Question

设a属于R 函数f(x)=ax^3-3x^2 ,x=2是函数y=f(x)的极值点(1)求a

2)若函数g(x)=f(x)+f'(x)x属于0到2闭区间x=0处取最大值求a得取值范围

Answer 1

(1) f'(x) = 3ax²-6x = 3x(ax - 2) = 0, 显然x = 2为ax - 2 = 0的解, 2a - 2 = 0, a = 1

(2) g(x) = f(x) + f'(x) = ax³ - 3x² + 3ax²-6x = ax³ + 3(a - 1)x²- 6x
g'(x) = 3ax² + 6(a -1)x - 6
在[0, 2]内，x = 0处g(x)取最大值，则g'(x)在[0, 2]内<0
g'(x)为抛物线
(i) a > 0 抛物线开口向上
[0, 2]在抛物线与x轴的两个交点之间(包括过(0, 0), (2, 0))
此时须下列2个条件同时成立:
(a) g'(0)≤0, g'(0) = -6 < 0总成立
(b) g'(2)≤0, g'(2) = 6(4a - 3) ≤0, a ≤ 3/4
结合前提a > 0, 得0 < a ≤ 3/4

(ii) a < 0, 抛物线开口向下
g'(x) = 3ax² + 6(a -1)x - 6的对称轴为x = (1 - a)/a
1-a>0, a < 0, x = (1 - a)/a < 0
对称轴在y轴左侧
在[0, 2]内, g(x)在x=0处取最大值, 须下列2个条件同时成立:
(a) g'(0)≤0, g'(0) = -6 < 0总成立
(b) g'(2)≤0, g'(2) = 6(4a - 3) ≤0, a ≤ 3/4
结合前提a < 0, 得a < 0

(iii) a = 0
g(x) = -3x² - 6x = -3x(x + 2)
抛物线开口向下, 对称轴x = -1
g(0) = 0, 在[0, 2]内, g(x)在x=0处取最大值

结合(i)(ii)(iii): a ≤ 3/4