一道关于函数增减性的问题已知函数f(x)=根号(x^2+1)-ax,a>0,求a的取值范围,使函数f(x)在区间0到正无穷大上是单调函数,答案是a≥1,答案说0到1的范围中,存在两点x1=0,x2=2a/1-a^2 f(x1)=f(x2).无
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![一道关于函数增减性的问题已知函数f(x)=根号(x^2+1)-ax,a>0,求a的取值范围,使函数f(x)在区间0到正无穷大上是单调函数,答案是a≥1,答案说0到1的范围中,存在两点x1=0,x2=2a/1-a^2 f(x1)=f(x2).无](/uploads/image/z/7312806-54-6.jpg?t=%E4%B8%80%E9%81%93%E5%85%B3%E4%BA%8E%E5%87%BD%E6%95%B0%E5%A2%9E%E5%87%8F%E6%80%A7%E7%9A%84%E9%97%AE%E9%A2%98%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%E6%A0%B9%E5%8F%B7%28x%5E2%2B1%29-ax%2Ca%3E0%2C%E6%B1%82a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%2C%E4%BD%BF%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%E5%8C%BA%E9%97%B40%E5%88%B0%E6%AD%A3%E6%97%A0%E7%A9%B7%E5%A4%A7%E4%B8%8A%E6%98%AF%E5%8D%95%E8%B0%83%E5%87%BD%E6%95%B0%2C%E7%AD%94%E6%A1%88%E6%98%AFa%E2%89%A51%2C%E7%AD%94%E6%A1%88%E8%AF%B40%E5%88%B01%E7%9A%84%E8%8C%83%E5%9B%B4%E4%B8%AD%2C%E5%AD%98%E5%9C%A8%E4%B8%A4%E7%82%B9x1%3D0%2Cx2%3D2a%2F1-a%5E2+f%EF%BC%88x1%EF%BC%89%3Df%EF%BC%88x2%EF%BC%89.%E6%97%A0)
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