函数 f(x)=ln(x+1) 曲线y=f(x)在点(x0,y0)处的切线方程为 y=g(x) ,证明,对所有x属于（-1,+∞）f(x)小于等于g(x)

题目

函数 f(x)=ln(x+1) 曲线y=f(x)在点(x0,y0)处的切线方程为 y=g(x) ,证明,对所有x属于（-1,+∞）f(x)小于等于g(x)

答案

f'(x) = 1/(x + 1)点(x0,y0)处的切线方程为:y - ln(x₀ + 1) = (x - x₀)/(x₀ + 1)y = g(x) = (x - x₀)/(x₀ + 1) + ln(x₀ + 1)h(x) = g(x) - f(x) = (x - x₀)/(x₀ + ...