曲线y=g(x)在点(2,g(2))处的切线方程为y=1-x,那么该点,也应该在切线上,所以将该点坐标带入切线方程有:g(2)=1-2=-1,所以曲线y=g(x)在点(2,-1)处的切线方程为y=1-x,那么f(2)=2g(2)=-2
则曲线y=f(x)在点(2,f(2))处的切线方程即为:曲线y=f(x)在点(2,-2)处的切线方程;
显然只需要知道法f(x)在点(2,-2)处的切线斜率即可
又f(x)=xg(x),那么f'(x)=g(x) + xg'(x),则f'(2)=g(2)+2g'(2)=-1+2g'(2),
又由于g'(2)=-1(y=1-x的斜率),那么f'(2)=-1+(-2)=-3,
所以曲线y=f(x)在点(2,f(2))处的切线方程为:
y-(-2)=-3(x-2),化简得:y=-3x+4