A function f : R R is defined as f(x) = x^3 + 4. Is it a bijectio… - Vidyadip

Question

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 $f^{-1}(3).$

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Answer

We have,
$f : R \rightarrow R$ in a function defined by
$f(x) = x^3 + 4$
Injectivity: Let f(x_1) = f(x_2) for $\text{x}_1,\text{x}_2\in\text{R}$
$\Rightarrow\ \text{x}_1^3+4=\text{x}_2^3+4$
$\Rightarrow\ \text{x}_1^3=\text{x}_2^3$
$\Rightarrow\ \text{x}_1=\text{x}_2$
$\Rightarrow f$ is one-one.
Surjectivity: Let $\text{y}\in\text{R}$ be artritrary such that
$f(x) = y$
$\Rightarrow x^3 + 4 = y$
$\Rightarrow x^3 + 4 - y = 0$
We know that an odd degree equation must have a real root.
$\Rightarrow\ \alpha^3+4=\text{y}$
$\Rightarrow\ \text{f}(\alpha)=\text{y}$
$\Rightarrow f$ is onto.
Since f is one-one and onto.
$\Rightarrow f$ is bijective.
finally, $f(x) = y$
$\Rightarrow x^3 + 4 = y$
$\Rightarrow x^3 = y - 4$
$\Rightarrow\ \text{x}=(\text{y}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}(\text{x})=(\text{x}-4)^\frac{1}{3}$
$\therefore\ \text{f}^{-1}(3)=(3-4)^\frac{1}{3}=-1$

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