Question

Determine the critical points for the function? f(x)=`sqrt(2x-x^2)`

Answer 1

There is one critical point at `(1,1)`

Explanation:

Start by calculating the domain of `f(x)`

Therefore,

`2x-x^2>=0`

`x(2-x)>=0`

To solve this inequality, let's make a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `0` `color(white)(aaaaaaaaa)` `2` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `2-x` `color(white)(aaaaa)` `+` `color(white)(aaaa)` ###color(white)(aaaaa)#`+` `color(white)(aaa)` `0` `color(white)(aaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaaa)` `+` `color(white)(aaa)` `0` `color(white)(aaa)` `-`

Therefore,

`f(x)>=0` when `x in [0, 2]`, this is the domain of `f(x)`

Calculate the first derivative

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

The critical point is when `f'(x)=0`, that is

`2-2x=0`, `=>`, `x=1`

graph{sqrt(2x-x^2) [-3.182, 4.613, -0.887, 3.01]}