f-x-x-3-3x-7-f-1-x-

Question Number 60287 by naka3546 last updated on 19/May/19

f (x) = x^{3} + 3 x - 7

f^{- 1} (x) = ?

Commented by mr W last updated on 19/May/19

f^{- 1} (x) = \sqrt[3]{\frac{\sqrt{x^{2} + 14 x + 53} + x + 7}{2}} - \sqrt[3]{\frac{2}{\sqrt{x^{2} + 14 x + 53} + x + 7}}

Commented by naka3546 last updated on 19/May/19

p l e a s e s h o w t h e w o r k i n g s, s i r .

Answered by mr W last updated on 19/May/19

x^{3} + 3 x - 7 = y

x^{3} + 3 x - (7 + y) = 0

l e t c = 7 + y

\Rightarrow x^{3} + 3 x - c = 0

d u e t o {(u + v)}^{3} - 3 u v (u + v) - (u^{3} + v^{3}) = 0

x = u + v

u v = - 1 \Rightarrow u^{3} v^{3} = - 1

u^{3} + v^{3} = c

\Rightarrow u^{3} a n d v^{3} a r e r o o t s o f e q n .

t^{2} - c t - 1 = 0

t = \frac{c \pm \sqrt{c^{2} + 4}}{2}

i . e . u^{3} = \frac{c + \sqrt{c^{2} + 4}}{2}, v^{3} = \frac{c - \sqrt{c^{2} + 4}}{2} = - \frac{1}{u^{3}}

x = u + v = u - \frac{1}{u} = \sqrt[3]{\frac{c + \sqrt{c^{2} + 4}}{2}} - \sqrt[3]{\frac{2}{c + \sqrt{c^{2} + 4}}}

c + \sqrt{c^{2} + 4} = 7 + y + \sqrt{y^{2} + 14 y + 49 + 4} = \sqrt{y^{2} + 14 y + 53} + y + 7

x = \sqrt[3]{\frac{\sqrt{y^{2} + 14 y + 53} + y + 7}{2}} - \sqrt[3]{\frac{2}{\sqrt{y^{2} + 14 y + 53} + y + 7}}

\Rightarrow f^{- 1} (x) = \sqrt[3]{\frac{\sqrt{x^{2} + 14 x + 53} + x + 7}{2}} - \sqrt[3]{\frac{2}{\sqrt{x^{2} + 14 x + 53} + x + 7}}

Commented by naka3546 last updated on 19/May/19

t h a n k y o u, s i r M r W

Answered by MJS last updated on 19/May/19

x^{3} + 3 x - 7 - y = 0

D = \frac{3^{3}}{27} + \frac{{(- 7 - y)}^{2}}{4} = \frac{1}{4} y^{2} + \frac{7}{2} y + \frac{53}{4} > 0 \forall y \in R

\frac{d f}{d x} = 3 x^{2} + 3 = 0 \Rightarrow x \notin R \Rightarrow f (x) has no “ {knee}^{″}

\Rightarrow we can use {Cardano}^{'} s formula to solve

y = x^{3} + 3 x - 7

x = u^{1 / 3} - v^{1 / 3}

u = \frac{y + 7}{2} + \frac{\sqrt{y^{2} + 14 y + 53}}{2}

v = - \frac{y + 7}{2} + \frac{\sqrt{y^{2} + 14 y + 53}}{2}

x = \sqrt[3]{\frac{y + 7}{2} + \frac{\sqrt{y^{2} + 14 y + 53}}{2}} - \sqrt[3]{- \frac{y + 7}{2} + \frac{\sqrt{y^{2} + 14 y + 53}}{2}}

\Rightarrow

f^{- 1} (x) = \sqrt[3]{\frac{x + 7}{2} + \frac{\sqrt{x^{2} + 14 x + 53}}{2}} - \sqrt[3]{- \frac{x + 7}{2} + \frac{\sqrt{x^{2} + 14 x + 53}}{2}}

which is the same as {MrW}^{'} s solution

Commented by naka3546 last updated on 19/May/19

t h a n k y o u, s i r M J S

Commented by MJS last updated on 19/May/19

f (x) = {a x}^{3} + {b x}^{2} + c x + d

f^{- 1} (x) = ?

we have to solve

{a x}^{3} + {b x}^{2} + c x + (d - y) = 0

x^{3} + \frac{b}{a} x^{2} + \frac{c}{a} x + \frac{d - y}{a} = 0

x = z - \frac{b}{3 a}

z^{3} + (\frac{c}{a} - \frac{b^{2}}{3 a^{2}}) z + (\frac{d}{a} - \frac{b c}{3 a^{2}} + \frac{2 b^{3}}{27 a^{3}} - \frac{y}{a}) = 0

p = \frac{c}{a} - \frac{b^{2}}{3 a^{2}}

q = \frac{d}{a} - \frac{b c}{3 a^{2}} + \frac{2 b^{3}}{27 a^{3}}

D = \frac{p^{3}}{27} + \frac{{(q - \frac{y}{a})}^{2}}{4} = \frac{1}{4 a^{2}} y^{2} - \frac{q}{2 a} y + \frac{4 p^{3} + 27 q^{2}}{108}

we have to decide

{\begin{cases} (1) D ⩾ 0 \Rightarrow 1 real solution for z \Rightarrow Cardano \\ (2) D < 0 \Rightarrow 3 real solutions for z \Rightarrow trigonometric method \end{cases}

\Rightarrow f^{- 1} (x) = {\begin{cases} (1) curve in 1 part \\ (2) curve in 3 parts \end{cases}

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