用Matlab求函数z＝x^4-8xy+2y^2-3的极值点和极值，这是我们的作业，我对matla 选择、提取和分析ATM交易状态的特征参数，用什么样的函数解析...
\u7528MATLAB\u5df2\u77e5\uff0c\u6c42\u5148\u753b\u56fe\u89c2\u5bdf\uff0c\u7136\u540e\u6c42\u51fd\u6570f(x)=x\uff3e3-x\uff3e2-x-b/3\u7684\u6781\u503c\u7528MATLAB\u6c42\u51fd\u6570f(x)=x\uff3e3-x\uff3e2-x-b/3\u7684\u6781\u503c\uff0c\u53ef\u4ee5\u5148\u7528\u56fe\u5f62\u6765\u5224\u65ad\uff0c\u662f\u5426\u6709\u6700\u5927\uff08\u5c0f\uff09\u503c\u5b58\u5728\u3002\u6c42\u89e3\u8fc7\u7a0b\u5982\u4e0b\uff1a
>> x=-10:10;b=6;>> f=x.^3-x.^2-x-b/3;>> plot(x,f) %\u7ed8\u56fe
>> func=@(x)x.^3-x.^2-x-6/3;>> [x,g]=fminsearch(func,0) %\u6781\u5c0f\u503c>> func=@(x)-(x.^3-x.^2-x-6/3);>> [x,g]=fminsearch(func,0) %\u6781\u5927\u503c
\u6781\u5c0f\u503c\uff1ax = 1.0000\uff1bg = -3
\u6781\u5927\u503c\uff1ax = -0.3333\uff1bg = 1.8148
\u7528\u6781\u503c\u51fd\u6570fminsearch\uff08\uff09\u6c42\u5f97\uff0c\u5176\u6781\u5c0f\u503c\u3002 x = 1.99999078443153 4.00001086059484 \u5373 x=1.99999078443153\uff1by=4.00001086059484 fval = -18.9999999969252 \u6781\u5c0f\u503c,z=18.9999999969252 \u5728\u547d\u4ee4\u7a97\u53e3\u4e0b\u6267\u884c\u4e0b\u5217\u4ee3\u7801
用极值函数fminsearch（）求得，其极小值。x =
1.99999078443153 4.00001086059484
即
x=1.99999078443153；y=4.00001086059484
fval =
-18.9999999969252
极小值,z=18.9999999969252
在命令窗口下执行下列代码：
clc,clear all,close all
syms x
fun=@(x)(x(1)^4-8*x(1).*x(2)+2.*x(2)^2-3);
[x,fval,exitflag] = fminsearch(fun,[0 0])
绛旓細鐢ㄦ瀬鍊鍑芥暟fminsearch锛堬級姹傚緱锛屽叾鏋佸皬鍊笺x = 1.99999078443153 4.00001086059484 鍗 x=1.99999078443153锛泍=4.00001086059484 fval = -18.9999999969252 鏋佸皬鍊,z=18.9999999969252 鍦ㄥ懡浠ょ獥鍙ｄ笅鎵ц涓嬪垪浠ｇ爜锛歝lc,clear all,close all syms x fun=@(x)(x(1)^4-8*x(1).*x(2)+2.*x(...
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