设f(x)在【a,b】上连续,(a,b)内可导,f(a)·f(b)>0,f(a)f【(a+b)/2】设f(x)在【a,b】上连续,(a,b)内可导,且f(a)·f(b)>0,f(a)·f【(a+b)/2】
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![设f(x)在【a,b】上连续,(a,b)内可导,f(a)·f(b)>0,f(a)f【(a+b)/2】设f(x)在【a,b】上连续,(a,b)内可导,且f(a)·f(b)>0,f(a)·f【(a+b)/2】](/uploads/image/z/11940047-71-7.jpg?t=%E8%AE%BEf%EF%BC%88x%EF%BC%89%E5%9C%A8%E3%80%90a%2Cb%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%EF%BC%88a%2Cb%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%2Cf%EF%BC%88a%EF%BC%89%C2%B7f%EF%BC%88b%EF%BC%89%3E0%2Cf%EF%BC%88a%EF%BC%89f%E3%80%90%EF%BC%88a%2Bb%EF%BC%89%2F2%E3%80%91%E8%AE%BEf%EF%BC%88x%EF%BC%89%E5%9C%A8%E3%80%90a%2Cb%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%EF%BC%8C%EF%BC%88a%EF%BC%8Cb%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%EF%BC%8C%E4%B8%94f%EF%BC%88a%EF%BC%89%C2%B7f%EF%BC%88b%EF%BC%89%3E0%EF%BC%8Cf%EF%BC%88a%EF%BC%89%C2%B7f%E3%80%90%EF%BC%88a%2Bb%EF%BC%89%2F2%E3%80%91)
设f(x)在【a,b】上连续,(a,b)内可导,f(a)·f(b)>0,f(a)f【(a+b)/2】设f(x)在【a,b】上连续,(a,b)内可导,且f(a)·f(b)>0,f(a)·f【(a+b)/2】
设f(x)在【a,b】上连续,(a,b)内可导,f(a)·f(b)>0,f(a)f【(a+b)/2】
设f(x)在【a,b】上连续,(a,b)内可导,且f(a)·f(b)>0,f(a)·f【(a+b)/2】
设f(x)在【a,b】上连续,(a,b)内可导,f(a)·f(b)>0,f(a)f【(a+b)/2】设f(x)在【a,b】上连续,(a,b)内可导,且f(a)·f(b)>0,f(a)·f【(a+b)/2】
令 G(x) = f(x) * e^(-λx) ,G(x)在【a,b】上连续,(a,b)可导,且 G(a) = G(b) = 0
G(x)在【a,b】上满足罗尔中值定理,至少存在一点 c ∈(a,b),使得 G'(c) = 0
即有 f '(c) = λf(c) 成立.
用罗尔定理
令 G(x) = f(x) -f'(x) G(x)在【a,b】上连续,(a,b)可导,因为f(a)·f(b)>0,f(a)f【(a+b)/2】<0 所以f(a)*f(b)<0
G(x)在【a,b】上满足罗尔中值定理, 至少存在一点 c ∈(a,b), 使得 G'(c) = 0
即有 f '(c) = f(c) 成立1.因为f(a)·f(b)>0,f(a)f【(a+b)/2】...
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令 G(x) = f(x) -f'(x) G(x)在【a,b】上连续,(a,b)可导,因为f(a)·f(b)>0,f(a)f【(a+b)/2】<0 所以f(a)*f(b)<0
G(x)在【a,b】上满足罗尔中值定理, 至少存在一点 c ∈(a,b), 使得 G'(c) = 0
即有 f '(c) = f(c) 成立
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