if f(x) has relative extrema, then they occur at critical values of x.
critical values occur where f'(x) = 0 or where f'(x) is undefined
you forgot one of the critical values ...
Find the x-values of all points where the functions defined as follows have any relative extrema. Find the value(s) of any relative extrema.
f(x)= 3x^(5/3) - 15x^(2/3)
I did diff(3*x^(5/3)-15*x^(2/3), x) = 5*x^(2/3)-10/x^(1/3)
solve(5*x^(2/3)-10/x^(1/3) = 0)
x = 2
y = 3*(2)^(5/3) - 15*(2)^(2/3) = - 14.28660947
Vertex = (2, - 14.28660947)
f''(x) = 10/(3x^(1/3)) + 10/(3x^(4/3))
f''(2) = 10/(3(2)^(1/3)) + 10/(3(2)^(4/3))
˜ 3.97
So f(2)= -14.286 is my relative minimum[COLOR=rgb(0, 0, 0)]?[/COLOR]