%ex1202.m用插值法求解最優(yōu)化問題例10-2 
clear all
f1202=inline('(x.*x - 2).^2/2-1', 'x');;
%%%設定區(qū)間端點
a = 0;
b = 5;
%%%分別設定x閾值和函數(shù)f閾值
TolX = 1e-5; 
TolFun = 1e-8; 
%%%設定迭代次數(shù)
MaxIter = 100;
%%%用二次插值函數(shù)求解
[xoq,foq] = Opt_Quadratic(f1202,[a b],TolX,TolFun,MaxIter)
%用matlab內置函數(shù)求最小值以及它的函數(shù)值
[xob,fob] = fminbnd(f1202,a,b) %MATLAB 內置函數(shù)求解