Lesson 15.pdf - 15 The Mean Value Theorem The mean value theorem Theorem 0.1 If f is differentiable on an open interval(a b and if f assumes a maximum

# Lesson 15.pdf - 15 The Mean Value Theorem The mean value...

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15. The Mean Value TheoremJuly 29, 2019The mean value theoremTheorem 0.1Iffis differentiable on an open interval(a, b), and iffassumes a maximumor minimum value at a pointc(a, b), thenf0(c) = 0.Proof:Supposefhas a maximum atc(a, b) withf0(c)6= 0 (We seek a contradiction).Supposef0(c)>0.Since limxcf(x)-f(c)x-x>0, There is a neighborhoodN(c, δ) on whichf(x)-f(c)x-c>0.Then on [c, c+δ),x-c >0,f(x)-f(c)>0, i.e.,f(x)> f(c) on[c, c+δ). This contracts with the fact thatfhas a maximum atc.Supposef0(c)<0.Since limxcf(x)-f(c)x-x<0, There is a neighborhoodN(c, δ) on whichf(x)-f(c)x-c<0.Then on (c-δ, c),x-c <0,f(x)-f(c)>0, i.e.,f(x)> f(c) on(c-δ, c]. This also contracts with the fact thatfhas a maximum atc.Therefore,f0(c) = 0. The minimum case forfhas the same proof.Theorem 0.2(Rolle’s theorem) Iffis continuous on[a, b]and is differentiable on(a, b)withf(a) =f(b), then there existsc(a, b), such thatf0(c) = 0.Proof:Letk:=f(a) =f(b). Iff(x) =kfor allx[a, b], thenf0(x) = 0 for anyx(a, b)so that we can takecto be any point in (a, b).Iff6=constant, thenfhas a maximum or minimum at some point in (a, b). Assumethatfhas a maximum at some pointc[a, b]. Then apply Theorem 0.1 to implyf0(c) = 0.128
Theorem 0.3(The mean value theorem) Letfbe continuous on[a, b]and differentiable on(a, b). Then there exists at least one numberc(a, b)such thatf0(c) =f(b)-f(a)b-a.Proof:Setg(x) :=f(x)-f(b)-f(a)b-a(x-a) +f(a).Thengsatisfies the hypothesis of the Rolle’s theorem:gis continuous on [a, b] and differ-entiable on (a, b) withg(a) =g(b) = 0. By Rolle’s theorem. there exists a pointc(a, b)such thatg0(c) = 0, i.e.,f0(c)-f(b)-f(a)b-a= 0which impliesf(b)-f(a)b-a=f0(c).[Examples and remarks]1. The conclusion of the Mean Value Theorem can be written equivalently as: there existsat least one numberc(a, b) such thatf(b)-f(a) =f0(c)(b-a).

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