f-x-x-3-2x-4-f-1-x-

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

f (x) = x^{3} - 2 x + 4

f^{- 1} (x) = ?

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

y = x^{3} - 2 x + 4

x^{3} - 2 x + 4 - y = 0

{(u + v)}^{3} - 3 u v (u + v) - (u^{3} + v^{3}) = 0

l e y x = 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} - 8 y + \frac{400}{27}}]

u = \frac{\sqrt[3]{y - 4 + \sqrt{y^{2} - 8 y + \frac{400}{27}}}}{\sqrt[3]{2}}

u = \frac{\sqrt[3]{y - 4 - \sqrt{y^{2} - 8 y + \frac{400}{27}}}}{\sqrt[3]{2}}

\Rightarrow x = u + v = \frac{\sqrt[3]{y - 4 + \sqrt{y^{2} - 8 y + \frac{400}{27}}} + \sqrt[3]{y - 4 - \sqrt{y^{2} - 8 y + \frac{400}{27}}}}{\sqrt[3]{2}}

\Rightarrow f^{- 1} (x) = \frac{\sqrt[3]{x - 4 + \sqrt{x^{2} - 8 x + \frac{400}{27}}} + \sqrt[3]{x - 4 - \sqrt{x^{2} - 8 x + \frac{400}{27}}}}{\sqrt[3]{2}}

Answered by MJS last updated on 15/Feb/19

f^{'} (x) = 0

3 x^{2} - 2 = 0 \Rightarrow x = \pm \frac{\sqrt{6}}{3}

f (- \frac{\sqrt{6}}{3}) = 4 + \frac{4}{9} \sqrt{6}

f (\frac{\sqrt{6}}{3}) = 4 - \frac{4}{9} \sqrt{6}

f^{″} (x) = 0

6 x = 0 \Rightarrow x = 0

f (0) = 4

\Rightarrow f^{- 1} (x) has got 3 parts which are real

\Rightarrow we have to use the trigonometric method

x = y^{3} - 2 y + 4

y^{3} - 2 y + (4 - x) = 0

y_{k} = 2 \sqrt{- \frac{p}{3}} \sin (\frac{1}{3} (2 π k + \arcsin (\frac{9 q}{2 p^{2}} \sqrt{- \frac{p}{3}})))

y_{1} = \frac{2 \sqrt{6}}{3} \sin (\frac{1}{3} (π + \arcsin \frac{3 \sqrt{6} (x - 4)}{8}))

y_{2} = - \frac{2 \sqrt{6}}{3} \cos (\frac{1}{3} (\frac{π}{2} + \arcsin \frac{3 \sqrt{6} (x - 4)}{8}))

y_{3} = - \frac{2 \sqrt{6}}{3} \sin (\frac{1}{3} \arcsin \frac{3 \sqrt{6} (x - 4)}{8})

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

l e t y = - 2 x + 4

y - 4 = - 2 x

\Rightarrow x_{1} = \frac{y - 4}{- 2}

t h e n l e t y = 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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