证明 二次函数Y=aX²+bX+c(a≠0)当a<0时函数在负无穷到-b/2a是单调递增的来位高人给我指出错误证明;已知a<0,取任意X1,X2,且X1<X2≤-b/2a,则 f(X2)-f(X1)=(aX2²+bX2+c)-(aX1²+bX1+c)
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/28 01:33:56
![证明 二次函数Y=aX²+bX+c(a≠0)当a<0时函数在负无穷到-b/2a是单调递增的来位高人给我指出错误证明;已知a<0,取任意X1,X2,且X1<X2≤-b/2a,则 f(X2)-f(X1)=(aX2²+bX2+c)-(aX1²+bX1+c)](/uploads/image/z/8800979-59-9.jpg?t=%E8%AF%81%E6%98%8E+%E4%BA%8C%E6%AC%A1%E5%87%BD%E6%95%B0Y%3DaX%26%23178%3B%2BbX%2Bc%EF%BC%88a%E2%89%A00%29%E5%BD%93a%EF%BC%9C0%E6%97%B6%E5%87%BD%E6%95%B0%E5%9C%A8%E8%B4%9F%E6%97%A0%E7%A9%B7%E5%88%B0-b%2F2a%E6%98%AF%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E7%9A%84%E6%9D%A5%E4%BD%8D%E9%AB%98%E4%BA%BA%E7%BB%99%E6%88%91%E6%8C%87%E5%87%BA%E9%94%99%E8%AF%AF%E8%AF%81%E6%98%8E%EF%BC%9B%E5%B7%B2%E7%9F%A5a%EF%BC%9C0%2C%E5%8F%96%E4%BB%BB%E6%84%8FX1%2CX2%2C%E4%B8%94X1%EF%BC%9CX2%E2%89%A4-b%2F2a%2C%E5%88%99+f%EF%BC%88X2%EF%BC%89-f%EF%BC%88X1%EF%BC%89%3D%EF%BC%88aX2%26%23178%3B%2BbX2%2Bc%EF%BC%89-%EF%BC%88aX1%26%23178%3B%2BbX1%2Bc%EF%BC%89)
xTAn@J"pR
pUuaWB&
X)m@v*cЙY
3C-'e]E,1JxP$3\_ϞJ_,jm59,oW-29ؽH~0>
_KFl2y?>%#h.m!Ҭz8q4u<1"|Nv-U*rU*/oLr"{Bͪr^MRJUG_?g12AA
,27DAT`N2i&l+H%_)tml"R_^fKb