一条涉及中值定理的高数题,已知函数f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=λb,f(b)=λa,其中
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let g(x) = f(x) - λξ,那么g(a) = λ(b-a),g(b) = λ(a-b),so g(a)= - g(b),

存在ξ∈(a,b),使得g(ξ)=0,that is ,f(ξ)=λξ.

let h(x) = f(x)/λ,h(a) = b>h(b) = a,h(ξ)=ξ∈(a,b).

h‘（α）= [h(ξ)-h(a)]/(ξ-a) =(ξ-h(a))/(ξ-a) ,

h’（β）= [h(ξ)-h(b)]/(b-ξ) = (ξ-a)/(b-ξ),

so f‘（α）*f’（β）=λ²

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