函数f(x)满足在[a,b]上连续,在(a,b)内可导、证明存在ξ∈(a,b),使f(ξ)—f(a)/（b—ξ）=f'(ξ)

Question

函数f(x)满足在[a,b]上连续,在(a,b)内可导、证明存在ξ∈(a,b),使f(ξ)—f(a)/（b—ξ）=f'(ξ)

Answer 1

令g(x) = (b-x)f(x)
则g(x)满足在[a,b]上连续,在(a,b)内可导,
由拉格朗日中值定理,
存在ξ∈(a,b)
使得g'(ξ) = (g(b)-g(a))/(b-a)
g(b) = (b-b)f(b) = 0
g(a) = (b-a)f(a)
g'(ξ) = -f(ξ) + (b-ξ)f'(ξ)
所以有
-f(ξ) + (b-ξ)f'(ξ) = -(b-a)f(a)/(b-a) = -f(a)
整理即得结果