Question Number 58971 by hovea cw last updated on 02/May/19
Commented by hovea cw last updated on 02/May/19
$$\mathrm{hd}\:\mathrm{plz} \\ $$
Answered by tanmay last updated on 02/May/19
$${y}={x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{px}+{q} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{6}{x}+\mathrm{2}{p} \\ $$$${let}\:{point}\:{N}\left(\alpha,\mathrm{0}\right)\:{at}\:{point}\:{N}\:\:{y}_{{minimum}} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\frac{{dy}}{{dx}}=\mathrm{0}\rightarrow\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{2}{p}\alpha+{q}=\mathrm{0}\:\:{for}\:{min} \\ $$$$\:\:\:\: \\ $$$$\left.\mathrm{2}\right)\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }>\mathrm{0}\:{for}\:{minimum}\: \\ $$$$\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{2}{p}\alpha+{q}=\mathrm{0} \\ $$$$\alpha=\frac{−\mathrm{2}{p}\pm\sqrt{\mathrm{4}{p}^{\mathrm{2}} −\mathrm{12}{q}}}{\mathrm{6}} \\ $$$$\alpha=\frac{−{p}\pm\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}}{\mathrm{3}} \\ $$$${from}\:{given}\:{figure}\:{it}\:{is}\:{clear}\:\alpha>\mathrm{0} \\ $$$${point}\:{N}\left(\frac{−{p}+\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}}{\mathrm{3}},\mathrm{0}\right) \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{x}=\alpha} =\mathrm{6}\alpha+\mathrm{2}{p} \\ $$$$\mathrm{6}\left(\frac{−{p}+\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}}{\mathrm{3}}\right)+\mathrm{2}{p}>\mathrm{0} \\ $$$$\mathrm{2}\left(\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}\:\right)>\mathrm{0}\:\:\:\left[{p}^{\mathrm{2}} >\mathrm{3}{q}\right] \\ $$$$ \\ $$$${now}\:{for}\:{point}\:{K}\left(\mathrm{0},\beta\right) \\ $$$${y}={x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r} \\ $$$$\beta={r} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{px}+{q} \\ $$$$\left(\frac{{dy}}{{dx}}\right)_{{x}=\mathrm{0}} \mathrm{3}×\mathrm{0}^{\mathrm{2}} +\mathrm{2}×{p}×\mathrm{0}+{q}=\mathrm{0}\:\:\left[{for}\:{max}\:\frac{{dy}}{{dx}}=\mathrm{0}\right] \\ $$$${q}=\mathrm{0} \\ $$$$ \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{x}=\mathrm{0}} \:\mathrm{6}×\mathrm{0}+\mathrm{2}{p}<\mathrm{0}\:\:\:{p}<\mathrm{0} \\ $$$${p}<\mathrm{0} \\ $$$${q}=\mathrm{0} \\ $$$${r}=\beta\:\rightarrow{K}\left(\mathrm{0},\beta\right)\:\:\beta\leftarrow{ordinate}\:{of}\:{point}\:{K} \\ $$$${i}\:{have}\:{tried}\:{to}\:{understand}\:{the}\:{question}… \\ $$$$ \\ $$