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Question Number 206616 by universe last updated on 20/Apr/24

Answered by aleks041103 last updated on 21/Apr/24

$$e^{3x}isincreasingandcontinuousforx>0$$$$ln\left(x\right)isincr.andcont.forx>0$$$$\Rightarrow f\left(x\right)isincr.andcont.forx>0$$$$\underset{x\to 0}{lim}f\left(x\right)\to -\mathrm{\infty }$$$$\underset{x\to +\mathrm{\infty }}{lim}f\left(x\right)\to +\mathrm{\infty }$$$$\Rightarrow \mathrm{\exists }{f}^{-1}:\mathbb{R}\to \mathbb{R}$$$$f\left(f^{-1}\left(x\right)\right)=x$$$$\Rightarrow {\left(\frac{df\left(y\right)}{dy}\right)}_{y=f^{-1}\left(x\right)}\frac{{df}^{-1}\left(x\right)}{dx}=1=\frac{d}{dx}x$$$$\Rightarrow \frac{d{f}^{-1}\left(x\right)}{dx}=\frac{1}{f{}^{\prime }\left(f^{-1}\left(x\right)\right)}$$$$f{}^{{}^{\prime }}\left(x\right)=3e^{3x}+\frac{1}{x}$$$$\Rightarrow {Df}^{-1}\left(e^3\right)={(3e^{3f^{-1}\left(e^3\right)}+\frac{1}{f^{-1}\left(e^3\right)})}^{-1}$$$$y=f^{-1}\left(e^3\right)\Rightarrow f\left(y\right)=e^{3y}+ln\left(y\right)=e^3$$$$\Rightarrow y=1isasoln.$$$$since\mathrm{\exists }!y:f\left(y\right)=e^3\Rightarrow y=1=f^{-1}\left(e^3\right)$$$$\Rightarrow {Df}^{-1}\left(e^3\right)=\frac{1}{3e^{3.1}+\frac{1}{1}}=\frac{1}{1+3e^3}$$

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