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Question Number 207082 by SEKRET last updated on 06/May/24

$$\mathbf{Let}\mathbf{g}\left(\mathbf{x}\right)\mathbf{be}\mathbf{the}\mathbf{inverse}\mathbf{function}\mathbf{of}$$$$$$$$-->\mathbf{f}\left(\mathbf{x}\right)={\mathbf{x}}^3+3{\mathbf{x}}^2+4\mathbf{x}+5<--$$$$$$$$$$$$\mathbf{Evaluate}\underset{\mathbf{n}\to \mathrm{\infty }}{\mathbf{Lim}}4\mathbf{n}\cdot (\mathbf{g}(1+\frac{1}{\mathbf{n}})-\mathbf{g}(1-\frac{2}{\mathbf{n}}))=?$$$$$$$$$$

Commented by Frix last updated on 06/May/24

$$\mathrm{I}\mathrm{get}-3$$

Commented by SEKRET last updated on 06/May/24

$$\mathbf{a})1\mathbf{b})2\mathbf{c})3\mathbf{d})4\mathbf{e})5$$

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

$$\mathrm{We}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{even}\mathrm{need}\mathrm{to}\mathrm{find}g\left(x\right)$$$$f\left(x\right)=x^3+3x^2+4x+5$$$$f\left(x\right)=\{\begin{array}{l}1+\frac{1}{n}\\ 1-\frac{2}{n}\end{array}$$$$n\to \mathrm{\infty }\Rightarrow f\left(x\right)=1\Rightarrow x=-2$$$$P=\left(\begin{array}{c}-2\\ 1\end{array}\right)$$$$\mathrm{The}\mathrm{slope}\mathrm{in}P\mathrm{is}{f}^{\prime }(-2)=4$$$$\Rightarrow$$$$\mathrm{The}\mathrm{slope}\mathrm{of}{f}^{-1}\left(x\right)=g\left(x\right)\mathrm{at}Q=\left(\begin{array}{c}1\\ -2\end{array}\right)$$$$\mathrm{is}-\frac{1}{4}$$$$\Rightarrow$$$$\mathrm{We}\mathrm{can}\mathrm{approximate}g\left(x\right)\mathrm{with}\mathrm{the}\mathrm{tangent}$$$$\mathrm{in}Q:y=-\frac{x+7}{4}$$$$y_1=-\frac{1+\frac{1}{n}+7}{4}\wedge y_2=\frac{1-\frac{2}{n}+7}{4}$$$$4n(y_1-y_2)=-3$$

Commented by SEKRET last updated on 06/May/24

$$\mathbf{thank}\mathbf{you}\mathbf{sir}$$

Answered by Berbere last updated on 06/May/24

$$g^3\left(x\right)+3g^2\left(x\right)+4g\left(x\right)+5=x$$$$\Rightarrow \underset{k=1}{\sum ^3}{g}^k(1+\frac{1}{n})-g^k(1-\frac{2}{n})=\frac{3}{n}$$$$a^3-b^3=(a-b)(a^2+b^2-ab)$$$$a^2-b^2=(a-b)(a+b)$$$$\Rightarrow (g(1+\frac{1}{n})-g(1-\frac{2}{n})){[g^2(1+\frac{1}{n})+g^2(1-\frac{2}{n})-g(1+\frac{1}{n})g(1-\frac{2}{n})+3(g(1+\frac{1}{n})+g(1-\frac{2}{n})+4)]}_B=\frac{3}{n}$$$$4n(g(1+\frac{1}{n})-g(1-\frac{2}{n}))=\frac{12}{B}$$$$\underset{n\to \mathrm{\infty }}{\lim }B=g^2\left(1\right)+6g\left(1\right)+4$$$$g\left(1\right)=a;\Rightarrow f\left(a\right)=1$$$$\Rightarrow a^3+3a^2+4a+5=1$$$$\Rightarrow {(a+1)}^3+a+1+2=0\Rightarrow a+1=-1\Rightarrow a=-2$$$$\Rightarrow B=-4$$$$\Rightarrow 4n(g(1+\frac{1}{n})-\mathit{g}(1-\frac{2}{\mathit{n}})]=\frac{12}{-4}=-3$$$$$$

Commented by SEKRET last updated on 06/May/24

$$\mathbf{thank}\mathbf{you}\mathbf{sir}$$

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