如果函数y=ax+1/x(a>0)在(0,2]上是减函数 在[2,正无穷）上是增函数求a

f(x)=ax+1/x
f(2)=2a+1/2
x:(0,2) f(x)>f(2)
x:(2,...) f(x)<f(2)
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x:(0,2) ax+1/x>2a+1/2
因为x>0 =>2ax^2+2>4ax+x
=> (2ax-1)(x-2)>0
=> x>1/2a and x>2 (不存在) or x<1/2a and x<2
=> 1/2a>0 =>a>0

x:(2,...) ax+1/x<2a+1/2
=> (2ax-1)(x-2)<0
=> x>1/2a and x<2 (不存在) or x<1/2a and x>2
=> 1/2a>=2 =>a<=1/4

=> 0<a<=1/4 a:(0,1/4]