一、复合函数复合函数的单调性,可利用“同增异减”来确定例1求函数y=(x~2-2008x)~(1/2)的单调递增区间.解:首先,由x~2-2008x≥0,得x≤0或x≥2008.所以函数的定义域是(-∞,0)∪[2008,+∞).①其次,由于函数y=n~(1/2)在[0,+∞)上是增函数,所以求函数y=(x~2-2008x)~(1/2)的单调递增 First, the monotonicity of compound function compound function, you can use “ same increase difference ” to determine the monotonic increment interval of function y=(x~2-2008x)~(1/2) in Example 1. Solution: First, From x~2-2008x≥0, we get x≤0 or x≥2008. So the domain of the function is (-∞,0)∪[2008,+∞).1 Secondly, because the function y=n~(1/ 2) It is an increasing function on [0,+∞), so find a monotonically increasing function y=(x~2-2008x)~(1/2)