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Question Number 173298 by mnjuly1970 last updated on 09/Jul/22

$$\mathrm{Q}:$$$$f\left(x\right)=e^x+x-4isgiven$$$$put:h\left(x\right)=ln(x-f^{-1}\left(x\right))$$$$find:{\mathrm{D}}_h=\left(domainofh\right)$$$$$$$$$$$$$$

Answered by floor(10²Eta[1]) last updated on 09/Jul/22

$$\mathrm{we}\mathrm{want}:\mathrm{x}-{\mathrm{f}}^{-1}\left(\mathrm{x}\right)>0\Rightarrow \mathrm{x}>{\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$$$\mathrm{note}\mathrm{that}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}+1>0\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$$$\Rightarrow \mathrm{f}\mathrm{is}\mathrm{increasing}\mathrm{i}.\mathrm{e}.,\mathrm{\forall }\mathrm{a},\mathrm{b}\in {\mathrm{D}}_{\mathrm{f}}\therefore \mathrm{a}>\mathrm{b}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)>\mathrm{f}\left(\mathrm{b}\right)$$$$$$$$\mathrm{now}\mathrm{back}\mathrm{to}\mathrm{x}>{\mathrm{f}}^{-1}\left(\mathrm{x}\right)$$$$\mathrm{since}\mathrm{f}\mathrm{is}\mathrm{incresing}\therefore \mathrm{f}\left(\mathrm{x}\right)>\mathrm{f}\left({\mathrm{f}}^{-1}\left(\mathrm{x}\right)\right)$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)>\mathrm{x}\Rightarrow {\mathrm{e}}^{\mathrm{x}}+\mathrm{x}-4>\mathrm{x}\Rightarrow {\mathrm{e}}^{\mathrm{x}}>4\Rightarrow \mathrm{x}>\ln 4$$$$\Rightarrow {\mathrm{D}}_{\mathrm{h}}=\{\mathrm{x}\in \mathbb{R}\mid \mathrm{x}>\ln 4\}$$

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