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A function f: R \rightarrow R is defined as f(x)=x^{3}+4 . Is it a bijection or not? In case it is a bijection, find \mathbf{f}^{-1} (3).
CBSE | Class 12 | Exercise 2.4 | Function | Maths | RD Sharma
A function f: R \rightarrow R is defined as f(x)=x^{3}+4 . Is it a bijection or not? In case it is a bijection, find \mathbf{f}^{-1} (3).

Given that f: R \rightarrow R is defined as f(x)=x^{3}+4 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}=v-4

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

\Rightarrow \mathrm{f} is onto.

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

Finding f^{-1} :

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

i

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