登录
14.探究函数$y = x + \frac{4}{x}$的图象与性质.
(1)函数$y = x + \frac{4}{x}$的自变量$x$的取值范围是
(2)下列四个函数图象中，函数$y = x + \frac{4}{x}$的图象大致是
A. B. C. D.
(3)对于函数$y = x + \frac{4}{x}$，当$x ＞ 0$时，求$y$的取值范围.
请将下列求解过程补充完整.
解：$\because x ＞ 0$，
$\therefore y = x + \frac{4}{x} = (\sqrt{x})^{2} + \left( \frac{2}{\sqrt{x}} \right)^{2} = \left( \sqrt{x} - \frac{2}{\sqrt{x}} \right)^{2} +$
$\because \left( \sqrt{x} - \frac{2}{\sqrt{x}} \right)^{2} \geq 0$，
$\therefore y \geq$
(4)【拓展运用】若函数$y = \frac{x^{2} - 5x + 9}{x}(x ＞ 0)$，则$y$的取值范围是
(1)函数$y = x + \frac{4}{x}$的自变量$x$的取值范围是
x≠0
.(2)下列四个函数图象中，函数$y = x + \frac{4}{x}$的图象大致是
C
.A. B. C. D.
(3)对于函数$y = x + \frac{4}{x}$，当$x ＞ 0$时，求$y$的取值范围.
请将下列求解过程补充完整.
解：$\because x ＞ 0$，
$\therefore y = x + \frac{4}{x} = (\sqrt{x})^{2} + \left( \frac{2}{\sqrt{x}} \right)^{2} = \left( \sqrt{x} - \frac{2}{\sqrt{x}} \right)^{2} +$
4
.$\because \left( \sqrt{x} - \frac{2}{\sqrt{x}} \right)^{2} \geq 0$，
$\therefore y \geq$
4
.(4)【拓展运用】若函数$y = \frac{x^{2} - 5x + 9}{x}(x ＞ 0)$，则$y$的取值范围是
y≥1
.答案
14.(1)x≠0 (2)C (3)4 4 (4)y≥1
解析
(1)$x\neq0$
(2)C
(3)解：$\because x＞0$，
$\therefore y=x+\frac{4}{x}=(\sqrt{x})^{2}+\left(\frac{2}{\sqrt{x}}\right)^{2}=\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{2}+4$．
$\because\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{2}\geq0$，
$\therefore y\geq4$．
(4)$y\geq1$
(2)C
(3)解：$\because x＞0$，
$\therefore y=x+\frac{4}{x}=(\sqrt{x})^{2}+\left(\frac{2}{\sqrt{x}}\right)^{2}=\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{2}+4$．
$\because\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{2}\geq0$，
$\therefore y\geq4$．
(4)$y\geq1$
