14.(2023春•泉州期末)已知函数f(x)=xln(x+1)−x2+12x3f(x)=xln(x+1)-{x}^{2}+\frac{1}{2}{x}^{3}f(x)=xln(x+1)−x2+21x3.(1)求曲线y=f(x)y=f(x)y=f(x)在点(1(1(1,fff(1))))处的切线方程;(2)证明:f(x)⩾0f(x)\geqslant 0f(x)⩾0.