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定理设g(x)、f(x)为集D上的单调递增函数,且 g(x)∈D,f(x)∈D,g(x)≤f(x),则 g[g(x)]≤f[f(x)]。证:∵ g(x)、f(x)在D上递增,且g(x)≤f(x),因g(x)∈D,f(x)∈D。∴ g[g(x)]≤g[f(x)],g[f(x)]≤f[f(x)], ∴ g[g(x)]≤f[f(x)]。上述定理在判断自身复合函数值大小时有着巧妙而简捷的功用,