设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=0证明 存在c∈(a,b)使f‘(c)+f(c)=0设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=0证明 存在c∈(a,b)使f ‘(c)+f(c)=0若要证f ‘(c)+[f(c)]^2=0
问题描述:
设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=0证明 存在c∈(a,b)使f‘(c)+f(c)=0
设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=0证明 存在c∈(a,b)使f ‘(c)+f(c)=0
若要证f ‘(c)+[f(c)]^2=0
答
令F(x)=e^x * f(x)
则F(a)=F(b)=0
由中值定理有
存在c∈(a,b),F'(c)= e^cf(c)+e^cf'(c)= e^c(f'(c)+f(c))=0
即f‘(c)+f(c)=0