# Given f(x)=xsqrt(2-x), how do you find the interval where f is decreasing?

Nov 13, 2016

$f \left(x\right)$ is decreasing when $x \in \left[\frac{4}{3} , 2\right]$

#### Explanation:

The domain of $f \left(x\right)$ is $\left(2 - x\right) \ge 0$
$x \le 2$

To calculate the derivative, we use

$\left(\sqrt{u}\right) ' = \frac{1}{2 \sqrt{u}}$

and $\left(g h\right) ' = g ' h + g h '$

Let's calculate the derivative of $f \left(x\right)$

$f ' \left(x\right) = 1 \cdot \sqrt{2 - x} + x \cdot \frac{1}{2 \sqrt{2 - x}} \cdot - 1$

$f ' \left(x\right) = \sqrt{2 - x} - \frac{x}{2 \sqrt{2 - x}}$

$= \frac{2 \left(2 - x\right) - x}{2 \sqrt{2 - x}} = \frac{4 - 2 x - x}{2 \sqrt{2 - x}}$

$f ' \left(x\right) = \frac{4 - 3 x}{2 \sqrt{2 - x}}$

$f ' \left(x\right) = 0$$\implies$$x = \frac{4}{3}$

let's do a sign chart

$\textcolor{w h i t e}{a a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$\frac{4}{3}$$\textcolor{w h i t e}{a a a a a}$$2$

$\textcolor{w h i t e}{a a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$-$

$\textcolor{w h i t e}{a a a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$\uparrow$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a}$$\downarrow$

Therefore, $f \left(x\right)$ is decreasing when $x \in \left[\frac{4}{3} , 2\right]$

graph{xsqrt(2-x) [-4.382, 4.386, -2.19, 2.192]}