作业帮函数y=f(x)满足对任意的x1x2属于R总有 [f(x1)-f(x2)] / [x1-x2] > 0则不等式f(m^2+1)>f(2m)的解集是?
来源：学生作业帮助网 编辑：作业帮 时间：2024/09/30 19:05:33
![函数y=f(x)满足对任意的x1x2属于R总有 [f(x1)-f(x2)] / [x1-x2] > 0则不等式f(m^2+1)>f(2m)的解集是?](/uploads/image/z/5266669-13-9.jpg?t=%E5%87%BD%E6%95%B0y%3Df%28x%29%E6%BB%A1%E8%B6%B3%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84x1x2%E5%B1%9E%E4%BA%8ER%E6%80%BB%E6%9C%89+%5Bf%28x1%29-f%28x2%29%5D+%2F+%5Bx1-x2%5D+%3E+0%E5%88%99%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28m%5E2%2B1%29%3Ef%282m%29%E7%9A%84%E8%A7%A3%E9%9B%86%E6%98%AF%3F)
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函数y=f(x)满足对任意的x1x2属于R总有 [f(x1)-f(x2)] / [x1-x2] > 0则不等式f(m^2+1)>f(2m)的解集是?
函数y=f(x)满足对任意的x1x2属于R总有 [f(x1)-f(x2)] / [x1-x2] > 0则不等式f(m^2+1)>f(2m)的解集是?
函数y=f(x)满足对任意的x1x2属于R总有 [f(x1)-f(x2)] / [x1-x2] > 0则不等式f(m^2+1)>f(2m)的解集是?
函数y=f(x)满足对任意的x1x2属于R总有 [f(x1)-f(x2)] / [x1-x2] > 0,所以
[f(x1)-f(x2)] 与 [x1-x2] 同号
即 y=f(x)是增函数
那么要满足f(m^2+1)>f(2m),必须
m^2 + 1 > 2m
m^2 - 2m + 1 > 0
(m - 1)^2 > 0
无论m取什么,都能满足
所以解集是任意实数