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^3 +px^2 +qx+r (dy/dx)=3x^2 +2px+q (d^2 y/dx^2 )=6x+2p let point N(α,0) at point N y_(minimum) 1)(dy/dx)=0→3α^2 +2pα+q=0 for min 2) (d^2 y/dx^2 )>0 for minimum 3α^2 +2pα+q=0 α=((−2p±(√(4p^2 −12q)))/6) α=((−p±(√(p^2 −3q)))/3) from given figure it is clear α>0 point N(((−p+(√(p^2 −3q)))/3),0) ((d^2 y/dx^2 ))_(x=α) =6α+2p 6(((−p+(√(p^2 −3q)))/3))+2p>0 2((√(p^2 −3q)) )>0 [p^2 >3q] now for point K(0,β) y=x^3 +px^2 +qx+r β=r (dy/dx)=3x^2 +2px+q ((dy/dx))_(x=0) 3×0^2 +2×p×0+q=0 [for max (dy/dx)=0] q=0 ((d^2 y/dx^2 ))_(x=0) 6×0+2p<0 p<0 p<0 q=0 r=β →K(0,β) β←ordinate of point K i have tried to understand the question...](Q58982.png)
$${y}={x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{px}+{q} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{6}{x}+\mathrm{2}{p} \\ $$$${let}\:{point}\:{N}\left(\alpha,\mathrm{0}\right)\:{at}\:{point}\:{N}\:\:{y}_{{minimum}} \\ $$$$ \\ $$$$\left.\mathrm{1}\right)\frac{{dy}}{{dx}}=\mathrm{0}\rightarrow\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{2}{p}\alpha+{q}=\mathrm{0}\:\:{for}\:{min} \\ $$$$\:\:\:\: \\ $$$$\left.\mathrm{2}\right)\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }>\mathrm{0}\:{for}\:{minimum}\: \\ $$$$\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{2}{p}\alpha+{q}=\mathrm{0} \\ $$$$\alpha=\frac{−\mathrm{2}{p}\pm\sqrt{\mathrm{4}{p}^{\mathrm{2}} −\mathrm{12}{q}}}{\mathrm{6}} \\ $$$$\alpha=\frac{−{p}\pm\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}}{\mathrm{3}} \\ $$$${from}\:{given}\:{figure}\:{it}\:{is}\:{clear}\:\alpha>\mathrm{0} \\ $$$${point}\:{N}\left(\frac{−{p}+\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}}{\mathrm{3}},\mathrm{0}\right) \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{x}=\alpha} =\mathrm{6}\alpha+\mathrm{2}{p} \\ $$$$\mathrm{6}\left(\frac{−{p}+\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}}{\mathrm{3}}\right)+\mathrm{2}{p}>\mathrm{0} \\ $$$$\mathrm{2}\left(\sqrt{{p}^{\mathrm{2}} −\mathrm{3}{q}}\:\right)>\mathrm{0}\:\:\:\left[{p}^{\mathrm{2}} >\mathrm{3}{q}\right] \\ $$$$ \\ $$$${now}\:{for}\:{point}\:{K}\left(\mathrm{0},\beta\right) \\ $$$${y}={x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r} \\ $$$$\beta={r} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{px}+{q} \\ $$$$\left(\frac{{dy}}{{dx}}\right)_{{x}=\mathrm{0}} \mathrm{3}×\mathrm{0}^{\mathrm{2}} +\mathrm{2}×{p}×\mathrm{0}+{q}=\mathrm{0}\:\:\left[{for}\:{max}\:\frac{{dy}}{{dx}}=\mathrm{0}\right] \\ $$$${q}=\mathrm{0} \\ $$$$ \\ $$$$\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{x}=\mathrm{0}} \:\mathrm{6}×\mathrm{0}+\mathrm{2}{p}<\mathrm{0}\:\:\:{p}<\mathrm{0} \\ $$$${p}<\mathrm{0} \\ $$$${q}=\mathrm{0} \\ $$$${r}=\beta\:\rightarrow{K}\left(\mathrm{0},\beta\right)\:\:\beta\leftarrow{ordinate}\:{of}\:{point}\:{K} \\ $$$${i}\:{have}\:{tried}\:{to}\:{understand}\:{the}\:{question}... \\ $$$$ \\ $$