# How do you find the domain of  f(x)=sqrt((x^2) -9x) ?

May 24, 2017

Domain of $x$ is $\left(- \infty , 0\right]$ and $\left[9 , \infty\right)$

#### Explanation:

Well we know that for $f \left(x\right) \in {\mathbb{R}}^{n}$:
${x}^{2} - 9 x \ge 0$

Since that's the only inequality, solving for that will give the domain.
${x}^{2} \ge 9 x$
$x \ge 9$

And when $x > 0$, ${x}^{2} - 9 x \ge 0$ will always hold true, so:
$x \le 0$

graph{sqrt(x^2-9x) [-28.42, 36.53, -1.3, 31.18]}

So, in interval notation, the domain of $x$ is:

$\left(- \infty , 0\right]$ and $\left[9 , \infty\right)$
And $f \left(x\right) \ge 0$