设f(x)在[a,b]上连续,在（a,b）可导,且f(a)=f(b)=0,证明存在c属于（a,b）,使f'(c)+f(c

令g(x)=f(x)+f³(x)/3,则g(a)=g(b)=0
由中值定理存在c∈(a,b)使得g'(c)=0
而g'(x)=f'(x)+f²(x)
即f'(c)+f²(c)=0