一道高数大题，求答案谢谢

Question

如题所述

Answer 1

证明：
因为f(x)在x_0 处可导，
lim┬(h→0)⁡〖(f(x_0+3h)-f(x_0+h))/h〗=lim┬(h→0)⁡〖(f(x_0+3h)-f(x_0+2h)+f(x_0+2h)-f(x_0+h))/h〗
=lim┬(h→0)⁡〖(f(x_0+3h)-f(x_0+2h))/h〗 +lim┬(h→0)⁡〖(f(x_0+2h)-f(x_0+h))/h〗=2f^,(x_0)

Answer 2

证明：因为f(x)在x0出可导
所以有：[f(x+h)-f(x)]/h=f(x0)'
同理可得：[f(x+3h)-f(x+2h)]/h=f(x0)'
[f(x+2h)-f(x+h ) ]/h=f(x0)'
两式相加可得：[f(x+3h)-f(x+h)]/h=2f(x0)'

Answer 3