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

Aug 11, 2018

The domain is $x \in \left[0 , \frac{1}{3}\right]$. The range is $y \in \left[0 , 0.289\right]$

#### Explanation:

The function is

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

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

Therefore,

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

$\implies$, $3 {x}^{2} - x \le 0$

$\implies$, $x \left(3 x - 1\right) \le 0$

The solution to this inequality (obtained with a sign chart) is

$x \in \left[0 , \frac{1}{3}\right]$

The domain is $x \in \left[0 , \frac{1}{3}\right]$

When $x = 0$, $\implies$, $y = 0$

When $x = \frac{1}{3}$, $\implies$, $y = 0$

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

$\implies$, ${y}^{2} = x - 3 {x}^{2}$

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

This is a quadratic equation in $x$ and in order to have solutions, the discriminant $\ge 0$

$\Delta = {\left(- 1\right)}^{2} - 4 \left(3\right) \left({y}^{2}\right) \ge 0$

$1 - 12 {y}^{2} \ge 0$

${y}^{2} \le \frac{1}{12}$

$y \le \pm \sqrt{\frac{1}{12}}$

We keep the positive solution

$y \le \frac{1}{\sqrt{12}} \le 0.289$

The range is $y \in \left[0 , 0.289\right]$

graph{sqrt(x-3x^2) [-0.192, 0.5473, -0.044, 0.3256]}