Answer
$\left[0,\dfrac{1}{3}\right]$
Work Step by Step
We are given the function:
$f(x)=\sqrt{x-3x^2}$
The function is defined for the values of $x$ which are solutions of the inequality:
$x-3x^2\geq 0$
Let $g(x)=-3x^2+x$.
Determine the intercepts:
$y$-intercept: $g(0)=-3(0^2)+0=0$
$x$-intercepts: $-3x^2+x=0$
$-x(3x-1)=0$
$x=0$ or $3x-1=0$
$x=0$ or $x=\dfrac{1}{3}$
The $y$-intercept is 0; the $x$-intercepts are 0 and $\dfrac{1}{3}$.
Determine the vertex:
$x_V=-\dfrac{b}{2a}=-\dfrac{1}{2(-3)}=\dfrac{1}{6}$
$y_V=f(x_V)=-3\left(\dfrac{1}{6}\right)^2+\dfrac{1}{3}=\dfrac{5}{18}$
The vertex is $V\left(\dfrac{1}{6},\dfrac{5}{18}\right)$
Graph the function.
The graph is above the $x$-axis for:
$x\in \left[0,\dfrac{1}{3}\right]$
The domain of $f$ is:
$\left[0,\dfrac{1}{3}\right]$
