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Question Number 205682 by naka3546 last updated on 27/Mar/24

Commented by naka3546 last updated on 28/Mar/24

$$\mathrm{Thank}\mathrm{you}$$

Commented by Frix last updated on 28/Mar/24

$$\mathrm{The}\mathrm{tangent}:y=46x-105$$$$p=-1-\sqrt{11}\wedge q=-1+\sqrt{11}$$$$pq=-10$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{easy}.\mathrm{We}\mathrm{have}\mathrm{the}\mathrm{polynomial}y=p\left(x\right)$$$$\mathrm{and}\mathrm{the}\mathrm{common}\mathrm{tangent}y=ax+b$$$$\Rightarrow p\left(x\right)-(ax+b)=0\mathrm{must}\mathrm{have}2\mathrm{double}\mathrm{zeros}$$$$\Rightarrow p\left(x\right)-(ax+b)={(x^2+\alpha x+\beta )}^2$$$$\mathrm{Now}\mathrm{just}\mathrm{match}\mathrm{the}\mathrm{constants}.$$

Answered by mr W last updated on 28/Mar/24

$$curve:f\left(x\right)=x^4+4x^3-16x^2+6x-5$$$$tangentline:h\left(x\right)=mx+c$$$$differencebetweenthem:$$$$g\left(x\right)=f\left(x\right)-h\left(x\right)$$$$g\left(x\right)=x^4+4x^3-16x^2+(6-m)x-(5+c)$$$$atx=p,q:g\left(x\right)=0$$$$g\left(x\right)=(x-p)(x-q)(x^2+ux+v)$$$$g^{\prime }\left(x\right)=(x-p)(x-q)(2x+u)+(2x-p-q)(x^2+ux+v)$$$$atx=p,q:g^{\prime }\left(x\right)=0$$$$g^{\prime }\left(p\right)=(p-q)(p^2+up+v)=0$$$$g^{\prime }\left(q\right)=(q-p)(q^2+uq+v)=0$$$$\Rightarrow p^2+up+v=0,q^2+qu+v=0$$$$p,qarerootsof{x}^2+ux+v=0$$$$\Rightarrow x^2+ux+v=(x-p)(x-q)$$$$\Rightarrow g\left(x\right)={\left[(x-p)(x-q)\right]}^2$$$$g^{\prime }\left(x\right)=4(x-p)(x-q)(x-\frac{p+q}{2})$$$$\Rightarrow p,qand\frac{p+q}{2}arerootsof{g}^{\prime }\left(x\right)=0$$$$ontheotherside:$$$$g^{\prime }\left(x\right)=4x^3+12x^2-32x+(6-m)$$$$p+q+\frac{p+q}{2}=-\frac{12}{4}=-3$$$$\Rightarrow p+q=-2$$$$pq+\frac{p+q}{2}\times (p+q)=-\frac{32}{4}$$$$pq+\frac{-2}{2}\times (-2)=-8$$$$\Rightarrow pq=-10✓$$$$$$$$pq\times \frac{p+q}{2}=-\frac{6-m}{4}$$$$(-10)\times \frac{-2}{2}=-\frac{6-m}{4}$$$$\Rightarrow m=46$$$$g\left(x\right)=x^4+4x^3-16x^2-40x-(5+c)$$$$g\left(x\right)={(x^2+2x-10)}^2$$$$\Rightarrow -5-c={(-10)}^2$$$$\Rightarrow c=-105$$$$\Rightarrow tangentline:y=46x-105$$

Commented by mr W last updated on 28/Mar/24

Commented by mr W last updated on 29/Mar/24

$$generally:$$$$curve:f\left(x\right)=x^4+{ax}^3+{bx}^2+cx+d$$$$tangentline:g\left(x\right)=mx+k$$$$h\left(x\right)=f\left(x\right)-g\left(x\right)$$$$h\left(x\right)=x^4+{ax}^3+{bx}^2+(c-m)x+d-k$$$$h^{\prime }\left(x\right)=4x^3+3{ax}^2+2bx+(c-m)$$$$h\left(x\right)={(x-p)}^2{(x-q)}^2$$$$p,qand\frac{p+q}{2}arerootsof{h}^{\prime }\left(x\right)=0$$$$p+q+\frac{p+q}{2}=-\frac{3a}{4}$$$$\Rightarrow p+q=-\frac{a}{2}$$$$pq+\frac{(p+q)(p+q)}{2}=\frac{2b}{4}$$$$\Rightarrow pq=\frac{b}{2}-\frac{a^2}{8}$$$$pq\times \frac{p+q}{2}=-\frac{c-m}{4}$$$$\Rightarrow m=c-a(\frac{b}{2}-\frac{a^2}{8})$$$${\left(pq\right)}^2=d-k$$$$\Rightarrow k=d-{(\frac{b}{2}-\frac{a^2}{8})}^2$$

Commented by naka3546 last updated on 30/Mar/24

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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