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Question Number 47915 by behi83417@gmail.com last updated on 16/Nov/18

Answered by MJS last updated on 16/Nov/18

$$f\left(x\right)\mathrm{is}\mathrm{kind}\mathrm{of}\mathrm{a}\mathrm{parabola}.\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}$$$$\mathrm{can}\mathrm{find}{f}^{-1}\left(x\right)\mathrm{but}\mathrm{of}\mathrm{course}\mathrm{we}\mathrm{can}\mathrm{approximately}$$$$\mathrm{calculate}\mathrm{the}\mathrm{area}.\mathrm{area}\mathrm{between}f\mathrm{and}{f}^{-1}=$$$$=2\times \mathrm{area}\mathrm{between}f\mathrm{and}y=x$$$$x^2-\sqrt{x^2+x+1}=x\Rightarrow x_1\approx -.557317;x_2\approx 2.27745$$$$2\mid \underset{x_1}{\stackrel{x_2}{\int }}{x}^2-x-\sqrt{x^2+x+1}dx\mid =$$$$=\mid 2{[\frac{x^3}{3}-\frac{x^2}{2}-\frac{(2x+1)\sqrt{x^2+x+1}}{4}-\frac{3\ln (2x+1+2\sqrt{x^2+x+1})}{8}]}_{x_1}^{x_2}\mid \approx$$$$\approx 6.47706$$

Commented by mr W last updated on 17/Nov/18

$$thisisonlyanapproximationasMJS$$$$sirsaid.$$$$tobeexactthecommonareaoff\left(x\right)$$$$and{f}^{-1}\left(x\right)isnotalways2\times areabtw.$$$$f\left(x\right)andy=x,asinthiscase.$$$$Thecommonareaisthegreenshaded$$$$area.Buttheareabetweenf\left(x\right)and$$$$y=xincludesalsotheblueshadedarea.$$

Commented by mr W last updated on 17/Nov/18

Commented by MJS last updated on 17/Nov/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}$$$$\mathrm{in}\mathrm{our}\mathrm{case},\mathrm{the}\mathrm{common}\mathrm{points}\notin y=x\mathrm{are}$$$$\left(\begin{array}{c}0\\ -1\end{array}\right)\mathrm{and}\left(\begin{array}{c}-1\\ 0\end{array}\right).{\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{try}\mathrm{to}\mathrm{find}\mathrm{this}\mathrm{small}\mathrm{area}...$$$$$$$$...\mathrm{because}\mathrm{of}\mathrm{the}\mathrm{symmetry}\mathrm{of}f\mathrm{and}{f}^{-1}\mathrm{the}$$$$\mathrm{small}\mathrm{area}\mathrm{is}$$$$2(\mid \underset{-.557317}{\stackrel{0}{\int }}f\left(x\right)dx\mid -\mid \underset{-1}{\stackrel{0}{\int }}f\left(x\right)dx-\underset{-.557317}{\stackrel{0}{\int }}(f\left(x\right)+.557317)dx\mid )=.0133537$$$$\mathrm{and}\mathrm{the}\mathrm{resulting}\mathrm{area}\mathrm{between}f\mathrm{and}{f}^{-1}\mathrm{is}$$$$\approx 6.46370$$

Commented by ajfour last updated on 17/Nov/18

$$Ithoughtofaway..$$

Commented by MJS last updated on 17/Nov/18

$$...\mathrm{still}\mathrm{trying}\mathrm{to}\mathrm{find}{f}^{-1}...$$

Answered by ajfour last updated on 17/Nov/18

$$let{f}^{-1}\left(x\right)=y,then$$$$x=y^2-\sqrt{y^2+y+1}$$$$\Rightarrow y^2=x+\sqrt{y^2+y+1}$$$$let{\left[z\left(y\right)\right]}^2=x-\sqrt{y^2+y+1}$$$$\Rightarrow y^2+z^2=2x...\left(i\right)$$$$\mathrm{\&}{(y^2-z^2)}^2=4(y^2+y+1)...\left(ii\right)$$$$Solety=2x\cos \theta ,z=2x\sin \theta$$$$ThenFrom\left(ii\right)$$$${\left[4x^2(2\cos {}^2\theta -1)\right]}^2=4(4x^2\cos {}^2\theta$$$$+2x\cos \theta +1)$$$$\Rightarrow 4x^4(4\cos {}^4\theta -4\cos {}^2\theta +1)$$$$=4x^2\cos {}^2\theta +2x\cos \theta +1$$$$\Rightarrow {\left(2x\cos \theta \right)}^4-(1+4x^2){\left(2x\cos \theta \right)}^2$$$$-2x\cos \theta -1=0$$$$backagainy=2x\cos \theta$$$$\Rightarrow y^4-\mathit{a}{y}^2-y-1=0$$$$where\mathit{a}=1+4x^2$$$$Let(y^2+Ay+B)(y^2-Ay+C)=0$$$$\iff y^4-{ay}^2-y-1=0$$$$\Rightarrow B+C-A^2=-a$$$$A(B-C)=1$$$$BC=-1$$$$\Rightarrow {(B-C)}^2=\frac{1}{A^2};{(B+C)}^2={(A^2-a)}^2$$$$\Rightarrow {(A^2-a)}^2-\frac{1}{A^2}=-4$$$$or{A}^2{(A^2-a)}^2+4A^2-1=0$$$$let{A}^2=s$$$$s{(s-a)}^2+4s-1=0$$$$........................................$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Commented by ajfour last updated on 17/Nov/18

$$Tworootsshouldgetrejected;$$$$consequenceofsquaring..$$

Commented by behi83417@gmail.com last updated on 17/Nov/18

$$thankyouallofmydears.godbless$$$$youall.$$

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