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Question Number 114878 by bobhans last updated on 21/Sep/20

$$Givenf\left(x\right)=\underset{0}{\stackrel{x}{\int }}\frac{dt}{\sqrt{1+t^3}}andg\left(x\right)bethe$$$$inversefunctionoff\left(x\right),theng{}^{″}\left(x\right)=\lambda g^2\left(x\right).$$$$thenthevalueof\lambda =$$

Answered by Olaf last updated on 21/Sep/20

$$g\mathrm{o}f\left(x\right)=x$$$$f^{\prime }\left(x\right)\times g^{\prime }\mathrm{o}f\left(x\right)=1$$$$g^{\prime }\mathrm{o}f\left(x\right)=\frac{1}{f^{\prime }\left(x\right)}=\sqrt{1+x^3}$$$$f^{\prime }\left(x\right)\times g^{″}\mathrm{o}f\left(x\right)=\frac{3x^2}{2\sqrt{1+x^3}}$$$$g^{″}\mathrm{o}f\left(x\right)=\frac{1}{f^{\prime }\left(x\right)}\left[\frac{3x^2}{2\sqrt{1+x^3}}\right]$$$$g^{″}\mathrm{o}f\left(x\right)=\sqrt{1+x^3}\left[\frac{3x^2}{2\sqrt{1+x^3}}\right]$$$$g^{″}\mathrm{o}f\left(x\right)=\frac{3}{2}{x}^2$$$$g^{″}\mathrm{o}f\left(x\right)=\frac{3}{2}{\left[g\mathrm{o}f\left(x\right)\right]}^2$$$$\mathrm{Let}y=f\left(x\right)$$$$g^{″}\left(y\right)=\frac{3}{2}{g}^2\left(y\right)$$$$g^{″}\left(y\right)=\lambda g^2\left(y\right)\mathrm{with}\lambda =\frac{3}{2}$$

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{\sqrt{1+{\mathrm{x}}^3}}\mathrm{now},{\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{f}}^{{}^{\prime }}\left\{\mathrm{g}\left(\mathrm{x}\right)\right\}}$$$${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=\sqrt{1+{\left\{\mathrm{g}\left(\mathrm{x}\right)\right\}}^3}$$$${\mathrm{g}}^{{}^{″}}\left(\mathrm{x}\right)=\frac{1}{2\sqrt{1+{\left\{\mathrm{g}\left(\mathrm{x}\right)\right\}}^3}}.{3\mathrm{g}}^2\left(\mathrm{x}\right).{\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)$$$$\lambda {\mathbf{g}}^2\left(\mathbf{x}\right)=\frac{1}{2{\mathbf{g}}^{\prime }\left(\mathbf{x}\right)}.3{\mathbf{g}}^2\left(\mathbf{x}\right).{\mathbf{g}}^{{}^{\prime }}\left(\mathbf{x}\right)$$$$\lambda =\frac{3}{2}$$

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