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题目
练习要求：
以下题目尽量不要先求导。先把目标翻译成“找一个统一上界”，再尝试用放缩完成。
-
1. 证明函数
\[ f(x)=\frac {x^2}{1+x^4} \]
在 \((-\infty ,+\infty )\) 内有界。
-
2. 证明函数
\[ f(x)=\frac {|x|}{1+x^2+x^4} \]
在 \((-\infty ,+\infty )\) 内有界。
-
3. 证明对任意 \(x\in \mathbb {R}\)，有
\[ 0<\sqrt {x^2+1}-|x|\le 1. \]
-
4. 证明函数
\[ f(x)=\frac {x}{\sqrt {x^2+1}+1} \]
在 \((-\infty ,+\infty )\) 内有界。
-
5. 设 \(x>0\)，证明
\[ \frac {x}{(1+x)^2}\le \frac 14. \]
题解
提示与答案：
-
1. 用
\[ 1+x^4\ge 2x^2 \]
得
\[ 0\le \frac {x^2}{1+x^4}\le \frac 12. \]
-
2. 分母越大，正分式越小。可先丢掉 \(x^4\)，用
\[ 1+x^2+x^4\ge 1+x^2\ge 2|x| \]
得
\[ 0\le \frac {|x|}{1+x^2+x^4}\le \frac 12. \]
注意 \(x=0\) 时原式为 \(0\)，不需要除以 \(|x|\)。
-
3. 用有理化：
\[ \sqrt {x^2+1}-|x| =\frac {1}{\sqrt {x^2+1}+|x|}. \]
分母大于等于 \(1\)，所以
\[ 0<\sqrt {x^2+1}-|x|\le 1. \]
补充：若在分情况时遇到
\[ \sqrt {x^2+1}+x<1, \]
必须先说明这是在 \(x<0\) 的情况下讨论。因为当 \(x<0\) 时，\(|x|=-x\)，所以
\[ \sqrt {x^2+1}+x=\sqrt {x^2+1}-|x|. \]
用有理化可得
\[ \sqrt {x^2+1}+x =\frac {1}{\sqrt {x^2+1}-x}. \]
此时 \(x<0\)，故 \(-x>0\)，分母
\[ \sqrt {x^2+1}-x=\sqrt {x^2+1}+|x|>1, \]
所以
\[ 0<\sqrt {x^2+1}+x<1. \]
注意：若不加 \(x<0\) 这个条件，原不等式不成立；例如 \(x=0\) 时左边等于 \(1\)。
-
4. 取绝对值：
\[ |f(x)|=\frac {|x|}{\sqrt {x^2+1}+1}. \]
因为 \(\sqrt {x^2+1}>|x|\)，所以
\[ |f(x)|<1. \]
因此 \(f(x)\) 有界。
-
5. 用
\[ (1+x)^2\ge 4x,\qquad x>0, \]
得
\[ \frac {x}{(1+x)^2}\le \frac 14. \]
复盘重点：
；第 3 题练“根式差有理化”；第 4 题练“根式分母天然大于 \(|x|\)”。这些都是有界性题中替代求导极值法的常用入口。
第 1、2、5 题练“分母放大，正分式变小”
易错点
易错点：
-
• 正分式中，分母放大后分式变小；如果分母或分子可能为负，必须先处理符号。
-
• 含 \(|x|\) 的放缩常要单独检查 \(x=0\)，避免非法约分。
-
• 根式差 \(\sqrt {x^2+1}-|x|\) 不要硬估，优先有理化。
复盘提醒：
做完后不要只看答案对不对，要给每题标出“触发信号”：分母有 \(1+x^4\)、分母有 \(1+x^2\)、根式差、根式分母、\((1+x)^2\)。以后看到这些结构，直接调用对应放缩模板。