Question

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

Answer 1

`f(x)` is decreasing when `x in[4/3, 2]`

Explanation:

The domain of `f(x)` is `(2-x)>=0`
`x<=2`

To calculate the derivative, we use

`(sqrtu)'=1/(2sqrtu)`

and `(gh)'=g'h+gh'`

Let's calculate the derivative of `f(x)`

`f'(x)=1*sqrt(2-x)+x*1/(2sqrt(2-x))*-1`

`f'(x)=sqrt(2-x)-x/(2sqrt(2-x))`

`=(2(2-x)-x)/(2sqrt(2-x))=(4-2x-x)/(2sqrt(2-x))`

`f'(x)=(4-3x)/(2sqrt(2-x))`

`f'(x)=0` `=>` `x=4/3`

let's do a sign chart

`color(white)(aaaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaa)` `4/3` `color(white)(aaaaa)` `2`

`color(white)(aaaaa)` `f'(x)` `color(white)(aaaaaa)` `+` `color(white)(aa)` `0` `color(white)(aaa)` `-`

`color(white)(aaaaaa)` `f(x)` `color(white)(aaaaaa)` `uarr` `color(white)(aa)` `0` `color(white)(aaa)` `darr`

Therefore, `f(x)` is decreasing when `x in[4/3, 2]`

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