If-x-1-f-x-f-1-x-1-x-1-find-f-x-

Question Number 86160 by jagoll last updated on 27/Mar/20

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

find f (x)

Commented by john santu last updated on 27/Mar/20

r e p l a c e x b y \frac{1}{x}

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

(\frac{1 - x}{x}) f (\frac{1}{x}) + f (x) = \frac{x}{1 - x} [\times x]

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

m u l t i p l y e q (i) b y (1 - x)

(1 - x) f (\frac{1}{x}) - {(1 - x)}^{2} f (x) = - 1 \leftarrow (i)

(i i) - (i)

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

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

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

Commented by jagoll last updated on 27/Mar/20

thank you mr john & tanmay

Commented by mathmax by abdo last updated on 27/Mar/20

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

w e g e t t h e s y s t e m {\begin{cases} (x - 1) f (x) + f (\frac{1}{x}) = \frac{1}{x - 1} \\ f (x) + \frac{1 - x}{x} f (\frac{1}{x}) = \frac{x}{1 - x} \end{cases}

Δ_{s} = | \begin{matrix} x - 1 1 \\ 1 \frac{1 - x}{x} \end{matrix} | = \frac{(x - 1) (1 - x)}{x} - 1 = \frac{x - x^{2} - 1 + x - x}{x}

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

= \frac{| \begin{matrix} \frac{1}{x - 1} 1 \\ \frac{x}{1 - x} \frac{1 - x}{x} \end{matrix} |}{\frac{- x^{2} + x - 1}{x}} = \frac{- \frac{1}{x} - \frac{x}{1 - x}}{\frac{- x^{2} + x - 1}{x}} = \frac{- 1 + x - x^{2}}{x (1 - x)} \times \frac{x}{- x^{2} + x - 1}

= \frac{1}{1 - x} \Rightarrow f (x) = \frac{1}{1 - x}

Answered by TANMAY PANACEA. last updated on 27/Mar/20

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

r e p l a c i n g x b y \frac{1}{x}

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

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

p u t t i n g t h e v a l u e o f f (x) i n g i v e n e q n

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

s o w e g e t

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

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

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

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

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

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

Commented by jagoll last updated on 27/Mar/20

thank you mister

Commented by Serlea last updated on 27/Mar/20

I think Something is wrong Sir

Line number 4 : How did u get ur Second f (\frac{1}{x})

2) Why {didn}^{'} t you multiply the both side by (x - 1) only the

left hand side of the equation .

Commented by john santu last updated on 27/Mar/20

y e s s i r .