Question

已知函数f(x)=(x^2-4)(x-a)的一个极值点为2

（1）求过点P（-4,0）且与函数f（x）的图像相切的直线方程
（2）求证：当x1,x2∈[1,3]时，总有|f(x1)-f(x2)|≤5成立
（3）若当x1,x2∈[m,m+1](m>0)时，总有|f(x1)-f(x2)|≤3成立，求实数m的取值范围

Answer 1

(1) f(x) = x³ - ax² - 4x + 4a
f'(x) = 3x² - 2ax - 4
x = 2, f'(x) = 12 - 4a - 4 = 0, a = 2
f'(x) = 3x² - 4x - 4 = (3x + 2)(x - 2)
f(x) = x³ - 2x² - 4x + 8 = (x + 2)(x - 2)²
P不在f(x)上, 设切点Q(b, (b + 2)(b - 2)²)
f'(b) = (3b + 2)(b - 2)
切线: y - (b + 2)(b - 2)² = (3b + 2)(b - 2)(x - b)
过点P: 0 - (b + 2)(b - 2)² = (3b + 2)(b - 2)(-4 - b)
(b - 2)(b² + 7b + b = 0
(b - 2)(b + 1)(b + 6) = 0
b = 2,Q(2, 0), 切线: y = 0

b = -1, Q(-1, 9), 切线: y = 3x + 12 = 3(x + 4)
b = -6, Q(-6, -256), 切线: y = 128x + 512 = 128(x + 4)

(2) 显然f(2) = 0为极小值点(1 < x < 2: f'(x)>0; 2 < x < 3: f'(x) > 0)
f(1) = 3, f(3) = 5
只须比较(f3), f(2), |f(3) - f(2)| = 5
|f(x1)-f(x2)|≤5

(3)不妨设x1 < x2

(i) m ≥ 2, f'(x) > 0
|f(x2) - f(x1)| = f(x2) - f(x1) ≤ f(m+1) - f(m) = 3m² - m - 5 ≤ 3
3m² - m - 8 ≤ 0
(1 - √97)/6 ≤ m ≤ (1 + √97)/6 < 2

结合前提: 无解

(ii)0 < m ≤ 1, f'(x) < 0
|f(x2) - f(x1)| = f(x1) - f(x2) ≤ f(m) - f(m + 1) = -3m² + m + 5 ≤ 3
3m² - m - 2 ≤ 0
(3m + 2)(m - 1) ≥ 0
m ≥ 1或m ≤ -2/3
结合前提: m= 1

(iii) 1 < m < 2
此时极小值点f(2) = 0在[m, m+1]内
此时只须f(m)≤ 3且f(m+1) ≤ 3
f(m) - 3 = m³ - 2m² - 4m + 5 = (m - 1)[m - (1 - √21)/2][m - (1 + √21)/2] ≤ 0
m ≤ (1 - √21)/2 < 0, 舍去
或1 < m < (1 + √21)/2 (a)
f(m+1) - 3 = m³ + m² - 5m = m[m - (-1 - √21)/2][m - (-1 + √21)/2] ≤ 0
m ≤ (-1 - √21)/2 < 0, 舍去
或0 < m < (-1 + √21)/2 (b)
结合(a)(b): 1 < m <(-1 + √21)/2

结合(i)(ii)(iii):
1 ≤ m <(-1 + √21)/2

Answer 2

f(x)=(x^2-4)(x-a)=x^3-ax^2-4x+4a

f'(x)=3x^2-2ax-4
一个极值点为2 => 12-4a-4=0 => a=2

f(x)=(x^2-4)(x-2)=x^3-2x^2-4x+8
f'(x)=3x^2-4x-4
（1）f'(-4)=60,所以设切线方程为y=60x+b,过(-4,0),解得b=240,所以y=60x+240
（2）f'(x)=3x^2-4x-4=0, x=2或x=-2/3后者不在[1,3]内
所以求得f(1),f(2),f(3)，函数在此区间的最大值、最小值就在其中
f(1)=3 f(2)=0 f(3)=5,所以最大值为5，最小值为0。所以在[1,3]区间内|f(x1)-f(x2)|≤5恒成立
（3）同上题，求出f(m)=m^3-2m^2-4m+8

f(m+1)=(m+1)^3-2(m+1)^2-4(m+1)+8

然后要讨论m，m+1，与2的大小关系，分段来求出m范围。过程略

Answer 3

首先因式分解及标出各段单调性，根据极值求a,