设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=1,证明存在c,d属于(a,b)使得e的(d-c)次方*[f(d)+f'(d)]=1
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![设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=1,证明存在c,d属于(a,b)使得e的(d-c)次方*[f(d)+f'(d)]=1](/uploads/image/z/5613291-27-1.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%28a%2Cb%29%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%28a%29%3Df%28b%29%3D1%2C%E8%AF%81%E6%98%8E%E5%AD%98%E5%9C%A8c%2Cd%E5%B1%9E%E4%BA%8E%28a%2Cb%29%E4%BD%BF%E5%BE%97e%E7%9A%84%28d-c%29%E6%AC%A1%E6%96%B9%2A%5Bf%28d%29%2Bf%27%28d%29%5D%3D1)
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设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=1,证明存在c,d属于(a,b)使得e的(d-c)次方*[f(d)+f'(d)]=1
设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=1,证明存在c,d属于(a,b)使得e的(d-c)次方*[f(d)+f'(d)]=1
设f(x)在[a,b]上连续,在(a,b)内可导,且f(a)=f(b)=1,证明存在c,d属于(a,b)使得e的(d-c)次方*[f(d)+f'(d)]=1
考虑函数G(x)=e^x*f(x)
G(a)=e^a,G(b)=e^b
G'(x)=e^x*(f(x)+f'(x)
由中值定理得存在一点d属于(a,b)使得
(G(b)-G(a))/(b-a)=(e^b-e^a)/(b-a)=G'(x)=e^d*(f(d)+f'(d))……式1
考虑J(x)=e^x
由中值定理得存在一点c属于(a,b)使得
(e^b-e^a)/(a-b)=e^c……式2
将式2代人式1,得e^(d-c)*[f(d)+f'(d)]=1
楼上的那位是对的
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