设函数f(x)=x3-3ax+b(a≠0)1.若函数f(x)在点(2,f(2))处与直线y=8相切,求a,b的值;2.求函数f(x)的单调区间.
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 03:48:50
![设函数f(x)=x3-3ax+b(a≠0)1.若函数f(x)在点(2,f(2))处与直线y=8相切,求a,b的值;2.求函数f(x)的单调区间.](/uploads/image/z/10478037-21-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dx3-3ax%2Bb%28a%E2%89%A00%291.%E8%8B%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9%EF%BC%882%2Cf%EF%BC%882%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%9B2.%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4.)
设函数f(x)=x3-3ax+b(a≠0)1.若函数f(x)在点(2,f(2))处与直线y=8相切,求a,b的值;2.求函数f(x)的单调区间.
设函数f(x)=x3-3ax+b(a≠0)
1.若函数f(x)在点(2,f(2))处与直线y=8相切,求a,b的值;
2.求函数f(x)的单调区间.
设函数f(x)=x3-3ax+b(a≠0)1.若函数f(x)在点(2,f(2))处与直线y=8相切,求a,b的值;2.求函数f(x)的单调区间.
f(x)=x^3-3ax+b
f'(x)=3x^2-3a ,12-3a=0 ,a=4
8=8-24+b ,b=24
f'(x)=3x^2-12=3(x^2-4)=3(x+2)(x-2)=0 ,
x=-2 ,x=2
x
f(x)=x^3-3ax+bf'(x)=3x^2-3a ,12-3a=0 ,a=48=8-24+b , b=24f'(x)=3x^2-12=3(x^2-4)=3(x+2)(x-2)=0 ,x=-2 ,x=2x<-2 ,f'(x)>0递增-2
f(x)=x^3-3ax+bf'(x)=3x^2-3a ,12-3a=0 ,a=48=8-24+b , b=24f'(x)=3x^2-12=3(x^2-4)=3(x+2)(x-2)=0 ,x=-2 ,x=2x<-2 ,f'(x)>0递增-2