设函数f(x)在[a,b]上连续,且a
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设函数f(x)在[a,b]上连续,且a
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设F(X)=f(x)-x
f(x)在(a.b)连续,则F(X)也连续
F(a)=f(a)-a
F(b)=f(b)-b
又a<f(x)<b
故F(a)>0,F(b)<0
连续函数的零点定理有存在ξ (a,b)使得F(x)=0
即为结果
F(x)=f(x)-x,Rolla定理
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