Question

已知f(x)=(根号下x^2+1)-ax(a>0）,在[0,+无穷)上是单调减函数，求a范围，

Answer 1

f'(x)=x/Sqrt(x^2+1)-a
monotonically decreasing on (0, infinity), so f'(x)<0
a>x/Sqrt(x^2+1) for any x>=0

the latter achieves its maximum when its own derivative is 0, which is when
f''(x)=0, or [1*Sqrt(x^2+1)-x*x/Sqrt(x^2+1)]/(x^2+1)=0
=> Sqrt(x^2+1)=x^2/Sqrt(x^2+1)=>x^2=x^2+1==>never!
truth is that f''(x)>0 always
so it is monotonically increasing==>f'(x) reaches maximum at x=infinity, which means f'(x)=1
but note that f'(x) will never reach 1 since x cannot reach infinity
所以 a>=1
中文不好打。。。谅解
monotonically increasing/decreasing:单调递增/递减
maximum：最大值
infinity：无穷
f'(x) 一阶导数
f''(x) 二阶导数
看在我这么辛苦的份上。。。