Question
Consider $\text{f}:\text{R}\rightarrow\text{R}_+\rightarrow[4,\infty)$ given by $f(x) = x^2 + 4$. Show that f is invertible with inverse of f given by $\text{f}^{-1}(\text{x})=\sqrt{\text{x}-4,}$ where $R^+$ is the set of all non-negative real numbers.

Answer

Injectivity of f: Let x and y be two elements of the domain (Q), such that $f(x) = f(y)$
$\Rightarrow x^2 + 4 = y^2 + 4$
$\Rightarrow x^2 = y^2$
$\Rightarrow x = y$ as co-domain as $R^+$
So, f is one-one.
Surjectivity of f: Let y be in the co-domain (Q), such that $f(x) = y$
$\Rightarrow x^2 + 4 = y$
$\Rightarrow x^2 = y - 4$
$\Rightarrow\ \text{x}=\text{y}-4\in\text{R}$
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding $f^{-1}$: Let $f^{-1}(x) = y .......(1)$
$\Rightarrow x = y^2 + 4$
$\Rightarrow x - 4 = y^2$
$\Rightarrow y = x - 4$
So, $f^{-1}(x) = x - 4$ [from $(1)]$

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