f(x)＝1/(x²-2)+x²+1
＝(x²-2)+1/(x²-2)+3.
(1)x²-2>0,即x∈(-∞,-√2)∪(√2,+∞)时,
f(x)≥2√[(x²-2)·1/(x²-2)]+3＝5,
此时,f(x)∈[5,+∞).
若x²-2
求函数f(x)=1/(x的平方-2)+x的平方+1的值域
f(x)＝1/(x²-2)+x²+1
＝(x²-2)+1/(x²-2)+3.
(1)x²-2>0,即x∈(-∞,-√2)∪(√2,+∞)时,
f(x)≥2√[(x²-2)·1/(x²-2)]+3＝5,
此时,f(x)∈[5,+∞).
若x²-2