高数 03.05函数的极值与最大值最小值_函数最大值和最小值的求法

$\color{blue}{第三章 第五节 函数的极值与最大值最小值}$

$\color{blue}{一、函数的极值及其求法}$

$定义:设函数f(x)在(a, b)内有定义,x_0 \in (a, b),\\ 若存在x_0的一个邻域,在x \neq x_0时, \\ (1)f(x) < f(x_0), 则称x_0为f(x)的极大点, 称f(x_0)为函数的极大值;\\ (2)f(x) > f(x_0), 则称x_0为f(x)的极小点,称f(x_0)为函数的极小值.$
$极大点与极小点统称为极值点$
$极大值与极小值点是一个局部问题，在某个邻域内。$
$例如，确定函数f(x) = 2x^3 - 9x^2 + 12x - 3的单调区间.\\ 解:这个函数的定义域为(-\infty, +\infty),求这个函数的导数：\\ f^{\prime}(x) = 6x^2 - 18x + 12 = 6(x - 1)(x -2).\\ 解方程f^{\prime}(x) = 0,得它在函数的定义域(-\infty, +\infty)内的两个根x_1 = 1, x_2 = 2.\\ 这两个根把(-\infty, +\infty)分成三个分区间(-\infty, 1], [1, 2], 及 [2, +\infty).$

$故f(x)的单调增区间为(-\infty, 1), (2, +\infty); \\ f(x)的单调减区间为(1, 2).$
$f(x) = 2x^3 - 9x^2 + 12x - 3 \\ x = 1为极大点,f(1) = 2是极大值,\\ x = 2为极小点,f(2) = 1是极小值.$
$注意：\\ 1)函数的极值是函数的局部性质. \\ 2)对常见函数,极值可能出现在导数为0或导数不存在的点.$

$定理1.(极值第一判别法)\\ 设函数f(x)在x_0的某邻域内连续,且在空心邻域内有导数,当x由小到大通过x_0时,\\ (1)f^{\prime}(x) “左正右负”,则f(x)在x_0取极大值.\\ (1)f^{\prime}(x) “左负右正”,则f(x)在x_0取极小值.\\ $
$例1.求函数f(x) = (x - 1)x^{\frac{2}{3}}的极值.$
$解：\\ 1)求导数\quad f^{\prime}(x) = (x - 1)^{\prime}x^{\frac{2}{3}} + (x - 1)(x^{\frac{2}{3}})^{\prime} \\ = 1 \cdot x^{\frac{2}{3}} +(x - 1) \cdot (\dfrac{2}{3} \dfrac{1}{\sqrt[3]{x}}) \\ =\dfrac{5}{3} \cdot \dfrac{x -\frac{2}{5}}{\sqrt[3]{x}} \\ 2)求极值可疑点\\ f^{\prime}(x) = \infty, x_1 = 0, f(0) = 0; \\ f^{\prime}(x) = 0, x_2 = \dfrac{2}{5}, f(\dfrac{2}{5}) = -0.33 \\ 3)列表判别$

$\therefore x = 0为极大点，其极大值为f(0) = 0\\ x = \dfrac{2}{5}为极小点，其极小值为f(\dfrac{2}{5}) = -0.33$

$定理2.(极值第二判别法)设函数f(x)在点x_0处具有二阶导数,\\ 且f^{\prime}(x_0) = 0, f^{\prime \prime}(x_0) \neq 0 \\ (1)若f^{\prime \prime}(x_0) < 0,则f(x)在点x_0取极大值;\\ (2)若f^{\prime \prime}(x_0) > 0,则f(x)在点x_0取极小值;$
$证：\\ (1) f^{\prime \prime}(x) = \lim_{x \rightarrow x_0}{\dfrac{f^{\prime}(x) - f^{\prime}(x_0)}{x - x_0}} = \lim_{x \rightarrow x_0}{\dfrac{f^{\prime}(x)}{x - x_0}} \\ 由f^{\prime \prime}(x) < 0知,存在\delta > 0,当0 < |x - x_0| < \delta时,\dfrac{f^{\prime}(x)}{x - x_0} < 0 \\ 故当x_0 - \delta < x < x_0时,f^{\prime}(x) > 0; \\ 当x_0 < x < x_0 + \delta时, f^{\prime}(x) < 0; \\ 根据第一判别法知f(x)在x_0取极大值. \\ 由f^{\prime \prime}(x) > 0知,存在\delta > 0,当0 < |x - x_0| < \delta时,\dfrac{f^{\prime}(x)}{x - x_0} > 0 \\ 故当x_0 - \delta < x < x_0时,f^{\prime}(x) < 0; \\ 当x_0 < x < x_0 + \delta时, f^{\prime}(x) > 0; \\ 根据第一判别法知f(x)在x_0取极小值. \\ $

$例2.求函数f(x) = (x^2 - 1)^3 + 1的极值.$
$解:\\ 1)求导数\\ f^{\prime}(x) = 3(x^2-1)^2(x^2 - 1)^{\prime} \\ = 6x(x^2 - 1)^2\\ f^{\prime \prime}(x) = 6[x^{\prime}(x^2 - 1)^2 + x[(x^2 - 1)^2]^{\prime}] \\ = 6[(x^2 - 1)^2 + x \cdot 2(x^2 - 1)(x^2 - 1)^{\prime}] \\ = 6[(x^2 - 1)^2 + 4x^2(x^2 - 1)] \\ = 6(x^2 - 1)(5x^2 - 1) \\ 2)求驻点\\ f^{\prime}(x) = 0\\ x_1 = -1, x_2 = 0, x_3 = 1 \\ 对应f(-1) = 1, f(0) = 0, f(1) = 1 \\ 3)判断\\ 因f^{\prime \prime}(0) = 6 > 0,故f(0) = 0为极小值;\\ 又f^{\prime \prime}(-1) = f^{\prime \prime}(1) = 0,故需要使用第一判别法判断.\\ 由于f^{\prime}(x)在\pm 1左右邻域内不变号,\\ \therefore f(x)在x = \pm 1点没有极值.$

$\color{blue}{二、最大值与最小值问题}$

$若函数f(x)在闭区间[a, b]上连续,则其最值只能在极值点或端点处达到.$
$求函数最值的方法：$
$(1)求f(x)在(a, b)内的极值可疑点(驻点或不可导点)x_1,x_2, \cdots, x_m$
$最大值\\ M = max[f(x_1), f(x_2), \cdots, f(x_m), f(a), f(b)]$
$最小值\\ m = min[f(x_1), f(x_2), \cdots, f(x_m), f(a), f(b)]$
$特别：\\ 当f(x)在[a, b]内只有一个极值可疑点时,\\ 若在此点取极大(小)值,则也是最大(小)值.\\ 当f(x)在[a, b]上单调时,最值必在端点处达到.\\ 对于应用问题,有时可根据实际意义判别求，\\ 出的可疑点是否为最大值点或最小值点$
$例3.求函数f(x) = |2x^3 - 9x^2 + 12x|在闭区间[-\frac{1}{4}, \frac{5}{2}]上的最大值和最小值.$
$解：显然f(x) \in C[-\frac{1}{4}, \frac{5}{2}],且$
$f(x) = \left \{ \begin{array}{l} -(2x^3 - 9x^2 + 12x), -\frac{1}{4} \leq x \leq 0 \\ 2x^3 - 9x^2 + 12x, 0 < x \leq \frac{5}{2} \end{array} \right.$
$f^{\prime}(x) = \left \{ \begin{array}{l} -6x^2 + 18x - 12 = -6(x - 1)(x - 2), -\frac{1}{4} \leq x \leq 0 \\ 6x^2 - 18x + 12 = 6(x - 1)(x - 2), 0 < x \leq \frac{5}{2} \end{array} \right.$
$f(x)在[-\frac{1}{4}, \frac{5}{2}]内有极值可疑点 x_1 = 0, x_2 = 1, x_3 = 2$
$f(-\frac{1}{4}) = 3\frac{19}{32}, f(0) = 0, f(1) = 5, f(2) = 4, f(\frac{5}{2}) = 5$
$故函数在x = 0取最小值0;在x = 1及\frac{5}{2}取最大值5.$

$例4.铁路上AB段的距离为100km, 工厂C距离A处20km,AC \perp AB,\\ 要在AB线上选定一点D向工厂修一条公路,已知铁路与公路每公里货运价\\ 之比为3:5，为使货物从B运到工厂C的运费最省,问D点应如何选取?$
$解：设AD = x(km)，则CD = \sqrt{20^2 + x^2}, 总运费\\ f(x) = 5k \sqrt{20^2 + x^2} + 3k(100 - x) \quad (0 \leq x \leq 100)\\ f^{\prime}(x) = k(\dfrac{5x}{\sqrt{400 + x^2}} - 3) \\ f^{\prime \prime}(x) = 5k \dfrac{400}{(400 + x^2)^{\frac{3}{2}}} > 0 \\ 令f^{\prime}(x) = 0,得x = 15 \\ f^{\prime \prime}(15) > 0,所以x = 15为极小点,\\ f(0) = 400k \\ f(15) = 330k \\ f(100) = 4k\sqrt{20^2 + 100^2} > 400k\\ f_{min}(x) = f(15) = 330k,\\ 故AD = 15km时，运费最省$

$例5.把一根直径为d的圆木锯成矩形粱,矩形截面的高h和宽b,\\ 问应该如何选择才能使梁的抗弯截面模量最大?$
$解:由力学分析知矩形梁的抗弯截面模量为w = \frac{1}{6}bh^2\\ w = \frac{1}{6} b(d^2 -b^2), b \in (0, d)\\ 令 w^{\prime} = \dfrac{1}{6}(d^2 - 3b^2) = 0 \\ 得 b = \dfrac{\sqrt{3}}{3}d, h = \dfrac{\sqrt{6}}{3} d, \\ w^{\prime \prime} = -b < 0,有极大值。\\ 即 d:h:b = \sqrt{3}: \sqrt{2}: 1时，抗弯截面模量最大. \\ $

内容小结
1.连续函数的极值
(1)极值的可疑点：使得导数为0或不存在的点
(2)第一充分条件
$f^{\prime}(x)过x_0由正变负 \Longrightarrow f(x_0)为极大值$
$f^{\prime}(x)过x_0由负变正 \Longrightarrow f(x_0)为极小值$
(3)第二充分条件
$f^{\prime}(x_0) = 0, f^{\prime \prime}(x_0) < 0 \Longrightarrow f(x_0)为极大值$
$f^{\prime}(x_0) = 0, f^{\prime \prime}(x_0) > 0 \Longrightarrow f(x_0)为极小值$
2.连续函数的最值
最值点应在极值点和边界点上找;
应用问题可根据实际问题判定.
思考与练习
$1.设y = f(x)是方程y^{\prime \prime} - 2y^{\prime} + 4y = 0的一个解,若f(x_0) > 0, 且f^{\prime}(x_0) = 0, 则f(x)在x_0(\ \ {\color{blue}{A} }\ \ ).$
(A).取得极大值;
(B).取得极小值;
(C).在某邻域内单调递增;
(D).在某邻域内单调递减;
$\color{blue}{解：\\ f^{\prime \prime}(x_0) = 2f^{\prime}(x_0) - 4f(x_0)\\ = -4f(x_0) < 0 \\ f^{\prime}(x_0) = 0, f^{\prime \prime}(x) < 0,f(x_0)在x_0取得极大值 }$