The domain of a function is the set of all x-values the function can take.
Since we have a radical symbol, (x² - 1) must not be equal to a negative number because it will make the function undefined.
For the function (x² - 1), the value of x must be greater than 1 or less than -1 in order for the function to be defined.
[tex]\begin{gathered} x^2-1>0 \\ x^2>1 \\ \sqrt{x^2}>\sqrt{1} \\ x>1 \\ x<-1 \end{gathered}[/tex]
If for instance, the value of x is 0.5, the value of (x² - 1) will be -0.75 and the square root of a negative number is undefined. Therefore, we must not have an x-value that is less than 1 or greater than -1.
Hence, the domain of this function in interval notation is:
[tex](-\infty,-1]\cup[1,\infty)[/tex]
In set notation, it is {x|x ≤ -1, x ≥ 1}.
In order to graph the function, let's assume some values of x that are found in the domain.
For the domain, x ≤ -1, we can assume x = -1, x = -2, and x = -3.
At x = -1, y = 1.
[tex]\begin{gathered} y=\sqrt{(-1)^2-1}-(-1) \\ y=\sqrt{1-1}+1 \\ y=\sqrt{0}+1 \\ y=1 \end{gathered}[/tex]
At x = -2, y = 3.73
[tex]\begin{gathered} y=\sqrt{(-2)^2-1}-(-2) \\ y=\sqrt{4-1}+2 \\ y=\sqrt{3}+2 \\ y=3.73 \end{gathered}[/tex]
At x = -3, y = 5.83.
[tex]\begin{gathered} y=\sqrt{(-3)^2-1}-(-3) \\ y=\sqrt{9-1}+3 \\ y=\sqrt{8}+3 \\ y=5.83 \end{gathered}[/tex]
For the domain x ≥ 1, we can assume x = 1, x = 2, and x = 3.
At x = 1, y = -1.
[tex]\begin{gathered} y=\sqrt{1^2-1}-1 \\ y=\sqrt{1-1}-1 \\ y=\sqrt{0}-1 \\ y=-1 \end{gathered}[/tex]
At x = 2, y = -0.268
[tex]\begin{gathered} y=\sqrt{2^2-1}-2 \\ y=\sqrt{4-1}-2 \\ y=\sqrt{3}-2 \\ y=-0.268 \end{gathered}[/tex]
At x = 3, y = -0.172.
[tex]\begin{gathered} y=\sqrt{3^2-1}-3 \\ y=\sqrt{9-1}-3 \\ y=\sqrt{8}-3 \\ y=-0.172 \end{gathered}[/tex]
Let's plot the 6 coordinates. The graph of this function, with limits indicated, is shown below.
