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Question Number 153252 by liberty last updated on 06/Sep/21

Answered by MJS_new last updated on 06/Sep/21

$$y^{″}=0$$$$\frac{2ax(x^2-3b)}{{(x^2+b)}^3}=0$$$$\Rightarrow x=0\vee x=\pm \sqrt{3b}\vee a=0\left(\mathrm{rejected}\right)$$$$2=\sqrt{3b}\Rightarrow b=\frac{4}{3}$$$$\frac{5}{2}=\frac{2a}{4+\frac{4}{3}}\Rightarrow a=\frac{20}{3}$$

Commented by liberty last updated on 06/Sep/21

$$thankyou.i{don}^{\prime }tunderstand$$$$aflexpoint.itissametosadle$$$$pointsir?$$

Commented by MJS_new last updated on 06/Sep/21

$$\mathrm{no}.$$$$\mathrm{at}\mathrm{a}\mathrm{flex}\mathrm{point}\mathrm{the}\mathrm{curvature}\mathrm{changes}$$$$y={ax}^3+{bx}^2+cx+d$$$$y^{\prime }=3{ax}^2+2bx+c$$$$2\mathrm{real}\mathrm{solutions}$$$$y\mathrm{has}1\min ,1\max \mathrm{and}1\mathrm{flex}\mathrm{point}$$$$1\mathrm{real}\mathrm{solution}$$$$y\mathrm{has}\mathrm{no}\min /\max \mathrm{but}1\mathrm{flex}\mathrm{point}\mathrm{with}$$$$y^{\prime }=0,\mathrm{sometimes}\mathrm{called}\mathrm{flat}\mathrm{point}$$$$2\mathrm{complex}\mathrm{solutions}$$$$y\mathrm{has}\mathrm{no}\min /\max \mathrm{but}1\mathrm{flex}\mathrm{point}$$$$y^{″}=6ax+2b$$$$y^{\prime }=0\wedge y^{″}>0\Rightarrow \min$$$$y^{\prime }=0\wedge y^{″}<0\Rightarrow \max$$$$y^{\prime }=0\wedge y^{″}=0\Rightarrow \mathrm{flat}\mathrm{point}$$$$y^{\prime }\ne 0\wedge y^{″}=0\Rightarrow \mathrm{flex}\mathrm{point}$$$$$$$$\mathrm{a}\mathrm{function}\mathrm{with}{y}^{\prime }=y^{″}=y^{‴}=0\mathrm{has}\mathrm{a}\mathrm{saddle}\mathrm{point}$$$${that}^{\prime }swhatIoncelearned,maybesomecall$$$$aflatpointasaddlepoint...$$

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