Menu Close

Question-101778

Tinku Tara 0 Comments
Pin




Question Number 101778 by mathocean1 last updated on 04/Jul/20
Answered by mr W last updated on 04/Jul/20
f(x)=P(x) ((x^2 +4x+1)/(x+1))=x^2 x^3 −4x−1=0 say g(x)=x^3 −4x−1 g(0)=−1<0 lim_(x→+∞) g(x)=+∞>0 ⇒there is at least one value x_A ∈(0,+∞) g(x_A )=0 i.e. f(x) and P(x) intersect at least at one point A(x_A ,y_A ) with x_A >0 and y_A =x_A ^2 >0. for roots of x^3 −4x−1=0 see Q101658
$${f}\left({x}\right)={P}\left({x}\right) \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{{x}+\mathrm{1}}={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\ $$$${say}\:{g}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{1} \\ $$$${g}\left(\mathrm{0}\right)=−\mathrm{1}<\mathrm{0} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{g}\left({x}\right)=+\infty>\mathrm{0} \\ $$$$\Rightarrow{there}\:{is}\:{at}\:{least}\:{one}\:{value}\:{x}_{{A}} \in\left(\mathrm{0},+\infty\right) \\ $$$${g}\left({x}_{{A}} \right)=\mathrm{0} \\ $$$${i}.{e}.\:{f}\left({x}\right)\:{and}\:{P}\left({x}\right)\:{intersect}\:{at}\:{least} \\ $$$${at}\:{one}\:{point}\:{A}\left({x}_{{A}} ,{y}_{{A}} \right)\:{with}\:{x}_{{A}} >\mathrm{0}\:{and} \\ $$$${y}_{{A}} ={x}_{{A}} ^{\mathrm{2}} >\mathrm{0}. \\ $$$$ \\ $$$${for}\:{roots}\:{of}\:{x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0}\:{see}\:{Q}\mathrm{101658} \\ $$
Commented by mathocean1 last updated on 07/Jul/20
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$
Answered by 1549442205 last updated on 07/Jul/20
The cordinates of the intersection points of two the curves are roots of the system: { ((y=((x^2 +4x+1)/(x+1)))),((y=x^2 )) :} ⇔ { ((((x^2 +4x+1)/(x+1))=x^2 (1))),((y=x^2 (2))) :} (1)⇔x^3 −4x−1=0⇔x=((−4(√3))/3) cos((arccos((−3(√3))/(16))+2kπ)/3) with k∈{0,1,2} (see to solve eqs. in the question 101658) For k∈{0,2} we get the negative roots which rejected For k=1 we get x=2.114907541 Thus,x_A =2.1149,y_A =4.4728
$$\mathrm{The}\:\mathrm{cordinates}\:\:\mathrm{of}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{points} \\ $$$$\mathrm{of}\:\mathrm{two}\:\mathrm{the}\:\mathrm{curves}\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\begin{cases}{\mathrm{y}=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{1}}{\mathrm{x}+\mathrm{1}}}\\{\mathrm{y}=\mathrm{x}^{\mathrm{2}} }\end{cases}\:\Leftrightarrow\begin{cases}{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{1}}{\mathrm{x}+\mathrm{1}}=\mathrm{x}^{\mathrm{2}} \:\left(\mathrm{1}\right)}\\{\mathrm{y}=\mathrm{x}^{\mathrm{2}} \:\left(\mathrm{2}\right)}\end{cases}\: \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\mathrm{x}^{\mathrm{3}} −\mathrm{4x}−\mathrm{1}=\mathrm{0}\Leftrightarrow\mathrm{x}=\frac{−\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{cos}\frac{\mathrm{arccos}\frac{−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{16}}+\mathrm{2k}\pi}{\mathrm{3}}\:\mathrm{with}\:\mathrm{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2}\right\} \\ $$$$\left(\mathrm{see}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{eqs}.\:\mathrm{in}\:\mathrm{the}\:\mathrm{question}\:\mathrm{101658}\right) \\ $$$$\mathrm{For}\:\mathrm{k}\in\left\{\mathrm{0},\mathrm{2}\right\}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{negative}\:\mathrm{roots} \\ $$$$\mathrm{which}\:\mathrm{rejected} \\ $$$$\mathrm{For}\:\mathrm{k}=\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\mathrm{x}=\mathrm{2}.\mathrm{114907541} \\ $$$$\boldsymbol{\mathrm{Thus}},\boldsymbol{\mathrm{x}}_{\boldsymbol{\mathrm{A}}} =\mathrm{2}.\mathrm{1149},\boldsymbol{\mathrm{y}}_{\boldsymbol{\mathrm{A}}} =\mathrm{4}.\mathrm{4728} \\ $$
Commented by mathocean1 last updated on 06/Jul/20
Thank you sirs! Please is there other way to solve x^3 −4x−1=0 ... i don′t know : arccos... i′m still learner in trigo.
$${Thank}\:{you}\:{sirs}! \\ $$$${Please}\:{is}\:{there}\:{other}\:{way}\:{to}\:{solve} \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{4x}−\mathrm{1}=\mathrm{0}\:…\:{i}\:{don}'{t}\:{know}\:: \\ $$$${arccos}…\:{i}'{m}\:{still}\:{learner}\:{in}\:{trigo}. \\ $$
Commented by 1549442205 last updated on 07/Jul/20
arccos or cos^(−1 ) is convert function of cos . Arccos m is a angle which 0≤Arccosm≤π so that if arccos m=ϕ then cosϕ=m.Example:If cosx=(1/3) then we have x=arccos(1/3).In other words, Arccos a is the root of the equation cosx=a.Sometimes ones write cos^(−1) m instead of arccos m
$$\mathrm{arccos}\:\mathrm{or}\:\mathrm{cos}^{−\mathrm{1}\:} \mathrm{is}\:\mathrm{convert}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{cos}\:\:.\:\mathrm{Arccos}\:\mathrm{m}\:\mathrm{is}\:\mathrm{a}\:\mathrm{angle}\:\mathrm{which}\: \\ $$$$\mathrm{0}\leqslant\mathrm{Arccosm}\leqslant\pi\:\mathrm{so}\:\mathrm{that}\:\mathrm{if}\:\mathrm{arccos}\:\mathrm{m}=\varphi \\ $$$$\mathrm{then}\:\mathrm{cos}\varphi=\mathrm{m}.\mathrm{Example}:\mathrm{If}\:\mathrm{cosx}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{have}\:\mathrm{x}=\mathrm{arccos}\frac{\mathrm{1}}{\mathrm{3}}.\mathrm{In}\:\mathrm{other}\:\mathrm{words}, \\ $$$$\mathrm{Arccos}\:\mathrm{a}\:\mathrm{is}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{cosx}=\mathrm{a}.\mathrm{Sometimes}\:\mathrm{ones}\:\mathrm{write}\:\mathrm{cos}^{−\mathrm{1}} \mathrm{m} \\ $$$$\mathrm{instead}\:\mathrm{of}\:\:\mathrm{arccos}\:\mathrm{m} \\ $$
Commented by mathocean1 last updated on 07/Jul/20
thank you very much sir!
$${thank}\:{you}\:{very}\:{much}\:{sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *