Find the function f(x) if 3f(x−1)−f(((1−x)/x)) = 2x


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Question Number 130894 by EDWIN88 last updated on 30/Jan/21

$F i n d t h e f u n c t i o n f (x) i f$ $3 f (x - 1) - f (\frac{1 - x}{x}) = 2 x$
Commented by benjo_mathlover last updated on 30/Jan/21
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Commented by EDWIN88 last updated on 30/Jan/21
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	Answered by mr W last updated on 30/Jan/21

	$3 f (x - 1) - f (\frac{1}{x} - 1) = 2 x . . . (i)$ $3 f (\frac{1}{x} - 1) - f (x - 1) = \frac{2}{x} . . . (i i)$ $3 \times (i) + (i i) :$ $8 f (x - 1) = 2 (3 x + \frac{1}{x})$ $f (x - 1) = \frac{1}{4} (3 x + \frac{1}{x})$ $\Rightarrow f (x) = \frac{1}{4} (3 (x + 1) + \frac{1}{x + 1})$
	Commented by EDWIN88 last updated on 30/Jan/21
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	Answered by benjo_mathlover last updated on 30/Jan/21

	$replace \frac{1 - x}{x} by t \Rightarrow x = \frac{1}{t + 1}$ $3 f (\frac{1}{t + 1} - 1) - f (t) = \frac{2}{t + 1}$ $3 f (\frac{- t}{t + 1}) - f (t) = \frac{2}{t + 1} . . . (i)$ $replace x - 1 by t \Rightarrow x = t + 1$ $3 f (t) - f (\frac{1 - t - 1}{t + 1}) = 2 t + 2$ $3 f (x) - f (\frac{- t}{t + 1}) = 2 t + 2 . . . (ii)$ $\Leftrightarrow 3 \times (ii) \Rightarrow 9 f (t) - 3 f (\frac{- t}{t + 1}) = 6 t + 6$ $\Leftrightarrow 1 \times (i) \Rightarrow - f (t) + 3 f (\frac{- t}{t + 1}) = \frac{2}{t + 1}$ $__________________________+$ $\Leftrightarrow 8 f (t) = 6 t + 6 + \frac{2}{t + 1}$ $\Leftrightarrow f (x) = \frac{3 x + 3}{4} + \frac{1}{4 x + 4} = \frac{12 {(x + 1)}^{2} + 4}{16 (x + 1)}$ $= \frac{{3 x}^{2} + 6 x + 4}{4 x + 4}$
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