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Question Number 38587 by tawa tawa last updated on 27/Jun/18
solve for x: e^x + e^x^2 + e^x^3 = 3 + x + x^2 + x^3
$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}:\:\:\:\:\:\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{e}^{\mathrm{x}^{\mathrm{2}} } \:+\:\mathrm{e}^{\mathrm{x}^{\mathrm{3}} } \:=\:\:\mathrm{3}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18
x=0
$${x}=\mathrm{0} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18
left hand side curve...exponential and right hand side curve...algebric...intersect...at one point(0,3)
$${left}\:{hand}\:{side}\:{curve}…{exponential}\:{and}\:{right} \\ $$$${hand}\:{side}\:{curve}…{algebric}…{intersect}…{at}\:{one} \\ $$$${point}\left(\mathrm{0},\mathrm{3}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18
Answered by MrW3 last updated on 29/Jun/18
let′s have a look at the function f(x)=e^x −x f′(x)=e^x −1=0 ⇒x=0⇒f(0)=1 f′′(x)=e^x ⇒f′′(0)=1>0 that means f(x)=e^x −x has its minimum at x=0 and the minimum is 1, or in other words: f(x)=e^x −x≥1 similarly e^x^2 −x^2 ≥1, e^x^3 −x^3 ≥1 ⇒(e^x −x)+(e^x^2 −x^2 )+(e^x^3 −x^3 )≥3 ⇒e^x +e^x^2 +e^x^3 ≥3+x+x^2 +x^3 the = sign is valid only at x=0. that is to say e^x +e^x^2 +e^x^3 =3+x+x^2 +x^3 has only one solution x=0.
$${let}'{s}\:{have}\:{a}\:{look}\:{at}\:{the}\:{function} \\ $$$${f}\left({x}\right)={e}^{{x}} −{x} \\ $$$${f}'\left({x}\right)={e}^{{x}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}''\left({x}\right)={e}^{{x}} \Rightarrow{f}''\left(\mathrm{0}\right)=\mathrm{1}>\mathrm{0} \\ $$$${that}\:{means}\:{f}\left({x}\right)={e}^{{x}} −{x}\:{has}\:{its}\:{minimum} \\ $$$${at}\:{x}=\mathrm{0}\:{and}\:{the}\:{minimum}\:{is}\:\mathrm{1},\:{or} \\ $$$${in}\:{other}\:{words}: \\ $$$${f}\left({x}\right)={e}^{{x}} −{x}\geqslant\mathrm{1} \\ $$$${similarly}\:{e}^{{x}^{\mathrm{2}} } −{x}^{\mathrm{2}} \geqslant\mathrm{1},\:{e}^{{x}^{\mathrm{3}} } −{x}^{\mathrm{3}} \geqslant\mathrm{1} \\ $$$$\Rightarrow\left({e}^{{x}} −{x}\right)+\left({e}^{{x}^{\mathrm{2}} } −{x}^{\mathrm{2}} \right)+\left({e}^{{x}^{\mathrm{3}} } −{x}^{\mathrm{3}} \right)\geqslant\mathrm{3} \\ $$$$\Rightarrow{e}^{{x}} +{e}^{{x}^{\mathrm{2}} } +{e}^{{x}^{\mathrm{3}} } \geqslant\mathrm{3}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \\ $$$${the}\:=\:{sign}\:{is}\:{valid}\:{only}\:{at}\:{x}=\mathrm{0}. \\ $$$${that}\:{is}\:{to}\:{say} \\ $$$${e}^{{x}} +{e}^{{x}^{\mathrm{2}} } +{e}^{{x}^{\mathrm{3}} } =\mathrm{3}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \:{has}\:{only} \\ $$$${one}\:{solution}\:{x}=\mathrm{0}. \\ $$
Commented by tawa tawa last updated on 29/Jun/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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