Question Number 100032 by bobhans last updated on 24/Jun/20

Given f((x/(x+1))) = x^2  . find minimum value  of function h(x)=f(x)−(3/(x−1))

Commented byjohn santu last updated on 24/Jun/20

let (x/(x+1)) = z ⇒xz+z = x  x = (z/(1−z)) ⇒f(z)= (z^2 /((1−z)^2 )) similar to  f(x)= (x^2 /((1−x)^2 )) .  h(x) = (x^2 /((1−x)^2 )) + (3/(1−x))  h(x) = ((x^2 −3x+3)/((1−x)^2 ))  h′(x)= (((2x−3)(1−x)^2 +2(x^2 −3x+3)(1−x))/((1−x)^4 )) = 0  ⇒(1−x){ (2x−3)(1−x)+2x^2 −6x+6 }=0  (1−x){−2x^2 +5x−3+2x^2 −6x+6 }=0  (1−x)(3−x)= 0 , critical point x= 3  min h(3) = (3/4)

Answered by mathmax by abdo last updated on 24/Jun/20

let (x/(x+1))=t ⇒x =tx +t ⇒(1−t)x=t ⇒x =(t/(1−t)) ⇒  f(t) =(t^2 /((1−t)^2 )) ⇒f(x) =(x^2 /((x−1)^2 )) =(x^2 /(x^2 −2x+1))  ⇒  g(x) =(x^2 /((x−1)^2 ))−(3/(x−1)) =((x^2 −3(x−1))/((x−1)^2 )) =((x^2 −3x+3)/((x−1)^2 )) g defined on R{1}  lim_(x→∞) g(x) =1  and lim_(x→1^− )  g(x) =+∞ =lim_(x→1^+ )   g(x)  g^′ (x) =(((2x−3)(x−1)^2 −2(x−1)(x^2 −3x+3))/((x−1)^4 ))  =^′ (((2x−3)(x−1)−2(x^2 −3x+3))/((x−1)^3 )) =((2x^2 −5x +3−2x^2 +6x−6)/((x−1)^3 ))  =(((x−3)(x−1))/((x−1)^4 ))  and g^′ (x) =0 ⇔x =3  x                  −∞                            1                       3                  +∞  g^′ (x)                              +             ∣∣       −            0          +  g(x)              1    inc           +∞    +∞decr  g(3)    inc     1  g(3) =(3/4)  is the minimum value for g