1、求函数y=根号x^2+6x+25 +根号x^2-10x+29的值域

3个回答

  • 1

    y=√(x^2+6x+25) + √(x^2-10x+29)

    =√[(x+3)^2+4^2]+ √[(x-5)^2+4]

    A(-3,4) B(5,-2)

    AB直线:y+2=[(-2-4)/(5+3)] (x-5)

    y=0,x=-8/3+5=7/3

    x=7/3

    y最小=√[(7/3+3)^2+16] + √[(7/3-5)^2+4] =(4/3)*5+(1/3)*10=10

    y≥10

    2

    y=√(x^2+6x+25)-√(x^2-10x+29)

    =√[(x+3)^2+16 - √[(x-5)^2+4]

    A1(-3,4) B1(5,2)

    直线A1B1:y-2=[(2-4)/(5+3)] (x-5)

    y=0,x=(-2)*(-4)+5=13

    x=13时,y最大=√[(13+3)^2+16] - √[(13-5)^2+4]

    =4√17 - √68

    =2√17

    y=√(x^2+6x+25)-√(x^2-10x+29)

    =[(x^2+6x+25)-(x^2-10x+29)] / [√(x^2+6x+25)+√(x^2-10x+29)

    lim(x→-∞)√(x^2+6x+25)-√(x^2-10x+29)

    =lim(x→-∞)(16x-4)/[√(x^2+6x+25)+√(x^2-10x+29)]

    =-16/2=-8

    2√17≥√(x^2+6x+25)-√(x^2-10x+29) >-8

    3

    y=(x-a)^2-2|x|

    =x^2-2ax+a^2-2|x|

    =x^2-2x(a+|x|/x)x+a^2

    =[x-(a+|a|/x)]^2+a^2-(a+|x|/x)^2

    x0

    a^2-(a+1)^2=-2a-1

    因此a=0时,y=(x-a)^2-2|x|最小值-2a-1