# What is the domain and range of f(x) =sqrt(( x- (3x^2)))?

Jun 9, 2018

Domain $\left\{x | x \in \mathbb{R} : 0 \le x \le \frac{1}{3}\right\}$

Range $\left\{y | y \in \mathbb{R} : 0 \le y \le \frac{\sqrt{3}}{6}\right\}$

#### Explanation:

$f \left(x\right) = \sqrt{\left(x - \left(3 {x}^{2}\right)\right)}$

Numbers under a radical must be greater than or equal to 0 or they are imaginary, so to solve the domain:

$x - \left(3 {x}^{2}\right) \ge 0$

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

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

$x \ge 0$

$1 - 3 x \ge 0$

$- 3 x \ge - 1$

$x \le \frac{1}{3}$

So our domain is:

$\left\{x | x \in \mathbb{R} : 0 \le x \le \frac{1}{3}\right\}$

Since the minimum input is $\sqrt{0} = 0$ the minimum in our range is 0.

To find the maximum we need to find the max of $- 3 {x}^{2} + x$

in the form $a {x}^{2} + b x + c$

$a o s = \frac{- b}{2 a} = \frac{- 1}{2 \cdot - 3} = \frac{1}{6}$

vertex (max) = $\left(a o s , f \left(a o s\right)\right)$

vertex (max) = $\left(\frac{1}{6} , f \left(\frac{1}{6}\right)\right)$

$f \left(x\right) = - 3 {x}^{2} + x$

$f \left(\frac{1}{6}\right) = - 3 {\left(\frac{1}{6}\right)}^{2} + \frac{1}{6} = \frac{1}{12}$

vertex (max) = $\left(\frac{1}{6} , \frac{1}{12}\right)$

Finally, don't forget the square root, we have a maximum at $x = \frac{1}{6}$ of $\sqrt{\frac{1}{12}} = \frac{\sqrt{3}}{6}$ so our range is:

$\left\{y | y \in \mathbb{R} : 0 \le y \le \frac{\sqrt{3}}{6}\right\}$