# What is the range of the function f(x)=x^2-1?

Apr 2, 2017

{y|y≥-1}

#### Explanation:

Basically, we need to find the values $y$ can take in $y = {x}^{2} - 1$.

One way to do this is to solve for $x$ in terms of $y$: $x = \pm \sqrt{y + 1}$.

Since $y + 1$ is under the square root sign, it must be the case that y+1≥0. Solving for $y$ here, we get y≥-1.

In other words, the range is {y|y≥-1}.