Differentiation

Let y = x4 - 2x3 + 2x, then
dy/dx = 4x3 - 6x2 + 2
= 2(x - 1)2(2x + 1)
So the stationary values of x4 - 2x3 + 2x are x = 1 and x = -1/2
When x < 1 for a little, dy/dx > 0 and when x > 1 for a little, dy/dx > 0
Therefore x = 1 is a resting point.
When x < -1/2, dy/dx < 0 and when x > -1/2, dy/dx > 0
Therefore x = -1/2 is a minimum point.

Let f(x) = x4 - 2x3 + 2x

f'(x) = 4x3 - 6x2 + 2

f"(x) = 12x2 - 12x

For stationary points, set f'(x) = 0

4x3 - 6x2 + 2 = 0

(x - 1)(4x2 - 2x - 2) = 0

(x - 1)(2x2 - x - 1) = 0

(x - 1)2(2x + 1) = 0

x = 1 or -1/2

2009-07-31 15:03:25 補充：
When x = 1, f"(x) = 12(1)2 - 12(1) = 0

y = (1)4 - 2(1)3 + 2(1) = 1

When x = -1/2, f"(x) = 12(-1/2)2 - 12(-1/2) = 9

y = (-1/2)4 - 2(-1/2)3 + 2(-1/2) = -11/16

So, the stationary points are (1 , 1) and (-1/2 , -11/16).

(1 , 1) is the inflexion point, and (-1/2 , -11/16) is the minimum point.