# How do you find the domain and range for y =sqrt(4x-1)?

May 25, 2015

Assuming we're working with real numbers, $\sqrt{a}$ is only defined for $a \ge 0$, but it is defined for all $a \ge 0$.

So we require $4 x - 1 \ge 0$.

Adding $1$ to both sides, then dividing both sides by $4$ we get:

$x \ge \frac{1}{4}$

So our domain is $\left\{x \in \mathbb{R} : x \ge \frac{1}{4}\right\}$ or $\left[\frac{1}{4} , \infty\right)$ in interval notation.

The range is $\left\{y \in \mathbb{R} : y \ge 0\right\}$ or $\left[0 , \infty\right)$ in interval notation.

To prove this:

Given any $y \ge 0$, let $x = \frac{{y}^{2} + 1}{4}$

Then $4 x - 1 = {y}^{2}$

and since $y \ge 0$ we have $y = \sqrt{4 x - 1}$ as required.