(1)]$

$f^{-1}(3)=\sqrt[3]{3-4}=\sqrt[3]{-1}=-1$