Help with differentiation - Determine coordinates of local extreme

Question

Need help with a question on differentiation.

 

*Determine the exact coordinates of the local extreme(s) of the following function. Also indicate whether it concerns a local minimum or a local maximum

 

k(x) = "the square root of 25-x^2

 

(I couldnt find the square root symbol :) )

 

-2 x/2 squareroot of 25-x^2 = -x/squreroot of 25-x^2 = 0

 

x=0

 

k(0) = square root of 25 = 5

 

Could someone explain this?

Thank you so much.

Answer 1

k(x) = (25-x²)^½

 

k'(x) = ½(25-x²)^-½(-2x)

 

        = -x/√(25-x²)

 

k'(x) = 0 when -x=0

 

So, the only critical point occurs when x = 0

 

When -5<x<0, k'(x) > 0.  So, k(x) is increasing on -5<x<0

 

When 0<x<5, k'(x)<0.  So, k(x) is decreasing on 0<x<5.

 

Therefore, there is a local (and absolute) maximum when x = 0.

 

NOTE:  The graph of y = √(25-x²) is the upper half of the circle with center (0,0) and radius 5.  The highest point on the graph occurs at (0,5).  The absolute minima occur at (5,0) and (-5,0).