Question Number 24598 by *D¬ B£$T* last updated on 22/Nov/17
$${if}\:{y}={x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{3}{x}…. \\ $$$${find}\:{its}\:{turning}\:{point} \\ $$
Answered by jota+ last updated on 23/Nov/17
$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{6}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\frac{\mathrm{1}}{\mathrm{3}}. \\ $$
Answered by mrW1 last updated on 23/Nov/17
$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}=\mathrm{3}\left({x}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{3}}>\mathrm{0} \\ $$$$\Rightarrow{there}\:{is}\:{no}\:{turning}\:{point}. \\ $$