设f(x)在[a,b]上l连续可导,且f(a)=f(b)=0,求证：存在η∈(a,b),使ηf(η)+f'(η)=0

Question

设f(x)在[a,b]上l连续可导,且f(a)=f(b)=0,求证：存在η∈(a,b),使ηf(η)+f'(η)=0

Answer 1

设F(x)=e^(1/2 * x^2) * f(x).
则F'(x)=e^(1/2 * x^2) *(f'(x)+xf(x)).
且F(a)=F(b)=0,在[a, b]上连续可导.
根据Rolle定理: 存在η∈(a,b), 使得F'(η)=0.
即ηf(η)+f'(η)=0.