Question

What is the domain and range of $y=\sqrt{{{x}^{2}}-1}$?

Answer 1

First of all find the domain of the function by considering the defined values of x. Use the fact that the term inside the square root must be greater than or equal to 0 and solve the inequality by using the formula if ${{x}^{2}}-{{a}^{2}}\ge 0$ then $x\le -a$ or $x\ge a$ to get the domain values. Now, use the set of domain values obtained to get the set of values of y which is the range.

Complete step-by-step solution:
Here we have been provided with the function $y=\sqrt{{{x}^{2}}-1}$ and we are asked to determine its domain and range. First let us see the definitions of the domain and range of a function one by one.
(1) Domain: - Domain of a function is defined as the set of values of x for which the function $y=f\left( x \right)$ is defined. As we can see that in the given function $y=\sqrt{{{x}^{2}}-1}$ we have the square root sign and for any term to be defined inside the square root sign it must be non – negative. The value can be greater than or equal to 0, so mathematically we have the inequality,
$\begin{aligned} & \Rightarrow {{x}^{2}}-1\ge 0 \\\ & \Rightarrow {{x}^{2}}-{{1}^{2}}\ge 0 \\\ \end{aligned}$
Using the formula if ${{x}^{2}}-{{a}^{2}}\ge 0$ then $x\le -a$ or $x\ge a$ we get,
$\Rightarrow x\le -1$ or $x\ge 1$
$\therefore x\in \left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$
Therefore, the above set of values of x represents the domain of the function.
(2) Range: - Range of a function $y=f\left( x \right)$ is defined as the set of all the possible values of y for the possible values of x. Now, we have the relation ${{x}^{2}}-1\ge 0$, so taking square root both the sides we get,
$\begin{aligned} & \Rightarrow \sqrt{{{x}^{2}}-1}\ge \sqrt{0} \\\ & \Rightarrow y\ge 0 \\\ & \therefore y\in \left[ 0,\infty \right) \\\ \end{aligned}$
Therefore, the above set of values of y represents the range of the function.

Note: Note that if you will square both the sides of the given function and then consider the range then you will get the inequality ${{y}^{2}}\ge 0$ for which you may think that y will be any real value but this is where mistakes happen because the original equal is $y=\sqrt{{{x}^{2}}-1}$ and not ${{y}^{2}}={{x}^{2}}-1$. Do not use square brackets in place of a small bracket towards the infinity sign because infinity is not a real number so we aren’t including it in the set. Square bracket is used when we include the boundary point value.