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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)=2x−3 and g(x)=−x^3 . .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\left(\mathrm{x},\mathrm{y}\right) \\ $$$$\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{3}\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right)=−\mathrm{x}^{\mathrm{3}} . \\ $$$$ \\ $$$$. \\ $$

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........ ======================

$$\begin{cases}{\boldsymbol{{y}}=\mathrm{2}\boldsymbol{{x}}−\mathrm{3}\left(\mathrm{1}\right)}\\{\boldsymbol{{y}}=−\boldsymbol{{x}}^{\mathrm{3}} \left(\mathrm{2}\right)}\end{cases}\left(\mathrm{2}\right)\boldsymbol{{dans}}\:\left(\mathrm{1}\right)=>\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{{x}}−\mathrm{3}=\mathrm{0} \\ $$$$\left(\boldsymbol{{x}}−\mathrm{1}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}+\mathrm{3}\right)=\mathrm{0}=>\boldsymbol{{x}}=\mathrm{1} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}+\mathrm{3}=\mathrm{0}\:\:\:\boldsymbol{{positive}}\:\boldsymbol{{c}}'\boldsymbol{{est}}\:\grave {\boldsymbol{{a}}}\:\boldsymbol{{dire}} \\ $$$$\boldsymbol{{n}}'\boldsymbol{{admet}}\:\boldsymbol{{pas}}\:\boldsymbol{{de}}\:\boldsymbol{{solution}} \\ $$$$\boldsymbol{{cas}}\:\boldsymbol{\Delta}=\mathrm{1}−\mathrm{16}=−\mathrm{16}<\mathrm{0} \\ $$$$\boldsymbol{{pour}}\:\boldsymbol{{x}}=\mathrm{1} \\ $$$$\boldsymbol{{y}}=−\mathrm{1} \\ $$$$\boldsymbol{{d}}'\boldsymbol{{o}}\grave {\boldsymbol{{u}}}\:\boldsymbol{{ce}}\:\boldsymbol{{point}}\:\boldsymbol{{est}} \\ $$$$\boldsymbol{{A}}\left(\mathrm{1};−\mathrm{1}\right) \\ $$$$===================== \\ $$$$......{le}\:{celebre}\:{cedric}\:{junior}........ \\ $$$$====================== \\ $$

Commented by mr W last updated on 30/Nov/22

you did it wrongly!

$${you}\:{did}\:{it}\:{wrongly}! \\ $$

Answered by mr W last updated on 30/Nov/22

2x−3=−x^3 x^3 +2x−3=0 (x−1)(x^2 +x+3)=0 ⇒x=1 ⇒y=−1 intersection (1,−1)

$$\mathrm{2}{x}−\mathrm{3}=−{x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{3}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}\:\Rightarrow{y}=−\mathrm{1} \\ $$$${intersection}\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$

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