设f(x)在[0,π/2]上连续,在(0,π/2)内可导,且f(π/2)=0,试证存在一点ζ∈(0,π/2)使f(ζ)+tanζ*f ‘(ζ)=0
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设f(x)在[0,π/2]上连续,在(0,π/2)内可导,且f(π/2)=0,试证存在一点ζ∈(0,π/2)使f(ζ)+tanζ*f‘(ζ)=0设f(x)在[0,π/2]上连续,在(0,π/2)内可导
设f(x)在[0,π/2]上连续,在(0,π/2)内可导,且f(π/2)=0,试证存在一点ζ∈(0,π/2)使f(ζ)+tanζ*f ‘(ζ)=0
设f(x)在[0,π/2]上连续,在(0,π/2)内可导,且f(π/2)=0,试证存在一点ζ∈(0,π/2)使f(ζ)+tanζ*
f ‘(ζ)=0
设f(x)在[0,π/2]上连续,在(0,π/2)内可导,且f(π/2)=0,试证存在一点ζ∈(0,π/2)使f(ζ)+tanζ*f ‘(ζ)=0
f(ζ)+tanζ * f ‘(ζ)=0
两边同乘以cosζ可得
cosζf(ζ) + sinζf'(ζ)=0
下面是证明:
设g(x)=f(x)sin(x)
所以g'(x)=f(x)cosx + f'(x)sinx
g(0)=0,g(π/2)=0
所以存在ζ∈(0,π/2)使得g'(ζ)=0
而g'(ζ)=f(ζ)cosx+f'(ζ)sinζ=0
因为cosζ≠0
所以f(ζ)+f'(ζ)tanζ=0
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