1)由题意,y=f(x)过(2,8),且在该点处切线斜率为0f(x)=x^3-3ax+bf(2)=8-6a+b=8①f'(x)=3x^2-3af'(2)=12-3a=0②解①②得 a=4,b=242)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&l 展开
1)由题意,y=f(x)过(2,8),且在该点处切线斜率为0f(x)=x^3-3ax+bf(2)=8-6a+b=8①f'(x)=3x^2-3af'(2)=12-3a=0②解①②得 a=4,b=242)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=40f(x)极小值为f(2)=8-24+24=8 收起
