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


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

$Given f (\frac{x}{x + 1}) = x^{2} . find minimum value$ $of function h (x) = f (x) - \frac{3}{x - 1}$
Commented by john santu last updated on 24/Jun/20
$let (x/(x+1)) = z ⇒xz+z = x x = (z/(1−z)) ⇒f(z)= (z^2 /((1−z)^2 )) similar to f(x)= (x^2 /((1−x)^2 )) . h(x) = (x^2 /((1−x)^2 )) + (3/(1−x)) h(x) = ((x^2 −3x+3)/((1−x)^2 )) h′(x)= (((2x−3)(1−x)^2 +2(x^2 −3x+3)(1−x))/((1−x)^4 )) = 0 ⇒(1−x){ (2x−3)(1−x)+2x^2 −6x+6 }=0 (1−x){−2x^2 +5x−3+2x^2 −6x+6 }=0 (1−x)(3−x)= 0 , critical point x= 3 min h(3) = (3/4)$
$let \frac{x}{x + 1} = z \Rightarrow xz + z = x$ $x = \frac{z}{1 - z} \Rightarrow f (z) = \frac{z^{2}}{{(1 - z)}^{2}} similar to$ $f (x) = \frac{x^{2}}{{(1 - x)}^{2}} .$ $h (x) = \frac{x^{2}}{{(1 - x)}^{2}} + \frac{3}{1 - x}$ $h (x) = \frac{x^{2} - 3 x + 3}{{(1 - x)}^{2}}$ $h^{'} (x) = \frac{(2 x - 3) {(1 - x)}^{2} + 2 (x^{2} - 3 x + 3) (1 - x)}{{(1 - x)}^{4}} = 0$ $\Rightarrow (1 - x) {(2 x - 3) (1 - x) + {2 x}^{2} - 6 x + 6} = 0$ $(1 - x) {- {2 x}^{2} + 5 x - 3 + {2 x}^{2} - 6 x + 6} = 0$ $(1 - x) (3 - x) = 0, critical point x = 3$ $\min h (3) = \frac{3}{4}$
	Answered by mathmax by abdo last updated on 24/Jun/20
	$let (x/(x+1))=t ⇒x =tx +t ⇒(1−t)x=t ⇒x =(t/(1−t)) ⇒ f(t) =(t^2 /((1−t)^2 )) ⇒f(x) =(x^2 /((x−1)^2 )) =(x^2 /(x^2 −2x+1)) ⇒ g(x) =(x^2 /((x−1)^2 ))−(3/(x−1)) =((x^2 −3(x−1))/((x−1)^2 )) =((x^2 −3x+3)/((x−1)^2 )) g defined on R{1} lim_(x→∞) g(x) =1 and lim_(x→1^− ) g(x) =+∞ =lim_(x→1^+ ) g(x) g^′ (x) =(((2x−3)(x−1)^2 −2(x−1)(x^2 −3x+3))/((x−1)^4 )) =^′ (((2x−3)(x−1)−2(x^2 −3x+3))/((x−1)^3 )) =((2x^2 −5x +3−2x^2 +6x−6)/((x−1)^3 )) =(((x−3)(x−1))/((x−1)^4 )) and g^′ (x) =0 ⇔x =3 x −∞ 1 3 +∞ g^′ (x) + ∣∣ − 0 + g(x) 1 inc +∞ +∞decr g(3) inc 1 g(3) =(3/4) is the minimum value for g$
	$let \frac{x}{x + 1} = t \Rightarrow x = tx + t \Rightarrow (1 - t) x = t \Rightarrow x = \frac{t}{1 - t} \Rightarrow$ $f (t) = \frac{t^{2}}{{(1 - t)}^{2}} \Rightarrow f (x) = \frac{x^{2}}{{(x - 1)}^{2}} = \frac{x^{2}}{x^{2} - 2 x + 1} \Rightarrow$ $g (x) = \frac{x^{2}}{{(x - 1)}^{2}} - \frac{3}{x - 1} = \frac{x^{2} - 3 (x - 1)}{{(x - 1)}^{2}} = \frac{x^{2} - 3 x + 3}{{(x - 1)}^{2}} g defined on R {1}$ $\lim_{x \to \infty} g (x) = 1 and \lim_{x \to 1^{-}} g (x) = + \infty = \lim_{x \to 1^{+}} g (x)$ $g^{^{'}} (x) = \frac{(2 x - 3) {(x - 1)}^{2} - 2 (x - 1) (x^{2} - 3 x + 3)}{{(x - 1)}^{4}}$ $=^{^{'}} \frac{(2 x - 3) (x - 1) - 2 (x^{2} - 3 x + 3)}{{(x - 1)}^{3}} = \frac{{2 x}^{2} - 5 x + 3 - {2 x}^{2} + 6 x - 6}{{(x - 1)}^{3}}$ $= \frac{(x - 3) (x - 1)}{{(x - 1)}^{4}} and g^{^{'}} (x) = 0 \Leftrightarrow x = 3$ $x - \infty 1 3 + \infty$ $g^{^{'}} (x) + ∣∣ - 0 +$ $g (x) 1 inc + \infty + \infty decr g (3) inc 1$ $g (3) = \frac{3}{4} is the minimum value for g$
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