Question

已知函数f(x)=(x^2-4)(x-a)(常数a∈R）（1）求f'(x)(2)若f'(-1)=0,求f(x)在[-2,

Answer 1

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

f'(-1)=3+2a-4=0

a=1/2 f'(x)=3x^2-x-4 f(x)=(x^2-4)(x-1/2)

令f'(x)=0 3x^2-x-4=0 即(x+1)(3x-4)=0

x=-1或x=4/3

易得x=-1为极大值,x=4/3为极小值

f(-1)=9/2

f(-2)=0

f(4)=42

比较得f(x)在[-2,4]的最大值为f(4)=42

不懂可问~