Question

What is the domain of the function$f\left(x\right)=\sqrt{x-3x^2}$?

Short answer

The domain of the function is$0\le x\le \frac{1}{3}$.

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=\sqrt{x-3x^2}$.

Step 2. Find the Intercepts and Vertex

• The square root function $\sqrt{x-3x^2}$is defined when $x-3x^2\ge 0$.
• Consider $g\left(x\right)=x-3x^2$and factor the expression.

$g\left(x\right)=x(1-3x)$

• $g\left(x\right)=0$implies $x=0$or $(1-3x)=0$.
• So, the x -intercepts of the curve are $(0,0),\left(\frac{1}{3},0\right)$.
• The value of $g\left(0\right)=0$, so, the y-intercept is $(0,0)$.
• Find the vertex of the curve as follows:

$\begin{array}{rcl}x& =& \frac{-b}{2a}\\ & =& \frac{-1}{-6}\\ & =& \frac{1}{6}\end{array}$

• So, the vertex is also at$\left(\frac{1}{6},\frac{1}{12}\right)$.

Step 3. Plot the function 

• Plot the graph of the function by using the obtained intercepts and vertex.

Step 4. Find the region above x-axis 

• From the graph, the region on and above the horizontal axis is $0\le x\le \frac{1}{3}$.
• The inequality is of the greater than or equal to type, so the solution set is $0\le x\le \frac{1}{3}$.