(1/2)设函数f(x)=x3-3ax+b,若曲线y=f(x)在点(2,f(2))处与直线y=8相切.求a.b的值?求函

4个回答

  • f(x)=x3-3ax+b

    y'=3x^2-3a

    曲线y=f(x)在点(2,f(2))处与直线y=8相切

    故y'(2)=0 f(2)=8

    即3x4-3a =0 a=4

    2^3-6a+b=8 b=24

    即y'=3x^2-3a =3x^2-12=3(x+2)(x-2)

    当y'>0时,f(x)为增函数,即此时x∈(-无穷,-2)或x∈(2,+无穷)

    当y'<0时,f(x)为减函数,即此时x∈(-2,2)

    当x∈[-3,3]时,

    即x∈[-3,-2]为增函数,f(x)极大=f(-2)=40

    x∈[-2,2]为减函数,f(x)极小=f(2)=8

    又f(-3)=33

    f(3)=15

    故在当x∈[-3,3]时,f(x)max=f(-2)=40

    f(x)min=f(2)=8