# How do you find the domain and range of f(x)= 2x^2-1?

Aug 14, 2015

Domain: $\left(- \infty , + \infty\right)$
Range: $\left[- 1 , + \infty\right)$

#### Explanation:

Your function is defined for any value of $x$, so you have no restrictions wehn it comes to its domain, which will be $x \in \mathbb{R}$, or $\left(- \infty , + \infty\right)$.

In order to determine the function's range, focus on the fact that you're dealing with the square of a value $x$. As you know, for real numbers, the square of any number will be positive.

This means that the minimum value this function can take will occur at $x = 0$

$f \left(0\right) = 2 \cdot {0}^{2} - 1 = - 1$

For any value of $x \ne 0$, $f \left(x\right) > f \left(0\right)$. This means that the function's range will be $\left[- 1 , + \infty\right)$.

graph{2x^2-1 [-10, 10, -5, 5]}