Closed Interval Method

[mmm4444bot]

Yes, that's better.

f(1/2) = [4(1/2)^2-4(1/2)]^(1/3) = -1

A graph of function f confirms the absolute min and max, over the restricted domain [-1,2].



Note why the derivative is zero at x=1/2 (the slope of the tangent line there is zero) and why the derivative is undefined at x=0 and x=1 (the tangent lines there are vertical -- that is, undefined slope).

Cheers :cool:

 

[stinajeana]

Yes, that's better.

f(1/2) = [4(1/2)^2-4(1/2)]^(1/3) = -1

A graph of function f confirms the absolute min and max, over the restricted domain [-1,2].

View attachment 2937

Note why the derivative is zero at x=1/2 (the slope of the tangent line there is zero) and why the derivative is undefined at x=0 and x=1 (the tangent lines there are vertical -- that is, undefined slope).

Cheers :cool:

Thank you very very much for helping me understand this!!!!!!! you have no idea how grateful I am