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Question Number 54975 by naka3546 last updated on 15/Feb/19

$$f\left(x\right)=x^3-2x+4$$$$f{}^{-1}\left(x\right)=?$$$$$$

Answered by mr W last updated on 15/Feb/19

$$y=x^3-2x+4$$$$x^3-2x+4-y=0$$$${(u+v)}^3-3uv(u+v)-(u^3+v^3)=0$$$$leyx=u+v$$$$\Rightarrow u^3{v}^3=\frac{8}{27}$$$$\Rightarrow u^3+v^3=y-4$$$$z^2-(y-4)z+\frac{8}{27}=0$$$$z=\frac{1}{2}[y-4\pm \sqrt{y^2-8y+\frac{400}{27}}]$$$$u=\frac{\sqrt[3]{y-4+\sqrt{y^2-8y+\frac{400}{27}}}}{\sqrt[3]{2}}$$$$u=\frac{\sqrt[3]{y-4-\sqrt{y^2-8y+\frac{400}{27}}}}{\sqrt[3]{2}}$$$$\Rightarrow x=u+v=\frac{\sqrt[3]{y-4+\sqrt{y^2-8y+\frac{400}{27}}}+\sqrt[3]{y-4-\sqrt{y^2-8y+\frac{400}{27}}}}{\sqrt[3]{2}}$$$$\Rightarrow f^{-1}\left(x\right)=\frac{\sqrt[3]{x-4+\sqrt{x^2-8x+\frac{400}{27}}}+\sqrt[3]{x-4-\sqrt{x^2-8x+\frac{400}{27}}}}{\sqrt[3]{2}}$$

Answered by MJS last updated on 15/Feb/19

$$f^{\prime }\left(x\right)=0$$$$3x^2-2=0\Rightarrow x=\pm \frac{\sqrt{6}}{3}$$$$f(-\frac{\sqrt{6}}{3})=4+\frac{4}{9}\sqrt{6}$$$$f\left(\frac{\sqrt{6}}{3}\right)=4-\frac{4}{9}\sqrt{6}$$$$f^{″}\left(x\right)=0$$$$6x=0\Rightarrow x=0$$$$f\left(0\right)=4$$$$$$$$\Rightarrow f^{-1}\left(x\right)\mathrm{has}\mathrm{got}3\mathrm{parts}\mathrm{which}\mathrm{are}\mathrm{real}$$$$\Rightarrow \mathrm{we}\mathrm{have}\mathrm{to}\mathrm{use}\mathrm{the}\mathrm{trigonometric}\mathrm{method}$$$$$$$$x=y^3-2y+4$$$$y^3-2y+(4-x)=0$$$$y_k=2\sqrt{-\frac{p}{3}}\sin \left(\frac{1}{3}(2\pi k+\arcsin \left(\frac{9q}{2p^2}\sqrt{-\frac{p}{3}}\right))\right)$$$$y_1=\frac{2\sqrt{6}}{3}\sin \left(\frac{1}{3}(\pi +\arcsin \frac{3\sqrt{6}(x-4)}{8})\right)$$$$y_2=-\frac{2\sqrt{6}}{3}\cos \left(\frac{1}{3}(\frac{\pi }{2}+\arcsin \frac{3\sqrt{6}(x-4)}{8})\right)$$$$y_3=-\frac{2\sqrt{6}}{3}\sin \left(\frac{1}{3}\arcsin \frac{3\sqrt{6}(x-4)}{8}\right)$$

Answered by Rio Mike last updated on 16/Feb/19

$$lety=-2x+4$$$$y-4=-2x$$$$\Rightarrow x_1=\frac{y-4}{-2}$$$$thenlety=x^3$$$$\Rightarrow x_2=\sqrt[3]{y}$$$$y=f^{-1}=x_1+x_2$$$$f^{-1}=\frac{x-4}{-2}+\sqrt[3]{x}$$$$f^{-1}=-\frac{x}{2}+2+\sqrt[3]{x}$$$$f^{-1}=\frac{-x+2\sqrt[3]{x}}{2}+2$$$$$$$$$$$$$$$$$$$$$$$$$$

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