求下列函数的值域 (1)Y=(5X-1)/(4X+2) (2)Y=(X^2-4X+3)/(2x^2-x-1) (3)Y=

1个回答

  • (1)Y=(5X-1)/(4X+2)

    =5/4 * (x+1/2-1/2-1/5)/(x-1/2) 【x≠1/2】

    =5/4[(x+1/2)-7/10]/(x-1/2)

    =5/4-(7/8)/(x-1/2)≠5/4

    值域:(-∞,5/4),(5/4,+∞)

    (2)Y=(X^2-4X+3)/(2x^2-x-1)

    =[(x-1)(x-3)]/[(2x+1)(x-1)] 【x≠-1/2,x≠1】

    =(x-3)/(2x+1)

    =1/2 (x+1/2-1/2-3)/(x+1/2)

    =1/2-(7/4)/(x+1/2)≠1/2

    又:x≠1

    y≠1/2-(7/4)/(3/2)=-2/3

    值域(-∞,-2/3),(-2/3,1/2),(1/2,+∞)

    (3)Y=(2X^2+4X-7)/(X^2+2X+3)

    =2[(x+1)^2-9/2]/[(x+1)^2+2]

    =2[(x+1)^2+2-13/2]/[(x+1)^2+2]

    =2-13/[(x+1)^2+2]

    [(x+1)^2+2]≥2

    13/[(x+1)^2+2]≤13/2

    2-13/[(x+1)^2+2]≥-11/12

    值域[-11/12,+∞)

    (4)Y=2X-√(X-1) 【x≥1】

    y'=2-1/[2√(x-1)]

    当1≤x≤17/16时单调减

    当x≥17/16时单调增

    x=17/16时,最小值ymin=2*17/16-√(17/16-1)=15/8

    值域[15/8,+∞)

    (5)Y=4-√(3+2X-X^2)

    =4-√[-(x+1)(x-3)] 【定义域-1<x<3】

    =4-√[-(x-1)^2+4]≥4-√4=2

    x=-1和时,y=4-0=4

    值域[2,4]