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

Jul 23, 2017

The domain is $x \in \left[\frac{1}{2} , + \infty\right)$
The range is $y \in \left[0 , + \infty\right)$

#### Explanation:

Our function is

$y = f \left(x\right) = \sqrt{2 x - 1}$

What's under the square root sign is $\ge 0$

Therefore,

$2 x - 1 \ge 0$

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

The domain is $x \in \left[\frac{1}{2} , + \infty\right)$

The minimum value of $y = 0$ and the maximum value is $+ \infty$

The range is $y \in \left[0 , + \infty\right)$

graph{sqrt(2x-1) [-6.244, 6.243, -3.12, 3.123]}