设函数f(x)和g(x)在闭区间[a,b]上连续,在(a,b)上可导,且f(a)=f(b)=0.证明:至少存在一点c,使f'(c)+f(c)g'(c)=0.

Question

设函数f(x)和g(x)在闭区间[a,b]上连续,在(a,b)上可导,且f(a)=f(b)=0.证明:至少存在一点c,使f'(c)+f(c)g'(c)=0.

Answer 1

证明：
作辅助函数F(x)＝f(x)e∧g(x)
则F'(x)＝f'(x)e∧g(x)＋f(x)g'(x)e∧g(x)
＝[f'(x)＋f(x)g'(x)]e∧g(x)
显然F(x)在[a,b]上连续,在(a,b)上可导,且F(a)＝F(b)＝0
由罗尔定理知,至少存在c∈(a,b),使F'(c)＝0
即 [f'(c)＋f(c)g'(c)]e∧g(c)＝0
而 e∧g(c)≠0
故 f'(c)＋f(c)g'(c)＝0.