Question

Find the domain and graph, indicate the limits in the graph.

Answer 1

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.

`\begin{gathered} x^2-1>0 \\ x^2>1 \\ √(x^2)>√(1) \\ x>1 \\ x<-1 \end{gathered}`

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:

`(-\infty,-1]\cup[1,\infty)`

In set notation, it is x.

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.

`\begin{gathered} y=√((-1)^2-1)-(-1) \\ y=√(1-1)+1 \\ y=√(0)+1 \\ y=1 \end{gathered}`

At x = -2, y = 3.73

`\begin{gathered} y=√((-2)^2-1)-(-2) \\ y=√(4-1)+2 \\ y=√(3)+2 \\ y=3.73 \end{gathered}`

At x = -3, y = 5.83.

`\begin{gathered} y=√((-3)^2-1)-(-3) \\ y=√(9-1)+3 \\ y=√(8)+3 \\ y=5.83 \end{gathered}`

For the domain x ≥ 1, we can assume x = 1, x = 2, and x = 3.

At x = 1, y = -1.

`\begin{gathered} y=√(1^2-1)-1 \\ y=√(1-1)-1 \\ y=√(0)-1 \\ y=-1 \end{gathered}`

At x = 2, y = -0.268

`\begin{gathered} y=√(2^2-1)-2 \\ y=√(4-1)-2 \\ y=√(3)-2 \\ y=-0.268 \end{gathered}`

At x = 3, y = -0.172.

`\begin{gathered} y=√(3^2-1)-3 \\ y=√(9-1)-3 \\ y=√(8)-3 \\ y=-0.172 \end{gathered}`

Let's plot the 6 coordinates. The graph of this function, with limits indicated, is shown below.