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Question Number 107461 by Don08q last updated on 10/Aug/20

Commented by john santu last updated on 11/Aug/20

$$\left(iii\right)y=x^3+(k+2)x^2-3kx+5$$$$\Rightarrow y^{\prime }=3x^2+2(k+2)x-3k$$$$\Rightarrow \mathrm{\Delta }=4{(k+2)}^2-4.3.(-3k)=0$$$$\Rightarrow {(k+2)}^2+9k=0$$$$\Rightarrow k^2+13k+4=0$$$$k=\frac{-13\pm \sqrt{169-4.1.4}}{2}=\frac{-13\pm \sqrt{153}}{2}$$$$k=\frac{-13\pm 3\sqrt{17}}{2}.$$

Answered by john santu last updated on 11/Aug/20

$$\divideontimes \mathcal{JS}\divideontimes$$$$\Rightarrow y={ax}^3+{bx}^2+cx+d\Rightarrow hasa$$$$vertexif{y}^{\prime }=0$$$$\Rightarrow y^{\prime }=3{ax}^2+2bx+c=0$$$$\Rightarrow x=\frac{-2b\pm \sqrt{4b^2-4.3a.c}}{6a}$$$$x=\frac{-2b\pm 2\sqrt{b^2-3ac}}{6a}=\frac{-b\pm \sqrt{b^2-3ac}}{3a}$$

Commented by john santu last updated on 11/Aug/20

$$\left(ii\right)novertexif{b}^2-3ac<0$$$$\Rightarrow (b-\sqrt{3ac})(b+\sqrt{3ac})<0$$$$\Rightarrow -\sqrt{3ac}<b<\sqrt{3ac}$$

Commented by Don08q last updated on 11/Aug/20

$$ThankyouSir$$

Answered by 1549442205PVT last updated on 11/Aug/20

$$\mathrm{We}\mathrm{need}\mathrm{to}\mathrm{have}\mathrm{the}\mathrm{hypothesis}\mathrm{that}$$$$\mathrm{a},\mathrm{b}\ne 0.\mathrm{Then}\mathrm{denote}\mathrm{by}\mathrm{C}\mathrm{the}\mathrm{graph}\mathrm{of}$$$$\mathrm{y}={\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}.\mathrm{We}\mathrm{have}$$$${\mathrm{y}}^{\prime }={3\mathrm{ax}}^2+2\mathrm{bx}+\mathrm{c}=0\iff {\mathrm{x}}_{1,2}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-3\mathrm{ac}}}{3\mathrm{a}}$$$${\mathrm{y}}^{″}=6\mathrm{ax}+2\mathrm{b}=0\iff {\mathrm{x}}_0=-\mathrm{b}/3\mathrm{a}$$$$\mathrm{i})\mathrm{if}\mathrm{\Delta }={\mathrm{b}}^2-3\mathrm{ac}>0\mathrm{the}\mathrm{eqn}.{\mathrm{y}}^{\prime }=0\mathrm{has}$$$$\mathrm{two}\mathrm{real}\mathrm{roots}\mathrm{and}{\mathrm{y}}^{\prime }\mathrm{change}\mathrm{the}\mathrm{sign}\mathrm{when}$$$$\mathrm{passes}\mathrm{through}\mathrm{its}\mathrm{roots},\mathrm{so}\mathrm{the}\mathrm{curve}$$$$\mathrm{C}\mathrm{has}\mathrm{two}\mathrm{extremums}\mathrm{points}\left(\mathrm{vertexes}\right)$$$$\mathrm{ii})\mathrm{If}{\mathrm{b}}^2-3\mathrm{ac}=0\mathrm{then}{\mathrm{x}}_0={\mathrm{x}}_1={\mathrm{x}}_2=-\mathrm{b}/3\mathrm{a}.$$$$\mathrm{Then}\mathrm{the}\mathrm{given}\mathrm{function}\mathrm{has}\mathrm{no}\mathrm{extremum}$$$$\left(\mathrm{vertex}\right)\mathrm{because}{\mathrm{y}}^{\prime }{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{change}\mathrm{the}$$$$\mathrm{sign}\mathrm{when}\mathrm{go}\mathrm{through}{\mathrm{x}}_0.\mathrm{Hence},\mathrm{the}$$$$\mathrm{condition}\mathrm{need}\mathrm{and}\mathrm{enough}\mathrm{for}\mathrm{the}\mathrm{curve}$$$$\mathrm{of}\mathrm{degree}3\mathrm{has}\mathrm{a}\mathrm{saddle}\mathrm{point}\mathrm{be}$$$$\mathrm{\Delta }={\mathrm{b}}^2-3\mathrm{ac}=0$$$$\mathrm{iii})\mathrm{The}\mathrm{curve}\mathrm{y}={\mathrm{x}}^3+(\mathrm{k}+2){\mathrm{x}}^2-3\mathrm{kx}+5\left({\mathrm{C}}_{\mathrm{k}}\right)$$$$\mathrm{has}\mathrm{the}\mathrm{saddle}\mathrm{point}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$$${\mathrm{y}}^{\prime }={3\mathrm{x}}^2+2(\mathrm{k}+2)\mathrm{x}-3\mathrm{k}=0\mathrm{has}\mathrm{unique}\mathrm{root}$$$$\iff {\mathrm{\Delta }}^{\prime }={(\mathrm{k}+2)}^2+9\mathrm{k}=0\iff {\mathrm{k}}^2+13\mathrm{k}+4=0$$$$\iff {\mathrm{k}}_{1,2}=\frac{-13\pm 3\sqrt{17}}{2}.\mathrm{Then}{\mathrm{x}}_0=\frac{-(\mathrm{k}+2)}{3}$$$$+)\mathrm{for}{\mathrm{k}}_1=\frac{-13+3\sqrt{17}}{2}\Rightarrow {\mathrm{x}}_0=-\frac{-9+3\sqrt{17}}{6}$$$$+)\mathrm{For}{\mathrm{k}}_2=\frac{-13-3\sqrt{17}}{2}\Rightarrow {\mathrm{x}}_0=\frac{9+3\sqrt{17}}{6}$$$$\mathrm{Thus},{\mathrm{C}}_{\mathrm{k}}\mathrm{has}\mathrm{the}\mathrm{saddle}\mathrm{point}\mathrm{if}\mathrm{and}$$$$\mathrm{only}\mathrm{if}\mathrm{k}=\frac{-13\pm 3\sqrt{17}}{2}$$

Commented by Don08q last updated on 11/Aug/20

$$Comprehensiveexplanation.Thankyou$$

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