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

Oct 8, 2017

Domain: $x \left[0 , \frac{1}{3}\right]$
Range: $y \ge 0$

#### Explanation:

First, we'll find the range. (it's easier in this problem)

Since the square root radical is already present, we can say right out of the gate that the range is $y \ge 0$. This is because the square root will only output numbers that are $\ge 0$.

Next, we'll find the domain.

Since you can't take the square root of a negative number, (over the set of real numbers) we'll see what values of $x$ make $x - \left(3 {x}^{2}\right) \ge 0$

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

Just to make it prettier, we'll multiply both sides by negative 1. It makes the coefficient of the squared term positive. Because we are multiplying by a negative, we'll also have to flip the inequality.

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

We factor out an $x$.

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

However, now we have to use critical number analysis. We'll test various values of $x$ to see if they make the inequality true.

To find critical numbers, we just use the values of $x$ that will make the multipled things equal to zero. If $x$ is less than a critical number, then that multiplied thing will be negative. If it is higher, then it will be positive.

Since we want the number to be less than or equal to zero, we want only one of the two multiplied things ($x$ and $3 x - 1$) to be negative.

$x = 0 \text{ when } x = 0$

$3 x - 1 = 0 \text{ when } x = \frac{1}{3}$

To start, we'll test a value of $x$ less than both of them.

When $x = - 1$, both $x$ and $3 x - 1$ are negative. A negative times a negative is a positive, meaning that we have a number which is not less than or equal to zero. Therefore, $x$ cannot be less than 0.

Next, we'll try a value of $x$ between our critical numbers, $x = .1$

$x = .1$ makes $x$ positive.

$x = .1$ means that $3 \times .1 - 1$ is negative. We have one positive number and one negative number. We have found some acceptable values of $x$ in the middle of our critical numbers.

Lastly, we have to try a value greater than both critical numbers, $x = 1$

$x = 1$ makes $x$ positive.

$x = 1$ makes $3 \times 1 - 1$ positive, too.

That means that our output will be positive, and therefore greater than zero. It does not work as a solution.

$x$ can therefore be greater than or equal to zero, as long as it's less than or equal to 1/3. That's our domain.