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Question-212067




Question Number 212067 by Spillover last updated on 28/Sep/24
Answered by Spillover last updated on 28/Sep/24
Commented by Ghisom last updated on 28/Sep/24
that′s not true. we can solve the equation  for y:  y=((x^4 −((1189)/(360))x^3 +3x^2 −((1189)/(180))x+1)/(x^5 −((1188)/(360))x^4 +4x^3 −((1189)/(120))x^2 +3x−((1189)/(360))))  and plug in any value for x except the  zero of the denominator which is very  close to 3 (x≈3+(1/(35947)))  we get .49<P<∞  (minimum at x≈1.47520852,  y≈.497303672, P≈.490429398)  x=3 ⇒ y=3 is maybe the only solution  which makes sense.
$$\mathrm{that}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{for}\:{y}: \\ $$$${y}=\frac{{x}^{\mathrm{4}} −\frac{\mathrm{1189}}{\mathrm{360}}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\frac{\mathrm{1189}}{\mathrm{180}}{x}+\mathrm{1}}{{x}^{\mathrm{5}} −\frac{\mathrm{1188}}{\mathrm{360}}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} −\frac{\mathrm{1189}}{\mathrm{120}}{x}^{\mathrm{2}} +\mathrm{3}{x}−\frac{\mathrm{1189}}{\mathrm{360}}} \\ $$$$\mathrm{and}\:\mathrm{plug}\:\mathrm{in}\:\mathrm{any}\:\mathrm{value}\:\mathrm{for}\:{x}\:\mathrm{except}\:\mathrm{the} \\ $$$$\mathrm{zero}\:\mathrm{of}\:\mathrm{the}\:\mathrm{denominator}\:\mathrm{which}\:\mathrm{is}\:\mathrm{very} \\ $$$$\mathrm{close}\:\mathrm{to}\:\mathrm{3}\:\left({x}\approx\mathrm{3}+\frac{\mathrm{1}}{\mathrm{35947}}\right) \\ $$$$\mathrm{we}\:\mathrm{get}\:.\mathrm{49}<{P}<\infty \\ $$$$\left(\mathrm{minimum}\:\mathrm{at}\:{x}\approx\mathrm{1}.\mathrm{47520852},\right. \\ $$$$\left.{y}\approx.\mathrm{497303672},\:{P}\approx.\mathrm{490429398}\right) \\ $$$${x}=\mathrm{3}\:\Rightarrow\:{y}=\mathrm{3}\:\mathrm{is}\:\mathrm{maybe}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution} \\ $$$$\mathrm{which}\:\mathrm{makes}\:\mathrm{sense}. \\ $$

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