# How do you find the domain and range of sqrt( x- (3x^2))?

Sep 7, 2017

Given: $y = \sqrt{x - 3 {x}^{2}}$

The argument of the square root must be greater than or equal to 0:

$x - 3 {x}^{2} \ge 0$

Becuase this quadratic is the equation of a parabola that opens downward we know that, if we find the roots of the quadratic, the quadratic will be greater than 0 between the roots.

$x - 3 {x}^{2} = 0$

$x \left(1 - 3 x\right) = 0$

$x = 0 \mathmr{and} x = \frac{1}{3}$

This gives us the domain $0 \le x \le \frac{1}{3}$

We know that the minimum for the range occurs at either root:

$0 \le y$

We know that the maximum for the range will occur halfway between the roots, $x = \frac{1}{6}$

$y = \sqrt{\frac{1}{6} - 3 {\left(\frac{1}{6}\right)}^{2}}$

$y = \sqrt{\frac{6}{36} - \frac{3}{36}}$

$y = \frac{\sqrt{3}}{6}$

This makes the range become $0 \le y \le \frac{\sqrt{3}}{6}$