# How do you find the range of f(x) =x^2-1 for the domain D={-2,-1,0,2}?

Apr 11, 2017

$R = \left\{- 1 , 0 , 3\right\}$

#### Explanation:

The range is the set of values of $f \left(x\right)$ you can get from your domain. So given $f \left(x\right) = {x}^{2} - 1$ and domain $D = \left\{- 2 , - 1 , 0 , 2\right\}$, all you have to do is plug in the elements of your domain into your function.

$\textcolor{red}{x = - 2}$

$\left[1\right] \text{ } f \left(- 2\right) = {\left(- 2\right)}^{2} - 1$

$\left[2\right] \text{ } f \left(- 2\right) = 4 - 1$

$\left[3\right] \text{ } \textcolor{red}{f \left(- 2\right) = 3}$

$\textcolor{b l u e}{x = - 1}$

$\left[1\right] \text{ } f \left(- 1\right) = {\left(- 1\right)}^{2} - 1$

$\left[2\right] \text{ } f \left(- 1\right) = 1 - 1$

$\left[3\right] \text{ } \textcolor{b l u e}{f \left(- 1\right) = 0}$

$\textcolor{g r e e n}{x = 0}$

$\left[1\right] \text{ } f \left(0\right) = {\left(0\right)}^{2} - 1$

$\left[2\right] \text{ } f \left(0\right) = 0 - 1$

$\left[3\right] \text{ } \textcolor{g r e e n}{f \left(0\right) = - 1}$

$\textcolor{\mathmr{and} a n \ge}{x = 2}$

$\left[1\right] \text{ } f \left(2\right) = {\left(2\right)}^{2} - 1$

$\left[2\right] \text{ } f \left(2\right) = 4 - 1$

$\left[3\right] \text{ } \textcolor{\mathmr{and} a n \ge}{f \left(2\right) = 3}$

Now that you have solved for all the possible values of $f \left(x\right)$, your range is:

$R = \left\{- 1 , 0 , 3\right\}$