# What is the range of the function f(x)= -sqrt((x^2) -9x) ?

Jul 8, 2017

Range of $f \left(x\right) = \left(- \infty , 0\right]$

#### Explanation:

$f \left(x\right) = - \sqrt{{x}^{2} - 9 x}$

First let's consider the domain of $f \left(x\right)$

$f \left(x\right)$ is defined where ${x}^{2} - 9 x \ge 0$

Hence where $x \le 0$ and $x \ge 9$

$\therefore$ Domain of $f \left(x\right) = \left(- \infty , 0\right] \cup \left[9 , + \infty\right)$

Now consider:

${\lim}_{x \to \pm \infty} f \left(x\right) = - \infty$

Also: $f \left(0\right) = 0$ and $f \left(9\right) = 0$

Hence the range of $f \left(x\right) = \left(- \infty , 0\right]$

This can be seen by the graph of #f(x) below.

graph{-sqrt(x^2-9x) [-21.1, 24.54, -16.05, 6.74]}

Jul 8, 2017

Range : $f \left(x\right) \le 0$, in interval notation : $\left(- \infty , 0\right]$

#### Explanation:

$f \left(x\right) = - \sqrt{{x}^{2} - 9 x}$

Range: Under root should be $\ge 0$ , So $f \left(x\right) \le 0$

Range : $f \left(x\right) \le 0$, in interval notation : $\left(- \infty , 0\right]$

graph{-(x^2-9x)^0.5 [-320, 320, -160, 160]} [Ans]