作业帮设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)在区间[0,+∞)上的单调函数
来源：学生作业帮助网 编辑：作业帮 时间：2024/07/13 23:36:57
![设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)在区间[0,+∞)上的单调函数](/uploads/image/z/11426807-47-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%5B%E6%A0%B9%E5%8F%B7%E4%B8%8B%28x%26%23178%3B%2B1%29%5D-ax+%2Ca%EF%BC%9E0.%E8%AF%81%E6%98%8E%3A%E5%BD%93a%E2%89%A71%E6%97%B6%2C%E5%87%BD%E6%95%B0f%28x%29%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3D%5B%E6%A0%B9%E5%8F%B7%E4%B8%8B%28x%26%23178%3B%2B1%29%5D-ax+%2Ca%EF%BC%9E0.%E8%AF%81%E6%98%8E%3A%E5%BD%93a%E2%89%A71%E6%97%B6%2C%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5B0%2C%2B%E2%88%9E%29%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E5%87%BD%E6%95%B0)
xՑJ@_T*-\ftoQDf2&jT,D.TD;]iUSXW)"n-t>s~3(ޢ~l3gi;6gg(4G|@it _9 ܠX]&+_>eX;ar7:i;1T (ӫVR
C+dNºmF#(`K~%F=!~l >t3"0#/Ы,k.(tl" ې!L 8
*)DP)#:C2u Qq&rQ q1&(Gcpeyle e"N蒼 A
设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)在区间[0,+∞)上的单调函数
设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)
设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)在区间[0,+∞)上的单调函数
设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)设函数f(x)=[根号下(x²+1)]-ax ,a＞0.证明:当a≧1时,函数f(x)在区间[0,+∞)上的单调函数