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

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Answer 1 · 2016-11-13T13:31:59

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

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]}

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