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Question Number 24598 by *D¬ B£$T* last updated on 22/Nov/17

if y=x^3 +x^2 +3x....  find its turning point

$${if}\:{y}={x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{3}{x}.... \\ $$$${find}\:{its}\:{turning}\:{point} \\ $$

Answered by jota+ last updated on 23/Nov/17

(d^2 y/dx^2 )=6x+2=0  ⇒ x=−(1/3).

$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{6}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\frac{\mathrm{1}}{\mathrm{3}}. \\ $$

Answered by mrW1 last updated on 23/Nov/17

(dy/dx)=3x^2 +2x+3=3(x+(1/3))^2 +(8/3)>0  ⇒there is no turning point.

$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}=\mathrm{3}\left({x}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{3}}>\mathrm{0} \\ $$$$\Rightarrow{there}\:{is}\:{no}\:{turning}\:{point}. \\ $$

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