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

Aug 20, 2017

See below.

#### Explanation:

The value under a radical cannot be negative, or else the solutions will be complex.

So,

$2 {x}^{2} - 1 \setminus \ge q 0$

$2 {x}^{2} \ge q 1$

${x}^{2} \ge q \frac{1}{2}$

$x \ge q \setminus \frac{\sqrt{2}}{2}$ and $x \le q - \setminus \frac{\sqrt{2}}{2}$. In interval notation, this is $\left(- \infty , - \setminus \frac{\sqrt{2}}{2}\right] \cup \left[\setminus \frac{\sqrt{2}}{2} , \infty\right)$

Since the radical value will always be greater than or equal to zero, the range is just $\left[0 , \infty\right)$.