Question

求函数f（x）=x³-3ax+b（a≠0）的极值

Answer 1

1)由题意,y=f(x)过(2,8),且在该点处切线斜率为0
f(x)=x^3-3ax+b
f(2)=8-6a+b=8①
f'(x)=3x^2-3a
f'(2)=12-3a=0②
解①②得
a=4,b=24
2)f(x)=x^3-12x+24,f'(x)=3x^2-12
令f'(x)>=0,
即3x^2-12>=0,
∴x>=2或x<=-2
令f'(x)<=0,
即3x^2-12<=0,
∴-2<=x<=2
∴f(x)在(-∞,-2]和[2,∞)上单调增,在[-2,2]上单调减
f(x)极大值为f(-2)=-8+24+24=40
f(x)极小值为f(2)=8-24+24=8
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