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Question Number 116167 by Ar Brandon last updated on 01/Oct/20
Show that ∀n≥2 the equation x^n =x+n admits a unique solution u_n ∈(1,2]
Showthat∀n⩾2theequationxn=x+nadmitsauniquesolutionun∈(1,2]
Answered by 1549442205PVT last updated on 01/Oct/20
Put f(x)=x^n −x−n.We have f(1)f(2)=(−n)(2^n −2−n)=−n(2^n −2−n)≤0(1) Indeed,put g(n)=2^n −n−2 then g′(n)=2^n ln2−1>4ln2−1>0.Hence, g(n) is an increase function on [2;∞) ⇒g(n)>g(2)=0∀n≥2 Thus,(1) is true ∀ n≥2,so by Rolle′s theorem we infer f(x) has a root u_n ∈(1,2).Furthermore,f ′(x)=nx^(n−1) −1 >0∀x∈(1,+∞) so u_n is unique root since f(x)is an icrease on [1,+∞)
Putf(x)=xn−x−n.Wehavef(1)f(2)=(−n)(2n−2−n)=−n(2n−2−n)⩽0(1)Indeed,putg(n)=2n−n−2theng′(n)=2nln2−1>4ln2−1>0.Hence,g(n)isanincreasefunctionon[2;∞)⇒g(n)>g(2)=0∀n⩾2Thus,(1)istrue∀n⩾2,sobyRolle′stheoremweinferf(x)hasarootun∈(1,2).Furthermore,f′(x)=nxn−1−1>0∀x∈(1,+∞)sounisuniquerootsincef(x)isanicreaseon[1,+∞)

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