Question: find the maximum and minimum values of the function, cube root of (x^3-9x).
f = (x^3 - 9x)^(1/3)
f' = df/dx = (1/3)(x^3 - 9x)^(-2/3)*(3x^2 - 9)
To find turning points, set f' = 0.
(1/3)(x^3 - 9x)^(-2/3)*(3x^2 - 9) = 0
(x^3 - 9x)^(-2/3)*(3x^2 - 9) = 0
(3x^2 - 9) / (x^3 - 9x)^(2/3) = 0
i.e. 3x^2 - 9 = 0
x^2 = 3
x = +/- sqrt(3)
To see which of these values for x gives a maximum or minimum, get f'' and see whether it is positive or negarive at x = =/- sqrt(3).
f'' = -(2/9)*(3*x^2-9)^2/(x^3-9*x)^(5/3)+2*x/(x^3-9*x)^(2/3)
at x = -sqrt(3), f'' = -(1/3)*6^(1/3)*3^(1/6)
Since f'' is negative, then f(x) is at a maximum at x = -sqrt(3),
at x = sqrt(3), f'' = 2*sqrt(3)/(6*sqrt(3))^(2/3)
Since f'' is negative, then f(x) is at a minimum at x = sqrt(3),
f_min = f(sqrt(3)) = -6^(1/3)*3^(1/6)
f_max = f(-sqrt(3)) = 6^(1/3)*3^(1/6)