f(x) = x^3 + 3x − 7 f^(−1) (x) = ?


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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
	$f(x)=ax^3 +bx^2 +cx+d f^(−1) (x)=? we have to solve ax^3 +bx^2 +cx+(d−y)=0 x^3 +(b/a)x^2 +(c/a)x+((d−y)/a)=0 x=z−(b/(3a)) z^3 +((c/a)−(b^2 /(3a^2 )))z+((d/a)−((bc)/(3a^2 ))+((2b^3 )/(27a^3 ))−(y/a))=0 p=(c/a)−(b^2 /(3a^2 )) q=(d/a)−((bc)/(3a^2 ))+((2b^3 )/(27a^3 )) D=(p^3 /(27))+(((q−(y/a))^2 )/4)=(1/(4a^2 ))y^2 −(q/(2a))y+((4p^3 +27q^2 )/(108)) we have to decide { (((1) D≥0 ⇒ 1 real solution for z ⇒ Cardano)),(((2) D<0 ⇒ 3 real solutions for z ⇒ trigonometric method)) :} ⇒ f^(−1) (x)= { (((1) curve in 1 part)),(((2) curve in 3 parts)) :}$
	$f (x) = {a x}^{3} + {b x}^{2} + c x + d$ $f^{- 1} (x) = ?$ $we have to solve$ ${a x}^{3} + {b x}^{2} + c x + (d - y) = 0$ $x^{3} + \frac{b}{a} x^{2} + \frac{c}{a} x + \frac{d - y}{a} = 0$ $x = z - \frac{b}{3 a}$ $z^{3} + (\frac{c}{a} - \frac{b^{2}}{3 a^{2}}) z + (\frac{d}{a} - \frac{b c}{3 a^{2}} + \frac{2 b^{3}}{27 a^{3}} - \frac{y}{a}) = 0$ $p = \frac{c}{a} - \frac{b^{2}}{3 a^{2}}$ $q = \frac{d}{a} - \frac{b c}{3 a^{2}} + \frac{2 b^{3}}{27 a^{3}}$ $D = \frac{p^{3}}{27} + \frac{{(q - \frac{y}{a})}^{2}}{4} = \frac{1}{4 a^{2}} y^{2} - \frac{q}{2 a} y + \frac{4 p^{3} + 27 q^{2}}{108}$ $we have to decide$ ${\begin{cases} (1) D ⩾ 0 \Rightarrow 1 real solution for z \Rightarrow Cardano \\ (2) D < 0 \Rightarrow 3 real solutions for z \Rightarrow trigonometric method \end{cases}$ $\Rightarrow f^{- 1} (x) = {\begin{cases} (1) curve in 1 part \\ (2) curve in 3 parts \end{cases}$
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