f(x)=2x^3-3ax^2+2(x∈R,a＞0）（1）若f(x)在点(1,f(1)）处的切线与Y=-1/3x+1垂直,求f(x)的单调区间.
问题描述：
f(x)=2x^3-3ax^2+2(x∈R,a＞0）（1）若f(x)在点(1,f(1)）处的切线与Y=-1/3x+1垂直,求f(x)的单调区间.
（2）试求f(x)在[0,2]上的最大值.
答
(1) f'(x) = 6x^2 - 6ax f'(1) = 6 -6a y = -x/3 + 1的斜率k = -1/3f(x)在点(1,f(1)）处的切线与y= -x/3 +1垂直,此切线的斜率= -(-1/3) = 3f'(1) = 6 -6a = 3a = 1/2f(x) = 2x^3 -3x^2/2 +2f'(x) = 6x^2 - 3x = 3x(2...第一问会，主要是第二问，我算的是两种情况，a＞4/3时，最大值是f(2)=18-2a.和a∈（0，4/3]时，最大值是f（0）=2,对么？(2) f'(x) = 6x^2 - 6ax= 6x(x - a) = 0x = 0, x = a (a > 0)x 0, f(x)递增0 0, x - a 0, f(x)递减x > a: x > 0: x - a > 0, x(x - a) > 0, f(x)递增 a > 2:f(x)在(0,2)上递减, 最大值是f(0) = 20 f(2): 2 > 18 - 12a, a > 4/3即4/3