Locate all critical points (both types) of h(x)= `sqrt(-10x-x^2+8x)`
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)
first step is to find domain of definition
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)=0` `rArr` `x=-1`
What kind of ?
We need to find second derivative:
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)
that is h(x).