一道高数大题,
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/13 13:55:57
![一道高数大题,](/uploads/image/z/10163523-3-3.jpg?t=%E4%B8%80%E9%81%93%E9%AB%98%E6%95%B0%E5%A4%A7%E9%A2%98%2C)
xn@_
*J^.Ľ[TgmhQqJ%T
ъF
R8F%a8%'^'NHSh|j3^1|=>
>4KbݹUV,k =|$ij2ȅz|qD1Z֊GqQV+0R< ]xԢ@KơքcK@92!4Jq$%QB>!#(D%ĩ=/.$P$$c5gzIK˧o}?^7FXIYN,{5O^5
)}xOeP@}3;턧څTQy*e2oF47vɩ|m2RW@ωk87LK a5ArZ0Dݔ O)FXc|e
=]BIM
dŜ^͘9slirSes Fދ$sLpSօoy8mM)zV=;$;GdMX罿 {
一道高数大题,
一道高数大题,
一道高数大题,
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)
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)'