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Soit $C$ une courbe d'équation $f(x)=\frac43x^3+2x²+3x+7$ et d une droite d'équation $y-2x+3=0$.

Translation

Title Determine the equation of the tangent to the curve $C$ which is parallel to the line $d$

Let $C$ be the curve of equation $f(x)=\frac43x^3+2x^2+3x+7$ and $d$ the line of equation $y-2x+3=0$.

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  • $\begingroup$ On Math Stack Exchange everyone writes in english, and you should do it too $\endgroup$ – Edoardo Lanari Oct 4 '14 at 11:51
  • $\begingroup$ questions en francais sont bienvenues ici meta.math.stackexchange.com/questions/1617/… / questions in French are welcome here. $\endgroup$ – Matthew Towers Oct 4 '14 at 12:12
  • $\begingroup$ @mt_. Everyone is welcome here but we have the chance to have a common language almost accepted everywhere : English. Why then should be avoid posts in Eskimo or in Papuan or written in hieroglyphs ? May I add that I am a Frenchman. $\endgroup$ – Claude Leibovici Oct 4 '14 at 14:12
  • $\begingroup$ With a translation, any language is welcome here (thanks @SamiBenRomdhane and others who contributed). The translation also fixed the problem statement appearing only in the title (it should be fully stated in the body of the Question as well). $\endgroup$ – hardmath Oct 4 '14 at 14:19
  • $\begingroup$ @ClaudeLeibovici since not everyone can write English, and since there are a large number of MSE users willing to translate non-English questions, I think (as the top-voted post on my link suggested) we ought to welcome translateable non-English posts. $\endgroup$ – Matthew Towers Oct 4 '14 at 16:31
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The equation of tangent to the curve $C$ on the point $(x_0,f(x_0))$ is

$$y=f(x_0)+(x-x_0)f'(x_0)\tag1$$ and the tangent is parallel to the line $d$ iff

$$f'(x_0)=4x_0^2+4x_0+3=2\iff(2x_0+1)^2=0\iff x_0=-\frac12$$ now replace $x_0$ by its value in $(1)$ and we are done.

Traduction

L'équation de la tangente à la courbe $C$ en le point $(x_0,f(x_0))$ est

$$y=f(x_0)+(x-x_0)f'(x_0)\tag1$$ et cette tangente est parallèle à la droie $d$ ssi

$$f'(x_0)=4x_0^2+4x_0+3=2\iff(2x_0+1)^2=0\iff x_0=-\frac12$$ maintenant remplace $x_0$ par sa valeur trouvée en $(1)$ et on a terminé.

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  • $\begingroup$ Wow formidable ! $\endgroup$ – Jo Mbayo Oct 4 '14 at 12:09
  • $\begingroup$ y=2x+41/6. Thanks! $\endgroup$ – Jo Mbayo Oct 4 '14 at 12:19
  • $\begingroup$ De rien;-)${}{}$ $\endgroup$ – user63181 Oct 4 '14 at 12:20
  • $\begingroup$ @SamiBenRomdhane. One more surprize from you ! What don't you know ? Cheers et merci. $\endgroup$ – Claude Leibovici Oct 4 '14 at 14:15
  • $\begingroup$ Well... I don't know everything:-) Merci! @ClaudeLeibovici $\endgroup$ – user63181 Oct 4 '14 at 14:19

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