Given-f-x-x-1-x-2-find-minimum-value-of-function-h-x-f-x-3-x-1-

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 \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 \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