设f (x)在[1,2]上连续,在(1,2)内可导且f(1) =1/2,f (2)= 2,求证:∃ξ ∈(1
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设f (x)在[1,2]上连续,在(1,2)内可导且f(1) =1/2,f (2)= 2,求证:∃ξ ∈(1,2)使 f ′(ξ ) = 2f (ξ )/ξ
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![设f (x)在[1,2]上连续,在(1,2)内可导且f(1) =1/2,f (2)= 2,求证:∃ξ ∈(1](/uploads/image/z/4193780-68-0.jpg?t=%E8%AE%BEf+%28x%29%E5%9C%A8%5B1%2C2%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%281%2C2%29%E5%86%85%E5%8F%AF%E5%AF%BC%E4%B8%94f%281%29+%3D1%2F2%2Cf+%282%29%3D+2%2C%E6%B1%82%E8%AF%81%3A%26%238707%3B%CE%BE+%E2%88%88%281)
设F(x)=f(x)/x^2,则F(x)在[1,2]上连续,在(1,2)内可导.
F(1)=f(1)=1/2、F(2)=f(2)/4=2/4=1/2=F(1).
F'(x)=[x^2f'(x)-2xf(x)]/x^4=[xf'(x)-2f(x)]/x^3.
由中值定理可得:
∃ξ ∈(1,2)使 F ′(ξ )=[ξf'(ξ)-2f(ξ)]/ξ^3=0.
即∃ξ ∈(1,2)使ξf'(ξ)-2f(ξ)=0、即f ′(ξ ) = 2f (ξ )/ξ.
F(1)=f(1)=1/2、F(2)=f(2)/4=2/4=1/2=F(1).
F'(x)=[x^2f'(x)-2xf(x)]/x^4=[xf'(x)-2f(x)]/x^3.
由中值定理可得:
∃ξ ∈(1,2)使 F ′(ξ )=[ξf'(ξ)-2f(ξ)]/ξ^3=0.
即∃ξ ∈(1,2)使ξf'(ξ)-2f(ξ)=0、即f ′(ξ ) = 2f (ξ )/ξ.
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