关于导数的极值和单调区间问题

2025-03-28 07:42:47
推荐回答(1个)
回答1:

函数取的极值的充分必要条件:
1,必要条件:
若f(x)在x0点可导,且在x0点取得极值,
则必有f(x)的导数=0。使导数=0的点称为驻点。
函数导数不存在的点,也可能取得极值。
上述两种点称为极值的嫌疑点
2,充分条件:
第一充分条件:在极值的嫌疑点的两端变号。
由左向右,当x经过x0时,
f(x)的导数由正变负,则在点x0取得极大值f(x0);
f(x)的导数由负变正,则在点x0取得极小值f(x0);
f(x)的导数不变号,则在点x0取不到极值。
第二充分条件:f(x0)的导数=0,f(x0)的二阶导数≠0,
若f(x0)的二阶导数< 0,则f(x)在点x0取得极大值f(x0);
若f(x0)的二阶导数> 0,则f(x)在点x0取得极小值f(x0);

(极值可以多于2个,
极值中最大的为最大值max,
极值中最小的为最小值min.
最值最多2个)

求f(x)的极大值和极小值
先利用所求函数的导数为0和导数不存在来求极值点。

y=f(x)求导数得:48x^2-40ax+8a^2……①
令48x^2-40ax+8a^2=0
即:6x^2-5ax+a^2=0
得x1=a/2, x2=a/3 (a≠0且a≠1, 则x≠0,x≠1/2且x≠1/3)
将x1,x2代入原函数f(x)
f(x1)=f(a/2)=0
f(x1)=f(a/3)=a^3/27
y=f(x)的二阶导数为96x-40a…… ②
将x1,x2代入②
y=f(x1)的二阶导= 8a,
当01时,y=f(x1)的二阶导= 8a>0,f(x)在点取得极小值f(a/2)=0
当a<0时, y=f(x1)的二阶导= 8a<0,f(x)在点取得极大值f(a/2)=0
y=f(x2)的二阶导= -8a
当01时,y=f(x1)的二阶导= -8a<0,f(x)在点取得极大值f(a/3)= a^3/27
当a<0时, y=f(x1)的二阶导= 8a>0,f(x)在点取得极小值f(a/3)= a^3/27

函数导数不存在的点,当a=0或a=1时:x=0,x=1/2,x=1/3,分别代入①和②
y=f(0)的导数=0
y=f(1/2)的导数=0
y=f(1/3)的导数=0
y=f(0)的二阶导=0
y=f(1/2)的二阶导=8 >0, f(x)在点取得极小值f(1/2)= 0
y=f(1/3)的二阶导= -8 <0, f(x)在点取得极大值f(1/3)=1/27

综上所述:
当a>0时,
f(x)在点取得极大值f(a/3)= a^3/27
f(x)在点取得极小值f(a/2)=0
当a<0时
f(x)在点取得极大值f(a/2)=0
f(x)在点取得极小值f(a/3)= a^3/27

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