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Question Number 32701 by caravan msup abdo. last updated on 31/Mar/18

$$letgivef\left(x\right)=\frac{x}{\sqrt{x+1}}$$$$1)calculate{f}^{-1}\left(x\right))$$$$2)calculate{\left(f^{-1}\right)}^{{}^{\prime }}\left(x\right).$$

Commented by Rio Mike last updated on 31/Mar/18

$$\mathrm{solution}$$$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y}$$$$\mathrm{y}=\frac{\mathrm{x}}{\sqrt{\mathrm{x}+1}}$$$$\frac{\mathrm{y}\sqrt{\mathrm{x}+1}}{\mathrm{y}}=\frac{\mathrm{x}}{\mathrm{y}}$$$$\mathrm{x}+1={\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^2$$$$\mathrm{x}=\frac{{\mathrm{x}}^2}{{\mathrm{y}}^2}-1$$$$\mathrm{replacing}\mathrm{y}\mathrm{by}\mathrm{x}.$$$${\mathrm{f}}^{-1}\left(\mathrm{x}\right)=\left(\frac{{\mathrm{x}}^{2_{}}}{{\mathrm{x}}^2}\right)-1$$

Commented by MJS last updated on 31/Mar/18

$$\mathrm{you}\mathrm{must}\mathrm{exchange}x\mathrm{and}y$$$$\mathbf{each}x\mathrm{must}\mathrm{be}\mathrm{replaced}\mathrm{by}y$$$$\mathrm{and}\mathbf{each}y\mathrm{must}\mathrm{be}\mathrm{replaced}\mathrm{by}x$$

Commented by abdo imad last updated on 01/Apr/18

$$letputf\left(x\right)=y\iff x=f^{-1}\left(y\right)\Rightarrow \frac{x}{\sqrt{x+1}}=y\Rightarrow x=y\sqrt{x+1}$$$$withconditionax>-1and\frac{x}{y}>0\Rightarrow x^2=y^2(x+1)=y^{2}x+y^2\Rightarrow$$$$x^2-y^{2}x-y^2=0equ.wthunknownx$$$$\mathrm{\Delta }=y^4+4y^2⩾0\Rightarrow x_1=\frac{y^2+\sqrt{y^4+4y^2}}{2}$$$$x_2=\frac{y^2-\sqrt{y^4+4y^2}}{2}butwemusthavex+1>0lettry$$$$with{x}_{2}becauseitsclearthat{x}_1+1>0$$$$x_2+1=\frac{y^2-\sqrt{y^4+4y^2}}{2}+1=\frac{y^2+2-\sqrt{y^4+4y^2}}{2}$$$${(y^2+2)}^2-(y^4+4y^2)=4conditionrealized$$$$\frac{x_2}{y}=\frac{y^2-\mid y\mid \sqrt{y^2+4}}{2y}=y+\xi \frac{\sqrt{4+y^2}}{2}with{\xi }^2=1butthesine$$$$of\frac{x_2}{y}isnotconstantsotheso<utionoftheequ.is{x}_1$$$$and{f}^{-1}\left(x\right)=\frac{1}{2}(x^2+\sqrt{x^4+4x^2})$$$$2){\left(f^{-1}\right)}^{{}^{\prime }}\left(x\right)=\frac{1}{2}(2x+\frac{4x^3+8x}{2\sqrt{x^4+4x^2}})$$$$=x+\frac{2x^3+4x}{\sqrt{x^4+4x^2}}.$$

Answered by MJS last updated on 31/Mar/18

$$x=\frac{y}{\sqrt{y+1}}$$$$x\sqrt{y+1}=y$$$$x^{2}y+x^2=y^2$$$$y^2-x^{2}y-x^2=0$$$$y=\frac{x^2}{2}\pm \sqrt{\frac{x^4}{4}+x^2}$$$$y=\frac{x^2}{2}\pm \sqrt{\frac{x^2}{4}(x^2+4)}$$$$y=f^{-1}\left(x\right)=\frac{x^2}{2}\pm \frac{x}{2}\sqrt{x^2+4}$$$$$$$$y^{\prime }=\frac{2x}{2}\pm (\frac{1}{2}\times {(x^2+4)}^{\frac{1}{2}}+\frac{x}{2}\times \frac{1}{2}\times {(x^2+4)}^{-\frac{1}{2}}\times 2x)$$$$y^{\prime }=x\pm (\frac{1}{2}{(x^2+4)}^{\frac{1}{2}}+\frac{x^2}{2}{(x^2+4)}^{-\frac{1}{2}})$$$$y^{\prime }=x\pm \frac{1}{2}{(x^2+4)}^{-\frac{1}{2}}({(x^2+4)}^1+x^2)$$$$y^{\prime }={\left(f^{-1}\right)}^{\prime }\left(x\right)=x\pm \frac{x^2+2}{\sqrt{x^2+4}}$$

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