求值域①y=1/(1+x²) ②y=1/(x²-x+1) ③y=(2x-1)/(x+3)
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求值域①y=1/(1+x²) ②y=1/(x²-x+1) ③y=(2x-1)/(x+3)
求值域①y=1/(1+x²) ②y=1/(x²-x+1) ③y=(2x-1)/(x+3)
求值域①y=1/(1+x²) ②y=1/(x²-x+1) ③y=(2x-1)/(x+3)
(1)因为 x^2 ≥ 0 ,所以 1+x^2 ≥ 1 ,因此 1/(1+x^2) ≤ 1 ,
又显然 1/(1+x^2) > 0 ,所以值域为(0,1] .
(2)因为 x^2-x+1 = (x-1/2)^2+3/4 ≥ 3/4 ,所以 1/(x^2-x+1) ≤ 4/3 ,
又因为 (x-1/2)^2+3/4>0,1/(x^2-x+1) >0 ,所以函数值域为(0,4/3] .
(3)y = 2-7/(x+3) ,由于 1/(x+3) ≠ 0 ,因此 y ≠ 2 ,
所以,函数值域为(-∞,2)U(2,+∞).
①y=1/(1+x²)值域为(0,1].
②y=1/(x²-x+1) 值域为(0,4/3],
③y=(2x-1)/(x+3)值域为(负无穷大,2)∪(2,正无穷大)。