证明:若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
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![证明:若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2](/uploads/image/z/6875077-13-7.jpg?t=%E8%AF%81%E6%98%8E%EF%BC%9A%E8%8B%A5g%28x%29%3Dx%5E2%2Bax%2Bb%2C%E5%88%99g%5B%28X1%2BX2%29%2F2%5D%E2%89%A4%5Bg%28x1%29%2Bg%28x2%29%5D%2F2)
证明:若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
证明:若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
证明:若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
[g(x1)+g(x2)]/2=[x1²+x2²+a(x1+x2)+2b]/2=(x1²+x2²)/2+a(x1+x2)/2+b
g[(x1+x2)/2]=(x1+x2)²/4+a(x1+x2)/2+b
∴g[(x1+x2)/2]-[g(x1)+g(x2)]/2
=(x1+x2)²/4-(x1²+x2²)/2
=(x1²+2x1x2+x2²-2x1²-2x2²)/4
=-(x1-x2)²/4≤0
∴g[(X1+X2)/2]≤[g(x1)+g(x2)]/2