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Question Number 121368 by abdullahquwatan last updated on 07/Nov/20
find minimun and maksimum  y=((x^3 −1)/(x^2 −3))
findminimunandmaksimumy=x31x23
Answered by MJS_new last updated on 07/Nov/20
x^3 −1=0 ⇒ zero at x=1  x^2 −3≠0 ⇒ asymptotes x=±(√3)  ⇒ −∞<y<+∞  ((x^3 −1)/(x^2 −3))=x+((3x−1)/(x^2 −3)) ⇒ asymptote y=x  y′=((x(x^3 −9x+2))/((x^2 −3)^2 ))  y′′=((6(x^3 −x^2 +9x−1))/((x^2 −3)^3 ))  solving y′=0 and testing y′′  { ((<0 (maximum))),((=0 (flat point))),((>0 (minimum))) :}  leads to  x_1 ≈−3.10548 local max with y≈−4.65822  x_2 =0 local minimum with y=(1/3)  x_3 ≈.223462 local maximum with y≈.335193  x_4 ≈2.88202 local minimum with y≈4.32303
x31=0zeroatx=1x230asymptotesx=±3<y<+x31x23=x+3x1x23asymptotey=xy=x(x39x+2)(x23)2y=6(x3x2+9x1)(x23)3solvingy=0andtestingy{<0(maximum)=0(flatpoint)>0(minimum)leadstox13.10548localmaxwithy4.65822x2=0localminimumwithy=13x3.223462localmaximumwithy.335193x42.88202localminimumwithy4.32303
Commented by abdullahquwatan last updated on 07/Nov/20
thank you sir
thankyousir
Commented by MJS_new last updated on 07/Nov/20
you are welcome
youarewelcome
Answered by TANMAY PANACEA last updated on 07/Nov/20
(dy/dx)=(((x^2 −3)(3x^2 )−(x^3 −1)2x)/((x^2 −3)^2 ))  (dy/dx)=((3x^4 −9x^2 −2x^4 +2x)/((x^2 −3)^2 ))=((x^4 −9x^2 +2x)/((x^2 −3)^2 ))  for max/min (dy/dx)=0  x(x^3 −9x+2)=0  x=0 and  0.223, 2.882  when x>2.882    (dy/dx)>0  when2.882 >x>0.223      (dy/dx)<0  when0.223>x>0      (dy/dx)>0  when x<0  (dy/dx)<0  sign of (dy/dx) changes from −ve to +ve  when curve cross x=0  so at x=0  given function minimum  y_(x=0) =(1/3)(minimum)=(1/3)  sign change from −ve to +ve when cross x=0.223  so maximum y_(x=0.223) =(((0.223)^3 −1)/((0.223)^2 −3))(maximum)=  when 2.882>x  (dy/dx)<0  when x>2.882  (dy/dx)>0   sign change from −ve to +ve   so minimum  y_(x=2.882)  =(((2.882)^3 −1)/((2.882)^2 −3))( minimum)=
dydx=(x23)(3x2)(x31)2x(x23)2dydx=3x49x22x4+2x(x23)2=x49x2+2x(x23)2formax/mindydx=0x(x39x+2)=0x=0and0.223,2.882whenx>2.882dydx>0when2.882>x>0.223dydx<0when0.223>x>0dydx>0whenx<0dydx<0signofdydxchangesfromveto+vewhencurvecrossx=0soatx=0givenfunctionminimumyx=0=13(minimum)=13signchangefromveto+vewhencrossx=0.223somaximumyx=0.223=(0.223)31(0.223)23(maximum)=when2.882>xdydx<0whenx>2.882dydx>0signchangefromveto+vesominimumyx=2.882=(2.882)31(2.882)23(minimum)=
Commented by TANMAY PANACEA last updated on 07/Nov/20
pls calculate the[value
plscalculatethe[value
Commented by abdullahquwatan last updated on 07/Nov/20
thank you sir
thankyousir
Commented by TANMAY PANACEA last updated on 07/Nov/20
most welcome
mostwelcome

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