# Locate all critical points (both types) of h(x)= `sqrt(-10x-x^2+8x)`

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### 2 Answers

The critical points of `h(x) = sqrt(-10x - x^2 + 8x)` have to be determined.

At the critical points `h'(x) = 0`

`h'(x) = (-2-2x)/(2*sqrt(-10x - x^2 + 8x))`

= `(-1-x)/sqrt(-10x - x^2 + 8x)`

`(-1-x)/sqrt(-10x - x^2 + 8x) = 0`

=> x = -1

At x = -1, h(x) = `sqrt(2 - 1)` = `+-1`

**The critical points are (-1, 1) and (-1, -1)**

`h(x)=sqrt(-10x-x^2+8x)` `=sqrt(-x^2-2x)`

first step is to find domain of definition

`h(x)=sqrt(-(x^2+2x))`

So h(x) is defined when `x^2+2x` has values less equal than zero.

Since `x^2+2x=x(x+2)` thus has negative-zero values for

`-2<= x <= 0` thus h(x) is defined in `-2<= x <=0`

(since square has taken always positive do'nt need search for sign).

`h('x)=-(x+1)/sqrt(-(x^2+2x))=`

`h'(x)=0` `rArr` `x=-1`

What kind of ?

We need to find second derivative:

`h''(x)=-1/((-(x^2+2x))^(3/2))<0`

Thus `x=-1` is Max point with value 1.

From the graph we can se it's an half circonference with radius 1. in center C(-1;0)

Indeed:

`(x+1)^2 +y^2=1`

`x^2+2x+1+y^2=1`

`x^2+2x +y^2=0`

`y^2=-x(x+2)`

`y=sqrt(-x(x+2))`

that is h(x).