Question

6．求函數(shù)f（k）=$\frac{\sqrt{{k}^{2}+2}}{{k}^{2}+6}$的最大值．

Answer 1

分析 化簡(jiǎn)可得f（k）$\frac{1}{\sqrt{{k}^{2}+2}+\frac{4}{\sqrt{{k}^{2}+2}}}$，由基本不等式和不等式的性質(zhì)可得．

解答 解：化簡(jiǎn)可得f（k）=$\frac{\sqrt{{k}^{2}+2}}{{k}^{2}+6}$
=$\frac{\sqrt{{k}^{2}+2}}{{k}^{2}+2+4}$=$\frac{1}{\sqrt{{k}^{2}+2}+\frac{4}{\sqrt{{k}^{2}+2}}}$
≤$\frac{1}{2\sqrt{\sqrt{{k}^{2}+2}•\frac{4}{\sqrt{{k}^{2}+2}}}}$=$\frac{1}{4}$，
當(dāng)且僅當(dāng)$\sqrt{{k}^{2}+2}$=$\frac{4}{\sqrt{{k}^{2}+2}}$即k=±$\sqrt{2}$時(shí)取等號(hào)．
∴函數(shù)f（k）=$\frac{\sqrt{{k}^{2}+2}}{{k}^{2}+6}$的最大值為$\frac{1}{4}$．

點(diǎn)評(píng) 本題考查基本不等式求最值，湊出可用基本不等式的形式是解決問(wèn)題的關(guān)鍵，屬基礎(chǔ)題．