Find the point of intersection (x,y) of f(x)=2x−3 and g(x)=−x^3 . .


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Question Number 181755 by Mastermind last updated on 29/Nov/22

$Find the point of intersection (x, y)$ $of f (x) = 2 x - 3 and g (x) = - x^{3} .$ $.$
Commented by CElcedricjunior last updated on 30/Nov/22
${ ((y=2x−3(1))),((y=−x^3 (2))) :}(2)dans (1)=>x^3 +2x−3=0 (x−1)(x^2 +x+3)=0=>x=1 x^2 +x+3=0 positive c′est a^� dire n′admet pas de solution cas 𝚫=1−16=−16<0 pour x=1 y=−1 d′ou^� ce point est A(1;−1) ===================== ......le celebre cedric junior........ ======================$
${\begin{cases} y = 2 x - 3 (1) \\ y = - x^{3} (2) \end{cases} (2) d a n s (1) => x^{3} + 2 x - 3 = 0$ $(x - 1) (x^{2} + x + 3) = 0 => x = 1$ $x^{2} + x + 3 = 0 p o s i t i v e c^{'} e s t \overset{`}{a} d i r e$ $n^{'} a d m e t p a s d e s o l u t i o n$ $c a s Δ = 1 - 16 = - 16 < 0$ $p o u r x = 1$ $y = - 1$ $d^{'} o \overset{`}{u} c e p o i n t e s t$ $A (1; - 1)$ $=====================$ $. . . . . . l e c e l e b r e c e d r i c j u n i o r . . . . . . . .$ $======================$
Commented by mr W last updated on 30/Nov/22

$y o u d i d i t w r o n g l y!$
	Answered by mr W last updated on 30/Nov/22

	$2 x - 3 = - x^{3}$ $x^{3} + 2 x - 3 = 0$ $(x - 1) (x^{2} + x + 3) = 0$ $\Rightarrow x = 1 \Rightarrow y = - 1$ $i n t e r s e c t i o n (1, - 1)$
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