14.已知f(x)=2lnx+ax+bxf(x)=2lnx+ax+\frac{b}{x}f(x)=2lnx+ax+xb在x=1x=1x=1处的切线方程为y=−3xy=-3xy=−3x.(1)求函数f(x)f(x)f(x)的解析式;(2)f′(x)f'(x)f′(x)是f(x)f(x)f(x)的导函数,证明:对任意x∈[1x\in [1x∈[1,+∞)+\infty )+∞),都有f(x)−f′(x)⩽−2x+1x+1f(x)-f'(x)\leqslant -2x+\frac{1}{x}+1f(x)−f′(x)⩽−2x+x1+1.