设f(x)=根号下(x—3)+根号下(6—x),若不等式f(x)

4个回答

  • 柯西不等式:∑(i=1,n)|ai×∑(i=1,n)|bi≥∑(i=1,n)|(aibi),即:

    (a1²+a2²+a3²+···+an²)(b1²+b2²+b3²+···+bn²)≥(a1b1+a2b2+a3b3+···+anbn)²

    此题不能直接或主要运用柯柯西不等式!

    f(x)=√(x-3)+√(6-x)

    先确定f(x)的定义域:

    x-3≥0,6-x≥;解得3≤x≤6.

    f(x)=√(x-3)+√(6-x)

    =√(√(x-3)+√(6-x))²

    =√(√(x-3)²+√(6-x)²+2√((x-3)(6-x)))

    =√((x-3+6-x)+2√(-x²+9x-18))

    =√(3+2√((9/4)-(x-(9/2))²))

    ∵3≤x≤6

    ∴0≤√((9/4)-(x-(9/2))²)≤√(9/4)=3/2

    ∴√3≤√(3+2√((9/4)-(x-(9/2))²))≤√(3+2×(3/2))=√6

    ∴√3≤f(x)≤√6

    当x=3或6时,f(x)min=√3;当x=9/2时,f(x)max=√6.

    ∴要f(x)√3即可!