Find the domain and range of the following real valued functions:… - Vidyadip

Question

Find the domain and range of the following real valued functions:
$\text{f(x)}=\sqrt{\text{x}^2-16}$

✓

Answer

Given,
$\text{f(x)}=\sqrt{\text{x}^2-16}$
$(\text{x}^2-16)\geq0$
$\Rightarrow\text{x}^2\geq16$
$\Rightarrow\text{x}\in(-\infty,-4)\cup[4,\infty)$
$\sqrt{\text{x}^2-16}$ is defined for all real numbers that are greater than or equal to 4 and less than or equal to -4.
Thus, domain of f(x) is $\{\text{x}:\text{x}\leq-4\text{ or x}\geq4\}$ or $(-\infty,-4]\cup[4,\infty)$
Range of f,
For $\text{x}\geq4,$ we have,
$\text{x}^2-16\geq0$
$\Rightarrow\sqrt{\text{x}^2-16}\geq0$
$\Rightarrow\text{f(x)}\geq0$
For $\text{x}\geq-4,$ we have
$\text{x}^2-16\geq0$
$\Rightarrow\sqrt{\text{x}^2-16}\geq0$
$\Rightarrow\text{f(x)}\geq0$
Thus, f(x) takes all real values greater than zero.
Hence, range $(\text{f})=[0,\infty)$

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