作业帮y=(3x²-1)/(x²+2) 求值域y=(3x²-1)/(x²+2) 求值域
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y=(3x²-1)/(x²+2) 求值域y=(3x²-1)/(x²+2) 求值域
y=(3x²-1)/(x²+2) 求值域
y=(3x²-1)/(x²+2) 求值域
y=(3x²-1)/(x²+2) 求值域y=(3x²-1)/(x²+2) 求值域
y=(3x²+6-7)/(x²+2)
=3-7/(x²+2)
x²+2>=2
0<1/(x²+2)<=1/2
-7/2<=-7/(x²+2)<0
3-7/2<=3-7/(x²+2)<3+0
所以值域[-1/2,3)
y=(3x²+6-7)/(x²+2)
=3-7/(x²+2)
∴x²+2>=2
∴0<7/(x²+2)<=7/2
0>-7/(x²+2)>=-7/2
3>3-7/(x²+2)>=3-7/2=-1/2
∴3>y>=-1/2
因此值域是[-1/2，3)