Question Number 207082 by SEKRET last updated on 06/May/24
$$\:\:\:\boldsymbol{\mathrm{Let}}\:\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)\:\:\boldsymbol{\mathrm{be}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{inverse}}\:\boldsymbol{\mathrm{function}}\:\:\:\boldsymbol{\mathrm{of}} \\ $$$$ \\ $$$$\:\:\:\:−−>\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{x}}+\mathrm{5}\:\:\:\:\:\:<−− \\ $$$$\:\: \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Evaluate}}\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\boldsymbol{\mathrm{Lim}}}\:\mathrm{4}\boldsymbol{\mathrm{n}}\centerdot\left(\:\boldsymbol{\mathrm{g}}\left(\mathrm{1}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}}\right)\:−\boldsymbol{\mathrm{g}}\left(\mathrm{1}−\frac{\mathrm{2}}{\boldsymbol{\mathrm{n}}}\right)\:\right)=? \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 06/May/24
$$\mathrm{I}\:\mathrm{get}\:−\mathrm{3} \\ $$
Commented by SEKRET last updated on 06/May/24
$$\left.\boldsymbol{\mathrm{a}}\left.\right)\left.\:\left.\mathrm{1}\left.\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}\right)\:\mathrm{2}\:\:\:\:\:\:\boldsymbol{\mathrm{c}}\right)\mathrm{3}\:\:\:\:\:\:\:\boldsymbol{\mathrm{d}}\right)\mathrm{4}\:\:\:\:\:\boldsymbol{\mathrm{e}}\right)\:\mathrm{5} \\ $$
Commented by Frix last updated on 06/May/24
$$\mathrm{All}\:\mathrm{given}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{wrong}. \\ $$
Commented by Frix last updated on 06/May/24
$$\mathrm{We}\:\mathrm{don}'\mathrm{t}\:\mathrm{even}\:\mathrm{need}\:\mathrm{to}\:\mathrm{find}\:{g}\left({x}\right) \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{5} \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\\{\mathrm{1}−\frac{\mathrm{2}}{{n}}}\end{cases} \\ $$$${n}\rightarrow\infty\:\Rightarrow\:{f}\left({x}\right)=\mathrm{1}\:\Rightarrow\:{x}=−\mathrm{2} \\ $$$${P}=\begin{pmatrix}{−\mathrm{2}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{The}\:\mathrm{slope}\:\mathrm{in}\:{P}\:\mathrm{is}\:{f}'\left(−\mathrm{2}\right)=\mathrm{4} \\ $$$$\Rightarrow \\ $$$$\mathrm{The}\:\mathrm{slope}\:\mathrm{of}\:{f}^{−\mathrm{1}} \left({x}\right)={g}\left({x}\right)\:\mathrm{at}\:{Q}=\begin{pmatrix}{\mathrm{1}}\\{−\mathrm{2}}\end{pmatrix} \\ $$$$\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{approximate}\:{g}\left({x}\right)\:\mathrm{with}\:\mathrm{the}\:\mathrm{tangent} \\ $$$$\mathrm{in}\:{Q}:\:{y}=−\frac{{x}+\mathrm{7}}{\mathrm{4}} \\ $$$${y}_{\mathrm{1}} =−\frac{\mathrm{1}+\frac{\mathrm{1}}{{n}}+\mathrm{7}}{\mathrm{4}}\wedge{y}_{\mathrm{2}} =\frac{\mathrm{1}−\frac{\mathrm{2}}{{n}}+\mathrm{7}}{\mathrm{4}} \\ $$$$\mathrm{4}{n}\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)=−\mathrm{3} \\ $$
Commented by SEKRET last updated on 06/May/24
$$\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{you}}\:\:\boldsymbol{\mathrm{sir}} \\ $$
Answered by Berbere last updated on 06/May/24
![g^3 (x)+3g^2 (x)+4g(x)+5=x ⇒Σ_(k=1) ^3 g^k (1+(1/n))−g^k (1−(2/n))=(3/n) a^3 −b^3 =(a−b)(a^2 +b^2 −ab) a^2 −b^2 =(a−b)(a+b) ⇒(g(1+(1/n))−g(1−(2/n)))[g^2 (1+(1/n))+g^2 (1−(2/n))−g(1+(1/n))g(1−(2/n))+3(g(1+(1/n))+g(1−(2/n))+4)]_B =(3/n) 4n(g(1+(1/n))−g(1−(2/n)))=((12)/B) lim_(n→∞) B=g^2 (1)+6g(1)+4 g(1)=a;⇒f(a)=1 ⇒a^3 +3a^2 +4a+5=1 ⇒(a+1)^3 +a+1+2=0⇒a+1=−1⇒a=−2 ⇒B=−4 ⇒4n(g(1+(1/n))−g(1−(2/n))]=((12)/(−4))=−3](https://www.tinkutara.com/question/Q207093.png)
$${g}^{\mathrm{3}} \left({x}\right)+\mathrm{3}{g}^{\mathrm{2}} \left({x}\right)+\mathrm{4}{g}\left({x}\right)+\mathrm{5}={x} \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{g}^{{k}} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)−{g}^{{k}} \left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)=\frac{\mathrm{3}}{{n}} \\ $$$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{ab}\right) \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({a}−{b}\right)\left({a}+{b}\right) \\ $$$$\Rightarrow\left({g}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)−{g}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)\right)\left[{g}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)+{g}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)−{g}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right){g}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)+\mathrm{3}\left({g}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)+{g}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)+\mathrm{4}\right)\right]_{{B}} =\frac{\mathrm{3}}{{n}} \\ $$$$\mathrm{4}{n}\left({g}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)−{g}\left(\mathrm{1}−\frac{\mathrm{2}}{{n}}\right)\right)=\frac{\mathrm{12}}{{B}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{B}={g}^{\mathrm{2}} \left(\mathrm{1}\right)+\mathrm{6}{g}\left(\mathrm{1}\right)+\mathrm{4} \\ $$$${g}\left(\mathrm{1}\right)={a};\Rightarrow{f}\left({a}\right)=\mathrm{1} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{5}=\mathrm{1} \\ $$$$\Rightarrow\left({a}+\mathrm{1}\right)^{\mathrm{3}} +{a}+\mathrm{1}+\mathrm{2}=\mathrm{0}\Rightarrow{a}+\mathrm{1}=−\mathrm{1}\Rightarrow{a}=−\mathrm{2} \\ $$$$\Rightarrow{B}=−\mathrm{4} \\ $$$$\Rightarrow\mathrm{4}{n}\left({g}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)−\boldsymbol{{g}}\left(\mathrm{1}−\frac{\mathrm{2}}{\boldsymbol{{n}}}\right)\right]=\frac{\mathrm{12}}{−\mathrm{4}}=−\mathrm{3} \\ $$$$ \\ $$
Commented by SEKRET last updated on 06/May/24
$$\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{sir}} \\ $$