设f(x)在[a,b]上连续,且f(a)b,试证:在(a,b)内至少有一点P,使得f(P)=P.
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![设f(x)在[a,b]上连续,且f(a)b,试证:在(a,b)内至少有一点P,使得f(P)=P.](/uploads/image/z/15002230-22-0.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E4%B8%94f%28a%29b%2C%E8%AF%95%E8%AF%81%EF%BC%9A%E5%9C%A8%EF%BC%88a%2Cb%EF%BC%89%E5%86%85%E8%87%B3%E5%B0%91%E6%9C%89%E4%B8%80%E7%82%B9P%2C%E4%BD%BF%E5%BE%97f%28P%29%3DP.)
设f(x)在[a,b]上连续,且f(a)b,试证:在(a,b)内至少有一点P,使得f(P)=P.
设f(x)在[a,b]上连续,且f(a)b,试证:在(a,b)内至少有一点P,使得f(P)=P.
设f(x)在[a,b]上连续,且f(a)b,试证:在(a,b)内至少有一点P,使得f(P)=P.
构造新函数F(x)=f(x)-a,由题意知此函数在[a,b]上连续
因为f(a)0
由零点存在性定理得在(a,b)内至少有一点P,使得F(p)=0
即f(P)=P
令F(x)=f(x)-x则F(a)F(b)<0.且F(x)在[a,b]上连续
y由零点定理,结论成立
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