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[不等式] 函数最值

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wzyl1860 posted 2018-1-9 13:19 |Read mode
Last edited by hbghlyj 2025-5-10 16:44函数$f(x) =x^2+1+\sqrt{x^4-8x^2-16x+52}$的最小值为

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kuing posted 2018-1-9 14:09
$f(x)=x^2+1+\sqrt{(x^2 - 6)^2 + (2 x - 4)^2}\geqslant x^2+1+(6-x^2)=7$ 且 $f(2)=7$。

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敬畏数学 posted 2018-1-10 08:34
设 $ y=x^2+1+\sqrt{x^4-8x^2-16x+52} $,两边平方有,
\[(10-2y)x^2+16x+y^2-2y-51=0\]显然$y\ne5$,判别式$\Delta\geqslant0$有;
\[y\geqslant 7,-2\leqslant y\leqslant 2, \]事实上,
\[x^2+1+\sqrt{x^4-8x^2-16x+52}>2,\]即证,
\[\sqrt{x^4-8x^2-16x+52}>1-x^2,\]当$x^2\geqslant 1$不等式显然成立;

当$x^2<1$,两边平方,\[ 6x^2+16x-51<0 ,\]此式显然成立。
所以所求函数的最小值为7,当$x=2$时等号成立。

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