作业帮设f(x)在[a,b]上连续,在（a,b）内可导,且f(a)=f(b)=0试证：在(a,b)内至少存在一点δ使得f(δ)+f'(δ)=0
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![设f(x)在[a,b]上连续,在（a,b）内可导,且f(a)=f(b)=0试证：在(a,b)内至少存在一点δ使得f(δ)+f'(δ)=0](/uploads/image/z/13945996-28-6.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%EF%BC%88a%2Cb%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%28a%29%3Df%28b%29%3D0%E8%AF%95%E8%AF%81%EF%BC%9A%E5%9C%A8%28a%2Cb%29%E5%86%85%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9%CE%B4%E4%BD%BF%E5%BE%97f%28%CE%B4%29%2Bf%27%28%CE%B4%29%3D0)
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设f(x)在[a,b]上连续,在（a,b）内可导,且f(a)=f(b)=0试证：在(a,b)内至少存在一点δ使得f(δ)+f'(δ)=0
设f(x)在[a,b]上连续,在（a,b）内可导,且f(a)=f(b)=0
试证：在(a,b)内至少存在一点δ使得f(δ)+f'(δ)=0
设f(x)在[a,b]上连续,在（a,b）内可导,且f(a)=f(b)=0试证：在(a,b)内至少存在一点δ使得f(δ)+f'(δ)=0
设 g(x) = f(x)*e^x
则：g(a) = g(b) = 0
由罗尔定理得：至少存在一点a < δ < b
使得：g'(δ) = 0
即：f'(δ)*e^δ + f(δ)*e^δ = 0
即：f(δ)+f'(δ)=0
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