设函数f(x)=x3-3ax+b(a≠0).(1)若曲线y=f(x)在点(2,f(x))处与直线y=8相切,求a,b的值;(2)求函数f(x)的单调区间与极值点.
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![设函数f(x)=x3-3ax+b(a≠0).(1)若曲线y=f(x)在点(2,f(x))处与直线y=8相切,求a,b的值;(2)求函数f(x)的单调区间与极值点.](/uploads/image/z/11728355-59-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3Dx3-3ax%2Bb%EF%BC%88a%E2%89%A00%EF%BC%89.%EF%BC%881%EF%BC%89%E8%8B%A5%E6%9B%B2%E7%BA%BFy%3Df%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%882%2Cf%EF%BC%88x%EF%BC%89%EF%BC%89%E5%A4%84%E4%B8%8E%E7%9B%B4%E7%BA%BFy%3D8%E7%9B%B8%E5%88%87%2C%E6%B1%82a%2Cb%E7%9A%84%E5%80%BC%EF%BC%9B%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%E4%B8%8E%E6%9E%81%E5%80%BC%E7%82%B9.)
设函数f(x)=x3-3ax+b(a≠0).(1)若曲线y=f(x)在点(2,f(x))处与直线y=8相切,求a,b的值;(2)求函数f(x)的单调区间与极值点.
设函数f(x)=x3-3ax+b(a≠0).(1)若曲线y=f(x)在点(2,f(x))处与直线y=8相切,求a,b的值;(2)求函数f(x)的单调区间与极值点.
设函数f(x)=x3-3ax+b(a≠0).(1)若曲线y=f(x)在点(2,f(x))处与直线y=8相切,求a,b的值;(2)求函数f(x)的单调区间与极值点.
答:
1)
f(x)=x^3-3ax+b
求导:f'(x)=3x^2-3a
x=2时,曲线与直线y=8相切,则:
切线斜率k=f'(2)=12-3a=0
解得:a=4
因为:切点处y=8,切点为(2,8)
所以:f(2)=8-6a+b=8
所以:b=6a=24
解得:a=4,b=24
2)
f(x)=x^3-12x+24
f'(x)=3x^2-12=3(x^2-4)
x2时,f'(x)>0,f(x)单调递增,单调递增区间为(-∞,-2)或者(2,+∞)
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