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Question Number 103322 by abony1303 last updated on 14/Jul/20

$$\mathrm{A}\mathrm{differentiable}\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)\mathrm{satisfies}$$$$\mathrm{f}({\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x})=2^{\mathrm{x}+1}\mathrm{for}\mathrm{every}\mathrm{real}\mathrm{number}x.$$$$\mathrm{When}\mathrm{g}\left(\mathrm{x}\right)\mathrm{is}\mathrm{the}\mathrm{inverse}\mathrm{function}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right),$$$$\mathrm{find}{\mathrm{g}}^{\prime }\left(4\right)?$$

Commented by abony1303 last updated on 14/Jul/20

$$\mathrm{pls}\mathrm{help}$$

Answered by Worm_Tail last updated on 14/Jul/20

$$f(x^3-x^2+x)=2^{x+1}\Rightarrow$$$$x^3-x^2+x=f^{-1}\left(2^{x+1}\right)$$$$x^3-x^2+x=g\left(2^{x+1}\right)$$$$3x^2-2x+1=2^{x+1}ln\left(2\right)g^{\prime }{\left(2^{x+1}\right)}_{x=1}$$$$3-2+1=4ln\left(2\right)g^{\prime }\left(4\right)$$$$2=4ln\left(2\right)g^{\prime }\left(4\right)\Rightarrow g^{\prime }\left(4\right)=\frac{1}{2ln\left(2\right)}$$

Commented by abony1303 last updated on 14/Jul/20

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}$$

Answered by mathmax by abdo last updated on 14/Jul/20

$$\mathrm{we}\mathrm{have}\mathrm{f}({\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x})=2^{\mathrm{x}+1}\Rightarrow {\mathrm{f}}^{-1}\left(2^{\mathrm{x}+1}\right)={\mathrm{x}}^3-{\mathrm{x}}^2+\mathrm{x}$$$$\mathrm{let}{2}^{\mathrm{x}+1}=\mathrm{t}\Rightarrow {\mathrm{e}}^{(\mathrm{x}+1)\ln 2}=\mathrm{t}\Rightarrow (\mathrm{x}+1)\ln 2=\ln \left(\mathrm{t}\right)\Rightarrow \mathrm{x}+1=\frac{\mathrm{lnt}}{\ln 2}\Rightarrow$$$$\mathrm{x}=\frac{\mathrm{lnt}}{\ln 2}-1\Rightarrow {\mathrm{f}}^{-1}\left(\mathrm{t}\right)={\left(\frac{\mathrm{lnt}}{\ln 2}\right)}^3-{\left(\frac{\mathrm{lnt}}{\ln 2}\right)}^2+\frac{lnt}{ln2}=g\left(t\right)\Rightarrow$$$${\mathrm{g}}^{{}^{\prime }}\left(\mathrm{t}\right)=\frac{1}{{\left(\ln 2\right)}^3}\times 3\frac{{\left(\mathrm{lnt}\right)}^2}{\mathrm{t}}-\frac{1}{{\left(\ln 2\right)}^2}\times 2\frac{\mathrm{lnt}}{\mathrm{t}}+\frac{1}{\mathrm{tln}2}\Rightarrow$$$${\mathrm{g}}^{{}^{\prime }}\left(4\right)=\frac{3}{4{\left(\ln 2\right)}^3}{\left(2\ln 2\right)}^2-\frac{2}{{\left(\ln 2\right)}^2}\times \frac{2\ln \left(2\right)}{4}+\frac{1}{4\ln 2}$$$$=\frac{3}{\ln 2}-\frac{1}{\ln 2}+\frac{1}{4\ln 2}=\frac{2}{\ln 2}+\frac{1}{4\ln 2}=(2+\frac{1}{4})\times \frac{1}{\ln 2}=\frac{9}{4\ln 2}$$

Commented by 1549442205 last updated on 15/Jul/20

$${\mathrm{Sir}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{idea}\mathrm{is}\mathrm{right},\mathrm{however}$$$$\mathrm{mistaked}\mathrm{at}\mathrm{third}\mathrm{line}:{\mathrm{f}}^{-1}\left(\mathrm{t}\right)={(\frac{\mathrm{lnt}}{\ln 2}-1)}^3-{(\frac{\mathrm{lnt}}{\ln 2}-1)}^2+\frac{\mathrm{lnt}}{\ln 2}-1$$$$={\left(\frac{\mathrm{lnt}}{\ln 2}\right)}^3-3{\left(\frac{\mathrm{lnt}}{\ln 2}\right)}^2+3\frac{\mathrm{lnt}}{\ln 2}-1-[{\left(\frac{\mathrm{lnt}}{\ln 2}\right)}^2-2\frac{\mathrm{lnt}}{\ln 2}+1]+\frac{\mathrm{lnt}}{\ln 2}-1$$$$={\left(\frac{\mathrm{lnt}}{\ln 2}\right)}^3-4{\left(\frac{\mathrm{lnt}}{\ln 2}\right)}^2+6\frac{\mathrm{lnt}}{\ln 2}-2$$$${\mathrm{g}}^{\prime }\left(\mathrm{t}\right)=\frac{1}{{\left(\ln 2\right)}^3}\times 3\frac{{\left(\mathrm{lnt}\right)}^2}{\mathrm{t}}-\frac{4}{{\left(\ln 2\right)}^2}\times 2\frac{\mathrm{lnt}}{\mathrm{t}}+\frac{6}{\mathrm{tln}2}$$$${\mathrm{g}}^{\prime }\left(4\right)=\frac{3}{\ln 2}-\frac{4}{\ln 2}+\frac{3}{2\ln 2}=\frac{1}{2\ln 2}$$

Commented by mathmax by abdo last updated on 14/Jul/20

$$\mathrm{yes}\mathrm{thank}\mathrm{you}\mathrm{sir}...$$

Commented by mathmax by abdo last updated on 14/Jul/20

$$\mathrm{i}\mathrm{forget}-1!$$

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