证明若f(x)=ax+b,则f(x1+x2/2)=f(x1)+f(x2)/2若g(x)=x+ax+b 则g(x1+x2/2)≤g(x1)+g(x2)/2过程

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证明若f(x)=ax+b,则f(x1+x2/2)=f(x1)+f(x2)/2若g(x)=x+ax+b 则g(x1+x2/2)≤g(x1)+g(x2)/2过程

f(x)=ax+b f((x1+x2)/2) =a((x1+x2)/2)+b =ax1/2+ax2/2+b [f(x1)+f(x2)]/2 =[ax1+b+ax2+b]/2 =ax1/2+ax2/2+b 所以 f(x1+x2/2)=[f(x1)+f(x2)]/2 2.g〔〔x1+x2〕/2〕-〔g〔x1〕+g〔x2〕〕/2 = [(x1 + x2)/2]^2 + a*(x...