证明若g(x)=x^2+ax+b,则g[(x1+x2)/2]小于等于(g(x1)+g(x2))/2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]小于等于(g(x1)+g(x2))/2证明 若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2](/uploads/image/z/5201827-43-7.jpg?t=%E8%AF%81%E6%98%8E%E8%8B%A5g%28x%29%3Dx%5E2%2Bax%2Bb%2C%E5%88%99g%5B%28x1%2Bx2%29%2F2%5D%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8E%28g%28x1%29%2Bg%28x2%29%29%2F2%E8%AF%81%E6%98%8E+%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]小于等于(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]小于等于(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]小于等于(g(x1)+g(x2))/2证明 若g(x)=x^2+ax+b,则g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
因为g[(X1+X2)/2]化简后是(gX1+gX2)/2
又因为[g(x1)+g(x2)]/2化简后是(gx1+gx2)/2
若x1>X1 x2>X2 则小于
若x1等于X1 x2等于X2 则等于
所以g[(X1+X2)/2]≤[g(x1)+g(x2)]/2
得证,
先给g(x)=x^2+ax+b求导
到到后面给后面作差