To solve x^3 =x+c Let y=(x−p)(x^3 −x−c) (dy/dx)=(x^3 −x−c)+(3x^2 −1)(x−p) let (dy/dx)=mx ⇒ (3x^2 −1)(x−p)=mx 3x^3 −3px^2 −(m+1)x+p=0 3(x+c)−3px^2 −(m+1)x+p=0 3px^2 +(m−2)x−(3c+p)=0 and since x^3 =x+c (m−2)x^2 +(2p−3c)x+3cp=0 [9cp^2 +(3c+p)(m−2)]x +[3cp(m−2)+(3c+p)(2p−3c)] =0 x=−(((3c+p)(2p−3c)+3cp(m−2))/(9cp^2 +(3c+p)(m−2))) (m−2)x^2 +(2p−3c)x+(3cp)=0 Now choosing suitable ′p′ value we find ′m′; and then x=f(p,m) ★


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Question Number 130089 by ajfour last updated on 22/Jan/21

$T o s o l v e x^{3} = x + c$ $L e t y = (x - p) (x^{3} - x - c)$ $\frac{d y}{d x} = (x^{3} - x - c) + (3 x^{2} - 1) (x - p)$ $l e t \frac{d y}{d x} = m x$ $\Rightarrow (3 x^{2} - 1) (x - p) = m x$ $3 x^{3} - 3 {p x}^{2} - (m + 1) x + p = 0$ $3 (x + c) - 3 {p x}^{2} - (m + 1) x + p = 0$ $3 {p x}^{2} + (m - 2) x - (3 c + p) = 0$ $a n d s i n c e x^{3} = x + c$ $(m - 2) x^{2} + (2 p - 3 c) x + 3 c p = 0$ $[9 {c p}^{2} + (3 c + p) (m - 2)] x$ $+ [3 c p (m - 2) + (3 c + p) (2 p - 3 c)]$ $= 0$ $x = - \frac{(3 c + p) (2 p - 3 c) + 3 c p (m - 2)}{9 {c p}^{2} + (3 c + p) (m - 2)}$ $(m - 2) x^{2} + (2 p - 3 c) x + (3 c p) = 0$ $N o w c h o o s i n g s u i t a b l e^{'} p^{'} v a l u e$ $w e f i n d^{'} m^{'};$ $a n d t h e n x = f (p, m) ★$
Commented by bobhans last updated on 22/Jan/21
$example : x^3 =x+6 y=(x−p)(x^3 −x−6) (dy/dx) = x^3 −x+6+(x−p)(3x^2 −1) (dy/dx)= 4x^3 −4px^2 −2x+p+6 (dy/dx)=4x+24−4px^2 −2x+p+6=2x−4px^2 +30+p let (dy/dx) = mx ⇒4px^2 +(m−2)x−(p+30)=0 x=((2−m±(√((m−2)^2 +16p(p+30))))/(8p)) now { ((p=?)),((m=?)) :}$
$e x a m p l e : x^{3} = x + 6$ $y = (x - p) (x^{3} - x - 6)$ $\frac{d y}{d x} = x^{3} - x + 6 + (x - p) (3 x^{2} - 1)$ $\frac{d y}{d x} = 4 x^{3} - 4 {p x}^{2} - 2 x + p + 6$ $\frac{d y}{d x} = 4 x + 24 - 4 {p x}^{2} - 2 x + p + 6 = 2 x - 4 {p x}^{2} + 30 + p$ $l e t \frac{d y}{d x} = m x \Rightarrow 4 {p x}^{2} + (m - 2) x - (p + 30) = 0$ $x = \frac{2 - m \pm \sqrt{{(m - 2)}^{2} + 16 p (p + 30)}}{8 p}$ $n o w {\begin{cases} p = ? \\ m = ? \end{cases}$
Commented by ajfour last updated on 23/Jan/21

	Answered by ajfour last updated on 23/Jan/21

	$x = - \frac{(3 c + p) (2 p - 3 c) + 3 c p (m - 2)}{9 {c p}^{2} + (3 c + p) (m - 2)}$ $(m - 2) x^{2} + (2 p - 3 c) x + (3 c p) = 0$ $s a y m - 2 = M$ $x = - \frac{3 c p M + (3 c + p) (2 p - 3 c)}{(3 c + p) M + 9 {c p}^{2}}$ $l e t \frac{M}{p} = s, \frac{1}{p} = q$ $x = - \frac{3 c s + (3 c q + 1) (2 - 3 c q)}{(3 c q + 1) s + 9 c}$ $s {[3 c s + (3 c q + 1) (2 - 3 c q)]}^{2}$ $- (2 - 3 c q) [3 c s + (3 c q + 1) (2 - 3 c q)] [(3 c q + 1) s + 9 c]$ $+ 3 c {[(3 c q + 1) s + 9 c]}^{2} = 0$ $9 c^{2} s^{3} + 6 c (3 c q + 1) (2 - 3 c q) s^{2}$ $+ \underline{{(3 c q + 1)}^{2} {(2 - 3 c q)}^{2} s}$ $- 6 c (3 c q + 1) s^{2} - 54 c^{2} s$ $+ 9 c^{2} q (3 c q + 1) s^{2} + 81 c^{3} q s$ $- \underline{{(3 c q + 1)}^{2} {(2 - 3 c q)}^{2} s}$ $- 9 c (3 c q + 1) {(2 - 3 c q)}^{2}$ $+ 3 c {(3 c q + 1)}^{2} s^{2} + 54 c (3 c q + 1) s$ $+ 243 c^{3} = 0$ $\Rightarrow$ $(9 c^{2}) s^{3} + [6 c (3 c q + 1) (2 - 3 c q)$ $- 6 c (3 c q + 1) + 9 c^{2} q (3 c q + 1)$ $+ 3 c {(3 c q + 1)}^{2}] s^{2} +$ $[54 c^{2} + 81 c^{3} q + 54 c (3 c q + 1)] s$ $+ [- 9 c (3 c q + 1) {(2 - 3 c q)}^{2} + 243 c^{3}] = 0$ $l e t q = 0$ $\Rightarrow s^{3} + \frac{1}{c} s^{2} + 6 (\frac{1}{c} + 1) s$ $- (\frac{4}{c} - 27 c) = 0$ $s = - 2.54409, - 0.95044$ $f o r c = \frac{30}{19 \sqrt{19}}$ $x = - \frac{3 c s + (3 c q + 1) (2 - 3 c q)}{(3 c q + 1) s + 9 c}$ $a n d f o r q = 0$ $x = - \frac{3 c s + 2}{s + 9 c} =$ $r i g h t a n s w e r i s x = 1.14708$
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