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

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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)

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oldnick | (Level 1) Valedictorian

Posted on

`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).      

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