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Question Number 162860 by Mathematification last updated on 01/Jan/22

Answered by MJS_new last updated on 01/Jan/22

$$f\left(x\right)=x^4-18x^2-x+83$$$$g\left(x\right)=c_{1}x+c_0$$$$h\left(x\right)=f\left(x\right)-g\left(x\right)=x^4-18x^2-(c_1+1)x+(83-c_0)$$$$\mathrm{we}\mathrm{know}h\left(x\right)\mathrm{must}\mathrm{have}2\mathrm{minima}{m}_1,m_2$$$$\mathrm{and}1\mathrm{maximum}\mathrm{at}x=\frac{m_1+m_2}{2}$$$$\Rightarrow$$$$h^{\prime }\left(x\right)=C(x-m_1)(x-m_2)(x-\frac{m_1+m_2}{2})=0$$$$\iff$$$$x^3-\frac{3(m_1+m_2)}{2}{x}^2+\frac{m_1^2+4m_1{m}_2+m_2^3}{2}x-\frac{m_1(m_1+m_2)m_2}{2}=0$$$$\mathrm{but}$$$$h^{\prime }\left(x\right)=4x^3-36x-(c_1+1)$$$$\Rightarrow$$$$x^3-9x-\frac{c_1+1}{4}=0$$$$\Rightarrow$$$$-\frac{3(m_1+m_2)}{2}=0\Rightarrow m_2=-m_1$$$$\Rightarrow$$$$h\left(x\right)=x^3-m_1^{2}x=0\Rightarrow m_1=\pm 3\wedge c_1=-1$$$$\Rightarrow g\left(x\right)\mathrm{touches}f\left(x\right)\mathrm{at}x=\pm 3$$$$\mathrm{now}\mathrm{we}\mathrm{easily}\mathrm{get}\mathrm{the}\mathrm{tangent}$$$$g\left(x\right)=-x+2$$

Commented by Tawa11 last updated on 02/Jan/22

$$\mathrm{Great}\mathrm{sir}.$$

Answered by mr W last updated on 02/Jan/22

$$curve:y=f\left(x\right)=x^4-18x^2-x+83$$$$saythetangentlineisy=g\left(x\right)=ax+b$$$$thedifferenceofthem,sayh\left(x\right),with$$$$h\left(x\right)=x^4-18x^2-(1+a)x+(83-b)$$$$mustbeacurvelikethis:$$

Commented by mr W last updated on 02/Jan/22

Commented by mr W last updated on 02/Jan/22

$${let}^{\prime }shaveacloselookatit.$$$$thecurvehastwozeros,saypandq.$$$$atthesetwopointsthecurvetangents$$$$alsothex-axis.thatmeansatthese$$$$twopointsthecurvehasminimum$$$$whichisalsozero.$$$$wecanassumethen$$$$h\left(x\right)=(x-p)(x-q)(x^2+ux+v)$$$$h^{\prime }\left(x\right)=(x-p)(x-q)(2x+u)+(2x-p-q)(x^2+ux+v)$$$$wehave$$$$h^{\prime }\left(p\right)=(p-q)(p^2+up+v)=0$$$$\Rightarrow p^2+up+v=0$$$$h^{\prime }\left(q\right)=(q-p)(q^2+uq+v)=0$$$$\Rightarrow q^2+uq+v=0$$$$thatmeansp,qarealsorootsof$$$$x^2+ux+v=0,i.e.$$$$x^2+ux+v=(x-p)(x-q).$$$$therefore$$$$h\left(x\right)=(x-p)(x-q)(x^2+ux+v)$$$$\Rightarrow h\left(x\right)={(x-p)}^2{(x-q)}^2$$$$h^{\prime }\left(x\right)=4(x-p)(x-q)(x-\frac{p+q}{2})$$$$thatmeans{h}^{\prime }\left(x\right)=0atx=r=\frac{p+q}{2}.$$$$thisisthepointwherethecurveh\left(x\right)$$$$haslocalmaximum.$$$$$$$$usingtheseknowledges:$$$$h\left(x\right)=x^4-18x^2-(1+a)x+(83-b)$$$$h^{\prime }\left(x\right)=4(x^3-9x-\frac{1+a}{4})=0$$$$(x^3-9x-\frac{1+a}{4})=(x-p)(x-q)(x-\frac{p+q}{2})$$$$p+q+\frac{p+q}{2}=0\Rightarrow p+q=0$$$$pq+(p+q)\frac{p+q}{2}=-9\Rightarrow pq=-9$$$$\frac{pq(p+q)}{2}=\frac{1+a}{4}\Rightarrow a=-1$$$$\Rightarrow p=3,q=-3$$$$h\left(3\right)=h(-3)=0$$$$\Rightarrow 3^4-18\times 3^2+83-b=0\Rightarrow b=2$$$$$$$$\Rightarrow thetangentlineisy=-x+2$$$$$$$$alternativeway:$$$$weknowthath\left(x\right)shouldbeofthe$$$$formh\left(x\right)={(x-p)}^2{(x-q)}^2.that$$$$meamsifithasno{x}^{3}term,thenit$$$$shouldalsohavenoxterm,$$$$i.e.1+a=0,ora=-1.$$$$h\left(x\right)=x^4-18x^2+(83-b)\stackrel{!}{=}{(x^2-9^2)}^2$$$$83-b=9^2\Rightarrow b=2$$

Commented by mr W last updated on 02/Jan/22

Commented by mr W last updated on 02/Jan/22

$$basically{it}^{\prime }sthesamesolutionas$$$$thatfromMJSsir,ijustputsome$$$$moreexplanation.$$

Commented by Tawa11 last updated on 02/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

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