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tarawa/sol.txt

Last active August 29, 2015 13:57
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sol
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sol.txt
f(0)=0
f(1)=1/3+(2a+1)/2+(a^2+a)
=a^2+2a+5/6
f(a)=a^3/3-(2a+1)a^2/2+(a^2+a)a
=a^3/3+a^2/2, a =[0,1]
f'(a)=a^2+a, max : a=1, maxval=5/6
f(a+1) = (a+1)^3/3-1/2(2a+1)(a+1)^2+(a^2+a)(a+1)
=a^3/3+a^2/2-1/6,a = (-1,0]
f'(a+1)=a^2+a, 值恒小于0，由于f(0)的存在舍去
分情况讨论四者取个max就可以了。。
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