Question

求值域 y=(x²-1)/(x²+x+1); y=2x-3+√(13-4x); y=(3x+2)/(

Answer 1

1.y=(x²-1)/(x²+x+1); y导=2x/(x²+x+1)-(x²-1)*(2x+1)/(x²+x+1)^2= [2x*(x²+x+1)-(x²-1)*(2x+1)]/(x²+x+1)^2=(x^2+4x+1)/(x²+x+1)^2,

容易得出：在（-2-√3,-∞）上递增,在（-2-√3,-2+√3）上递减,在（-2+√3,+∞）上递增

x=-2-√3,y有极大值2√3/3,x=-2+√3,y有最小值-2√3/3,x趋近+∞时,y趋近1

所以值域为[-2√3/3,2√3/3]

2.y=2x-3+√(13-4x)；y导=2+（1/2）*（-4）/√(13-4x)=2-2/√(13-4x)；

当0