![](https://static.youtibao.com/asksite/comm/h5/images/m_q_title.png)
[主观题]
设f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=0,试证在(0,1)内至少存在一点c,使 cf'(c)+kf(c)=f'(c)
设f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=0,试证在(0,1)内至少存在一点c,使
cf'(c)+kf(c)=f'(c)
查看答案
![](https://static.youtibao.com/asksite/comm/h5/images/solist_ts.png)
设f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=0,试证在(0,1)内至少存在一点c,使
cf'(c)+kf(c)=f'(c)
设函数f(x)在[0,1]上连续,在(0,1)内可导且f(0)=f(1)=0,f(1/2)=1,试证明至少存在一点ξ∈(0,1),使得f`(ξ)=1.
求证:在(a,b)内至少存在一点ξ,使f'(ξ)=0.
求证:在(a,b)内至少存在一点ξ,使f'(ξ)=0.
|f'(ξ)|≥2M,|f'(η)|≤2M其中M=max{|f(x)|}
设f(x)在[a,b]上连续,在(a,b)内二阶可导,f(a)=f(b)=0,且有c(a<c<b),使f(c)>0.证明存在c∈(a,b)使f‘(c)+f(c)=0.