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f-x-x-a-x-b-x-a-x-b-Find-minimum-and-maximum-




Question Number 53732 by ajfour last updated on 25/Jan/19
f(x)=(((x+a)(x+b))/((x−a)(x−b)))  Find minimum and maximum.
f(x)=(x+a)(x+b)(xa)(xb)Findminimumandmaximum.
Commented by mr W last updated on 25/Jan/19
at x=a and x=b: f(x)→±∞
atx=aandx=b:f(x)±
Commented by ajfour last updated on 25/Jan/19
There is a definite answer given,  maximum and minimum are  asked. Its from a good book, Sir.  must then be local minimum and  maximum.
Thereisadefiniteanswergiven,maximumandminimumareasked.Itsfromagoodbook,Sir.mustthenbelocalminimumandmaximum.
Commented by mr W last updated on 25/Jan/19
since f(x)→±∞ when x→a or →b,  there is no min. or max. but  there can be a local minimum and a  local maximum.  f(x)=((x^2 +(a+b)x+ab)/(x^2 −(a+b)x+ab))  f(x)=1+((2(a+b)x)/(x^2 −(a+b)x+ab))  f′(x)=0  (1/(x^2 −(a+b)x+ab))−((x(2x−(a+b)))/((x^2 −(a+b)x+ab)^2 ))=0  x^2 −(a+b)x+ab=x(2x−(a+b))  x^2 =ab  ⇒x=±(√(ab))    i.e. if ab≥0 there are local min. and  local max. at x=±(√(ab))  f(x)_(min/max) =(((a±(√(ab)))(b±(√(ab))))/((a∓(√(ab)))(b∓(√(ab)))))    if ab<0 there is no local min. or max.    but we don′t need to use calculus to  get this result, just look at  f(x)=1+((2(a+b)x)/(x^2 −(a+b)x+ab))=1+((2(a+b))/(x+((ab)/x)−(a+b)))  if ab≥0,  x+((ab)/x)≥2(√(ab))  x+((ab)/x)≤−2(√(ab))  f(x)=1+((2(a+b))/(x+((ab)/x)−(a+b)))≤1+((2(a+b))/(2(√(ab))−(a+b)))=−((((√a)+(√b))/( (√a)−(√b))))^2 =f_(max)   f(x)=1+((2(a+b))/(x+((ab)/x)−(a+b)))≥1+((2(a+b))/(−2(√(ab))−(a+b)))=−((((√a)−(√b))/( (√a)+(√b))))^2 =f_(min)
sincef(x)±whenxaorb,thereisnomin.ormax.buttherecanbealocalminimumandalocalmaximum.f(x)=x2+(a+b)x+abx2(a+b)x+abf(x)=1+2(a+b)xx2(a+b)x+abf(x)=01x2(a+b)x+abx(2x(a+b))(x2(a+b)x+ab)2=0x2(a+b)x+ab=x(2x(a+b))x2=abx=±abi.e.ifab0therearelocalmin.andlocalmax.atx=±abf(x)min/max=(a±ab)(b±ab)(aab)(bab)ifab<0thereisnolocalmin.ormax.butwedontneedtousecalculustogetthisresult,justlookatf(x)=1+2(a+b)xx2(a+b)x+ab=1+2(a+b)x+abx(a+b)ifab0,x+abx2abx+abx2abf(x)=1+2(a+b)x+abx(a+b)1+2(a+b)2ab(a+b)=(a+bab)2=fmaxf(x)=1+2(a+b)x+abx(a+b)1+2(a+b)2ab(a+b)=(aba+b)2=fmin
Commented by ajfour last updated on 25/Jan/19
answer:  −((((√a)−(√b))/( (√a)+(√b))))^2  &  −((((√a)+(√b))/( (√a)−(√b))))^2  .
answer:(aba+b)2&(a+bab)2.
Commented by mr W last updated on 25/Jan/19
this is the same as my result (see below).  answer in book is not the best, since  it requests that a≥0 and b≥0, but in  fact this is not necessary. it is only  necessary that ab≥0, e.g. a=−2, b=−5.
thisisthesameasmyresult(seebelow).answerinbookisnotthebest,sinceitrequeststhata0andb0,butinfactthisisnotnecessary.itisonlynecessarythatab0,e.g.a=2,b=5.
Commented by mr W last updated on 25/Jan/19
(((a+(√(ab)))(b+(√(ab))))/((a−(√(ab)))(b−(√(ab)))))  =(((√a)((√a)+(√b))(√b)((√b)+(√a)))/( (√a)((√a)−(√b))(√b)((√b)−(√a))))  =−((((√a)+(√b))((√a)+(√b)))/(((√a)−(√b))((√a)−(√b))))  =−((((√a)+(√b))/( (√a)−(√b))))^2 =as given in book    (((a−(√(ab)))(b−(√(ab))))/((a+(√(ab)))(b+(√(ab)))))  =(((√a)((√a)−(√b))(√b)((√b)−(√a)))/( (√a)((√a)+(√b))(√b)((√b)+(√a))))  =−((((√a)−(√b))((√a)−(√b)))/(((√a)+(√b))((√a)+(√b))))  =−((((√a)−(√b))/( (√a)+(√b))))^2 =as given in book
(a+ab)(b+ab)(aab)(bab)=a(a+b)b(b+a)a(ab)b(ba)=(a+b)(a+b)(ab)(ab)=(a+bab)2=asgiveninbook(aab)(bab)(a+ab)(b+ab)=a(ab)b(ba)a(a+b)b(b+a)=(ab)(ab)(a+b)(a+b)=(aba+b)2=asgiveninbook
Commented by ajfour last updated on 25/Jan/19
yes, sir. too well did you explain.  Thank you too much.
yes,sir.toowelldidyouexplain.Thankyoutoomuch.
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19
y=(((x+a)(x+b))/((x−a)(x−b)))  lny=ln(x+a)+ln(x+b)−ln(x−a)−ln(x−b)  (1/y)(dy/dx)=(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b))  for max/min (dy/dx)=0  (1/(x+a))−(1/(x−b))+(1/(x+b))−(1/(x−a))=0  ((x−b−x−a)/((x+a)(x−b)))+((x−a−x−b)/((x+b)(x−a)))=0  ((−(a+b))/((x+a)(x−b)))+((−(a+b))/((x+b)(x−a)))=0  (1/((x+a)(x−b)))+(1/((x+b)(x−a)))=0  x^2 −ax+bx−ab+x^2 −bx+ax−ab=0  2x^2 −2ab=0  x=±(√(ab))   (1/y)(dy/dx)=(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b))  (dy/dx)=(((x+a)(x+b))/((x−a)(x−b)))[(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b))]  ((wait...)/)wl  now i am going to find change of sign  using first derivativd method to find max/min  ((dy/dx))=(((x+a)(x+b))/(x^2 −x(a+b)+ab))×[(1/(x+a))−(1/(x−b))+(1/(x+b))−(1/(x−a))]  =((x^2 +x(a+b)+ab)/(x^2 −x(a+b)+ab))×[((x−b−x−a)/(x^2 −xb+ax−ab))+((x−a−x−b)/(x^2 −ax+bx−ab))]  =(N/D)×[((−(a+b))/(x^2 −ab+x(a−b)))+((−(a+b))/(x^2 −ab−x(a−b)))]  =(N/D)×−(a+b)×[((x^2 −ab−x(a−b)+x^2 −ab+x(a−b))/((x^2 −ab)^2 −x^2 (a−b)^2 ))]  nowD<N  so(N/D)>1   (N/D)×−(a+b)=−ve  [((2(x^2 −ab))/((x^2 −ab)^2 −x^2 {(a+b)^2 −4ab))]  when x>(√(ab))  x^2 =h+ab  =(N/D)×−(a+b)×[((2(x^2 −ab))/((x^2 −ab)^2 −x^2 {(a+b)^2 −4ab))]  =(−ve)×[((2h)/(h^2 −(h+ab)(a−b)^2 ))]=(−ve)×((+ve)/(−ve))  when x>(√(ab)) sign change from −ve to +ve  so at x=(√(ab)) f(x) min  y_ =(((x+a)(x+b))/((x−a)(x−b))) (at x=(√(ab)) )    =(((√a) ((√b) +(√a) )×(√b) ((√a) +(√b) ))/( (√a) ((√b) −(√a) )×(√b) ((√a) −(√b) )))  =(−)((((√a) +(√b) )^2 )/(((√(b−))(√a) )^2 )) →min value  so at x=(√(ab))  min value    at x=−(√(ab)) max value  y=(((−(√(ab)) +a)(−(√(ab)) +b))/((−(√(ab)) −a)(−(√(ab)) −b)))  =(((√a) ((√a) −(√b) )×−(√b) ((√a) −(√b) ))/(−(√a) ((√a) +(√b) )×−(√b) ((√a) +(√b) )))   =(−1)×((((√a) −(√b) )^2 )/(((√a) +(√b) )))←max vzlud...
y=(x+a)(x+b)(xa)(xb)lny=ln(x+a)+ln(x+b)ln(xa)ln(xb)1ydydx=1x+a+1x+b1xa1xbformax/mindydx=01x+a1xb+1x+b1xa=0xbxa(x+a)(xb)+xaxb(x+b)(xa)=0(a+b)(x+a)(xb)+(a+b)(x+b)(xa)=01(x+a)(xb)+1(x+b)(xa)=0x2ax+bxab+x2bx+axab=02x22ab=0x=±ab1ydydx=1x+a+1x+b1xa1xbdydx=(x+a)(x+b)(xa)(xb)[1x+a+1x+b1xa1xb]waitwlnowiamgoingtofindchangeofsignusingfirstderivativdmethodtofindmax/min(dydx)=(x+a)(x+b)x2x(a+b)+ab×[1x+a1xb+1x+b1xa]=x2+x(a+b)+abx2x(a+b)+ab×[xbxax2xb+axab+xaxbx2ax+bxab]=ND×[(a+b)x2ab+x(ab)+(a+b)x2abx(ab)]=ND×(a+b)×[x2abx(ab)+x2ab+x(ab)(x2ab)2x2(ab)2]nowD<NsoND>1ND×(a+b)=ve[2(x2ab)(x2ab)2x2{(a+b)24ab]whenx>abx2=h+ab=ND×(a+b)×[2(x2ab)(x2ab)2x2{(a+b)24ab]=(ve)×[2hh2(h+ab)(ab)2]=(ve)×+vevewhenx>absignchangefromveto+vesoatx=abf(x)miny=(x+a)(x+b)(xa)(xb)(atx=ab)=a(b+a)×b(a+b)a(ba)×b(ab)=()(a+b)2(ba)2minvaluesoatx=abminvalueatx=abmaxvaluey=(ab+a)(ab+b)(aba)(abb)=a(ab)×b(ab)a(a+b)×b(a+b)=(1)×(ab)2(a+b)maxvzlud
Commented by ajfour last updated on 25/Jan/19
welcome back Sir..
welcomebackSir..
Commented by ajfour last updated on 25/Jan/19
thanks sir, do i post the answer?
thankssir,doiposttheanswer?
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19
i went to see the news...
iwenttoseethenews

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