How do you find domain and range for f(x) =sqrt( x- (3x^2))?

Refer to explanation

Explanation:

We have that for the domain is

$x - 3 {x}^{2} \ge 0 \implies x \cdot \left(1 - 3 x\right) \ge 0$ which is true for $0 \le x \le \frac{1}{3}$

Hence the domain is $\left[0 , \frac{1}{3}\right]$

For the range we set $y = f \left(x\right) \ge 0$ so

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

The last is a trinomial with respect to x so the discriminant must be greater or equal to zero hence

${\left(- 1\right)}^{2} - 4 \cdot 3 \cdot \left({y}^{2}\right) \ge 0 \implies \frac{1}{12} \ge {y}^{2} \implies \frac{1}{2 \sqrt{3}} \ge y$

Hence the range is $R \left(f\right) = \left[0 , \frac{1}{2 \sqrt{3}}\right]$