differentiation(2)

It is given that a differentiable function f(x)=x3-2x2+kx-1(where k is a constant) is decreasing on 1/3≦x≦1 and increasing on 1≦x≦2. a) find the value of kf'(x) = 3x^2 - 4x + kf'(1) = 0 since decreasing on 1/3≦x≦1 and increasing on 1≦x≦23 - 4 + k = 0k = 1b) find the turning point(s) of the curve y=f(x)
f'(x) = 3x^2 - 4x + 1
f'(x) = 0 => 3x^2 - 4x + 1 = 0
(3x - 1)(x - 1) = 0
x = 1/3 or 1
f(1/3) = (1/3)^3 - 2(1/3)^2 + (1/3) - 1
= 1/27 - 2/9 + 1/3 - 1
= -5/27 - 2/3
= -23/27
f(1) = 1 - 2 + 1 - 1
= -1
Turning points are
(1,-1) and (1/3,-23/27)


2011-01-17 19:34:31 補充：
點解會代f '(1)?
原因好簡單
decreasing on 1/3≦x≦1 and increasing on 1≦x≦2
指出 1/3 <= x <= 1 果時 f(x) 係向下落緊
1<= x <=2 果時 f(x) 向上升緊
由此可見, 當 x = 1 時, f(x) 處於轉折點
所版 f'(1) = 0

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