What is the domain of h(x)=sqrt(( x- (3x^2)))?

Sep 9, 2015

Domain: $\left(0 , \frac{1}{3}\right)$

Explanation:

Right from the start, you know that the domain of the function must only include values of $x$ that will make the expression under the square root positive.

In other words, you need to exclude from the function's domain any value of $x$ will result in

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

The expression under the square root can be factored to give

$x - 3 {x}^{2} = x \cdot \left(1 - 3 x\right)$

Make this expression equal to zero to find the values of $x$ that make it negative.

$x \cdot \left(1 - 3 x\right) = 0 \implies \left\{\begin{matrix}x = 0 \\ x = \frac{1}{3}\end{matrix}\right.$

So, in order for this expression to be positive, you need to have
$x > 0$ and $\left(1 - 3 x\right) > 0$, or $x < 0$ and $\left(1 - 3 x\right) < 0$.

Now, for $x < 0$, you have

$\left\{\begin{matrix}x < 0 \\ 1 - 3 x > 0\end{matrix}\right. \implies x \cdot \left(1 - 3 x\right) < 0$

Likewise, for $x > \frac{1}{3}$, you have

$\left\{\begin{matrix}x > 0 \\ 1 - 3 x > 0\end{matrix}\right. \implies x \cdot \left(1 - 3 x\right) < 0$

This means that the only values of $x$ that will make that expression positive can be found in the interval $x \in \left(0 , \frac{1}{3}\right)$.

Any other value of $x$ will cause the expression under the square root to be negative. The domain of the function will thus be $x \in \left(0 , \frac{1}{3}\right)$.

graph{sqrt(x-3x^2) [-0.466, 0.866, -0.289, 0.377]}