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Question Number 95150 by bobhans last updated on 23/May/20

how do you solve f (x) + 2 f (\frac{1}{1 - x}) = x

for f ?

Answered by mr W last updated on 23/May/20

f (x) + 2 f (\frac{1}{1 - x}) = x \dots (i)

f (\frac{1}{1 - x}) + 2 f (\frac{x - 1}{x}) = \frac{1}{1 - x} \dots (i i)

f (\frac{x - 1}{x}) + 2 f (x) = \frac{x - 1}{x} \dots (i i i)

(i) - 2 (i i) :

f (x) - 4 f (\frac{x - 1}{x}) = x - \frac{2}{1 - x} \dots (i v)

4 (i i i) + (i v) :

9 f (x) = \frac{4 (x - 1)}{x} + x - \frac{2}{1 - x}

\Rightarrow f (x) = \frac{x^{3} + 3 x^{2} - 6 x + 4}{9 x (x - 1)}

Commented by bobhans last updated on 23/May/20

ok . i compare to my way

f (x) + 2 f (\frac{1}{1 - x}) = x

f (\frac{1}{1 - x}) + 2 f (\frac{1}{1 - \frac{1}{1 - x}}) = \frac{1}{1 - x}

f (\frac{1}{1 - x}) + 2 f (\frac{x - 1}{x}) = \frac{1}{1 - x} \dots (2)

f (\frac{1}{1 - \frac{1}{1 - x}}) + 2 f (\frac{\frac{1}{1 - x} - 1}{\frac{1}{1 - x}}) = \frac{1}{1 - \frac{1}{1 - x}}

f (\frac{x - 1}{x}) + 2 f (x) = \frac{x - 1}{x} \dots (3)

(1) - 2 \times (2) + 4 \times (3)

9 f (x) = \frac{x^{3} + 3 x^{2} - 6 x + 4}{x (x - 1)} \Rightarrow f (x) = \frac{x^{3} + 3 x^{2} - 6 x + 4}{9 x (x - 1)}

Commented by I want to learn more last updated on 23/May/20

Weldone sir