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Question Number 121175 by Ar Brandon last updated on 05/Nov/20

Answered by MJS_new last updated on 05/Nov/20

x<1  g(x)<1

$${x}<\mathrm{1} \\ $$$${g}\left({x}\right)<\mathrm{1} \\ $$

Commented by Ar Brandon last updated on 05/Nov/20

Any calculations or explanations, Sir ?��

Commented by MJS_new last updated on 05/Nov/20

e^x +e^(g(x)) =e ⇔ g(x)=ln (e−e^x )  e−e^x >0 ⇒ e>e^x  ⇒ x<1  lim_(x→1^− )  g(x) =−∞  lim_(x→−∞)  g(x) =1  ⇒ g(x)<1

$$\mathrm{e}^{{x}} +\mathrm{e}^{{g}\left({x}\right)} =\mathrm{e}\:\Leftrightarrow\:{g}\left({x}\right)=\mathrm{ln}\:\left(\mathrm{e}−\mathrm{e}^{{x}} \right) \\ $$$$\mathrm{e}−\mathrm{e}^{{x}} >\mathrm{0}\:\Rightarrow\:\mathrm{e}>\mathrm{e}^{{x}} \:\Rightarrow\:{x}<\mathrm{1} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:{g}\left({x}\right)\:=−\infty \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{g}\left({x}\right)\:=\mathrm{1} \\ $$$$\Rightarrow\:{g}\left({x}\right)<\mathrm{1} \\ $$

Commented by Ar Brandon last updated on 05/Nov/20

It's clear now. Thanks Sir

Commented by MJS_new last updated on 05/Nov/20

you′re welcome

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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