[{"createTime":1735734952000,"id":1,"img":"hwy_ms_500_252.jpeg","link":"https://activity.huaweicloud.com/cps.html?fromacct=261f35b6-af54-4511-a2ca-910fa15905d1&utm_source=V1g3MDY4NTY=&utm_medium=cps&utm_campaign=201905","name":"华为云秒杀","status":9,"txt":"华为云38元秒杀","type":1,"updateTime":1735747411000,"userId":3},{"createTime":1736173885000,"id":2,"img":"txy_480_300.png","link":"https://cloud.tencent.com/act/cps/redirect?redirect=1077&cps_key=edb15096bfff75effaaa8c8bb66138bd&from=console","name":"腾讯云秒杀","status":9,"txt":"腾讯云限量秒杀","type":1,"updateTime":1736173885000,"userId":3},{"createTime":1736177492000,"id":3,"img":"aly_251_140.png","link":"https://www.aliyun.com/minisite/goods?userCode=pwp8kmv3","memo":"","name":"阿里云","status":9,"txt":"阿里云2折起","type":1,"updateTime":1736177492000,"userId":3},{"createTime":1735660800000,"id":4,"img":"vultr_560_300.png","link":"https://www.vultr.com/?ref=9603742-8H","name":"Vultr","status":9,"txt":"Vultr送$100","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":5,"img":"jdy_663_320.jpg","link":"https://3.cn/2ay1-e5t","name":"京东云","status":9,"txt":"京东云特惠专区","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":6,"img":"new_ads.png","link":"https://www.iodraw.com/ads","name":"发布广告","status":9,"txt":"发布广告","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":7,"img":"yun_910_50.png","link":"https://activity.huaweicloud.com/discount_area_v5/index.html?fromacct=261f35b6-af54-4511-a2ca-910fa15905d1&utm_source=aXhpYW95YW5nOA===&utm_medium=cps&utm_campaign=201905","name":"底部","status":9,"txt":"高性能云服务器2折起","type":2,"updateTime":1735660800000,"userId":3}]
<>求最大值和最小值
求 f ( x ) f(x) f(x)在 [ a , b ] [a,b] [a,b]上最大值和最小值的方法：
* 求出所有驻点和不可导点 x 1 , x 2 , x 3 … x_1,x_2,x_3\dots x1,x2,x3…
* 计算 f ( x 1 ) , f ( x 2 ) , f ( x 3 ) … f(x_1),f(x_2),f(x_3)\dots f(x1),f(x2
),f(x3)…和 f ( a ) , f ( b ) f(a),f(b) f(a),f(b)的值
* 比较大小
<>例1
求 f ( x ) = ( x − 5 ) x 2 3 f(x)=(x-5)\sqrt[3]{x^2} f(x)=(x−5)3x2 在 [ − 2 , 3
] [-2,3][−2,3]上的最值
解： f ′ ( x ) = x 2 3 + 2 3 ( x − 5 ) 1 x 3 = 5 ( x − 2 ) 3 x 3
f'(x)=\sqrt[3]{x^2}+\dfrac
23(x-5)\dfrac{1}{\sqrt[3]{x}}=\dfrac{5(x-2)}{3\sqrt[3]{x}}f′(x)=3x2 +32(x−5)3x
1=33x 5(x−2)
可能极值点： x 1 = 0 , x 2 = 2 x_1=0,x_2=2 x1=0,x2=2
f ( 0 ) = 0 , f ( 2 ) = − 3 4 3 , f ( − 2 ) = − 7 4 3 , f ( 3 ) = − 2 9 3
f(0)=0,f(2)=-3\sqrt[3]4,f(-2)=-7\sqrt[3]4,f(3)=-2\sqrt[3]9f(0)=0,f(2)=−334 ,f(−
2)=−734 ,f(3)=−239
所以最大值为 0 0 0，最小值为 − 7 4 3 -7\sqrt[3]4 −734
<>例2
函数 f ( x ) = 1 3 x 3 − 2 x 2 + 5 f(x)=\dfrac 13x^3-2x^2+5 f(x)=31x3−2x2+5在 [
− 2 , 2 ] [-2,2][−2,2]的最大值为 ‾ \underline{\qquad}
解： f ′ ( x ) = x 2 − 4 x = x ( x − 4 ) f'(x)=x^2-4x=x(x-4) f′(x)=x2−4x=x(x−4)
可能的极值点： x 1 = 0 , x 2 = 4 x_1=0,x_2=4 x1=0,x2=4
f ( 0 ) = 5 , f ( 4 ) = − 17 3 , f ( − 2 ) = − 17 3 , f ( 2 ) = − 1 3
f(0)=5,f(4)=-\dfrac{17}{3},f(-2)=-\dfrac{17}{3},f(2)=-\dfrac 13f(0)=5,f(4)=−317
,f(−2)=−317,f(2)=−31
所以最大值为 5 5 5
