设函数f(x)在闭区间[0 1]上连续 在开区间(0 1)内可导 且f(0)＝f(1)＝0 f(1／

正确答案：证：(1)令φ(x)＝f(x)－x则φ(x)在[01上连续又φ(1)＝－1＜0φ(1／2)＝1／2＞0故由闭区间上连续函数的介值定理知存在η∈(1／21)使得φ(η)＝f(η)－η＝0即f(η)＝η． (2)设F(x)＝e^－λφ(x)＝e^－λx[f(x)－x则F(x)在[0η上连续在(xη)内可导且 F(0)＝0F(η)＝e^－ληφ(η)＝0 即F(x)在[xη上满足罗尔定理的条件故存在ε∈(xη)使得 Fˊ(ε)＝0即e^－λε｛fˊ(ε)－λ[f(ε)－ε－1｝＝0 从而fˊ(ε)－λ[fˊ(ε)－ε＝1
证：(1)令φ(x)＝f(x)－x,则φ(x)在[0,1上连续,又φ(1)＝－1＜0,φ(1／2)＝1／2＞0,故由闭区间上连续函数的介值定理知,存在η∈(1／2,1),使得φ(η)＝f(η)－η＝0,即f(η)＝η．(2)设F(x)＝e－λφ(x)＝e－λx[f(x)－x,则F(x)在[0,η上连续,在(x,η)内可导,且F(0)＝0,F(η)＝e－ληφ(η)＝0即F(x)在[x,η上满足罗尔定理的条件,故存在ε∈(x,η),使得Fˊ(ε)＝0,即e－λε｛fˊ(ε)－λ[f(ε)－ε－1｝＝0从而fˊ(ε)－λ[fˊ(ε)－ε＝1