(P)如果函数f(x)在[α,b]上连续,在(α,b)内可导,则在(α,b)内至少存在一点P,使得f(b)-f(α)=______.
f′(p)(b-α) 。 解析: 由拉格朗日中值定理可得f′(p)=[f(b)-f(α)]/(b-α).
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