Consider the function f(x)=xsqrt(36−x^2), −1≤x≤6.
This function has an absolute minimum value equal to what?
f(x) = √(36-x^2) = (36-x^2)^(1/2)
f ' (x) = (1/2)(36-x^2)^(-1/2) (-2x)
= 0 for a max/min
thus x = 0 , f(0) = 6
but f(1) = √(36-1)
= √35 which is < 6
f(6) = √0 = 0
So the absolute minimum is 0