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

Commented bymathocean1 last updated on 07/Jul/20

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

Commented bymathocean1 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.

Commented by1549442205 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

Commented bymathocean1 last updated on 07/Jul/20

thank you very much sir!