Find the domain and the range of the given function: f(x)= x^ 2 -…

Question

Find the domain and the range of the given function: $\mathrm{f}(\mathrm{x})=\sqrt{x^{2}-4}$

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Answer

Given $\mathrm{f}(\mathrm{x})=\sqrt{x^{2}-4}$
For $\mathrm{D}_{\mathrm{f}}, \mathrm{f}(\mathrm{x})$ must be a real number $\Rightarrow \sqrt{x^{2}-4}$ must be a real number
$x^{2}-4 \geq 0 \Rightarrow(x+2)(x-2) \geq 0$
either $\mathrm{x} \leq-2$ or $\mathrm{x} \geq 2$
$\Rightarrow \mathrm{D}_{\mathrm{f}}=(-\infty,-2] \cup[2, \infty)$
For $\mathrm{R}_{\mathrm{f}}$, let $\mathrm{y}=\sqrt{x^{2}-4} \ldots$ (i)
As square root of a real number is always non-negative, $\mathrm{y} \geq 0$
On squaring (i), we get $y^{2}=x^{2}-4$
$\Rightarrow x^{2}=y^{2}+4$ but $x^{2} \geq 0$ for all $x \in D_{f}$
$\Rightarrow \mathrm{y}^{2}+4 \geq 0 \Rightarrow \mathrm{y}^{2} \geq-4$, which is true for all $\mathrm{y} \in \mathbf{R}$. Also $\mathrm{y} \geq 0$
$\Rightarrow \mathrm{R}_{\mathrm{f}}=[0, \infty)$

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