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Question Number 129320 by ajfour last updated on 14/Jan/21

Commented by gaferrafie last updated on 14/Jan/21

سلام چطور میشه تو خروجی آدرس سایت را حذف کرد

Answered by ajfour last updated on 14/Jan/21

y=(x−p)(x^3 −x−c) =x^4 −px^3 −x^2 +(p−c)x+cp (dy/dx)=4x^3 −3px^2 −2x+p−c (d^2 y/dx^2 )=12x^2 −6px−2 y=p−kx^2 let at x=α, (d^2 y/dx^2 )=0 and y=0 ⇒ 12x^2 −6kx^3 −2=0 ⇒ 6x^2 −3k(x+c)−1=0 ⇒ 6x^2 −3kx−(3ck+1)=0 ..(i) 6(x+c)−3kx^2 −(3ck+1)x=0 ⇒ 3kx^2 +(3ck−5)x−6c=0 ..(ii) ⇒ 2(ii)−k(i) gives (6ck−10+3k^2 )x=12c−k(3ck+1) ⇒ x=((12c−k(3ck+1))/(3k^2 +6ck−10)) 6{12c−k(3ck+1)}^2 −3k{12c−k(3ck+1)}(3k^2 +6ck−10) −(3ck+1)(3k^2 +6ck−10)^2 =0 ......

$${y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−{c}\right) \\ $$$$\:\:={x}^{\mathrm{4}} −{px}^{\mathrm{3}} −{x}^{\mathrm{2}} +\left({p}−{c}\right){x}+{cp} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{px}^{\mathrm{2}} −\mathrm{2}{x}+{p}−{c} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{12}{x}^{\mathrm{2}} −\mathrm{6}{px}−\mathrm{2} \\ $$$${y}={p}−{kx}^{\mathrm{2}} \\ $$$${let}\:\:{at}\:{x}=\alpha,\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0}\:\:{and}\:\:{y}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{12}{x}^{\mathrm{2}} −\mathrm{6}{kx}^{\mathrm{3}} −\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{6}{x}^{\mathrm{2}} −\mathrm{3}{k}\left({x}+{c}\right)−\mathrm{1}=\mathrm{0} \\ $$$$\:\Rightarrow\:\:\mathrm{6}{x}^{\mathrm{2}} −\mathrm{3}{kx}−\left(\mathrm{3}{ck}+\mathrm{1}\right)=\mathrm{0}\:\:..\left({i}\right) \\ $$$$\:\:\:\:\mathrm{6}\left({x}+{c}\right)−\mathrm{3}{kx}^{\mathrm{2}} −\left(\mathrm{3}{ck}+\mathrm{1}\right){x}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{3}{kx}^{\mathrm{2}} +\left(\mathrm{3}{ck}−\mathrm{5}\right){x}−\mathrm{6}{c}=\mathrm{0}\:\:..\left({ii}\right) \\ $$$$\Rightarrow\:\mathrm{2}\left({ii}\right)−{k}\left({i}\right)\:\:\:{gives} \\ $$$$\:\left(\mathrm{6}{ck}−\mathrm{10}+\mathrm{3}{k}^{\mathrm{2}} \right){x}=\mathrm{12}{c}−{k}\left(\mathrm{3}{ck}+\mathrm{1}\right) \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{12}{c}−{k}\left(\mathrm{3}{ck}+\mathrm{1}\right)}{\mathrm{3}{k}^{\mathrm{2}} +\mathrm{6}{ck}−\mathrm{10}} \\ $$$$\mathrm{6}\left\{\mathrm{12}{c}−{k}\left(\mathrm{3}{ck}+\mathrm{1}\right)\right\}^{\mathrm{2}} \\ $$$$−\mathrm{3}{k}\left\{\mathrm{12}{c}−{k}\left(\mathrm{3}{ck}+\mathrm{1}\right)\right\}\left(\mathrm{3}{k}^{\mathrm{2}} +\mathrm{6}{ck}−\mathrm{10}\right) \\ $$$$−\left(\mathrm{3}{ck}+\mathrm{1}\right)\left(\mathrm{3}{k}^{\mathrm{2}} +\mathrm{6}{ck}−\mathrm{10}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$...... \\ $$

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