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单词 ProofOfRollesTheorem
释义

proof of Rolle’s theorem


Because f is continuousMathworldPlanetmathPlanetmath on a compact (closed and bounded) interval I=[a,b], it attains itsmaximum and minimum values. In case f(a)=f(b) is both the maximum andthe minimum, then there is nothing more to say, for then f is a constant function andf0 on the whole interval I. So suppose otherwise, and f attains an extremumMathworldPlanetmathin the open interval(a,b), and without loss of generality, let this extremum be a maximum, considering -f inlieu of f as necessary. We claim that at this extremum f(c) we have f(c)=0, with a<c<b.

To show this, note that f(x)-f(c)0 for allxI, because f(c) is the maximum. By definition of the derivativePlanetmathPlanetmath, we have that

f(c)=limxcf(x)-f(c)x-c.

Looking at the one-sided limits, we note that

R=limxc+f(x)-f(c)x-c0

because the numerator in the limit is nonpositive in the interval I,yet x-c>0, as x approaches c from the right. Similarly,

L=limxc-f(x)-f(c)x-c0.

Since f is differentiableMathworldPlanetmathPlanetmath at c, the left and right limits must coincide, so0L=R0, that is to say, f(c)=0.

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更新时间:2025/5/5 0:02:32