f-x-3-x-1-f-x-3-1-x-x-x-1-f-x-

Question Number 213000 by golsendro last updated on 28/Oct/24

f (\frac{x - 3}{x + 1}) + f (\frac{x + 3}{1 - x}) = x, x \neq \pm 1

f (x) = ?

Answered by Ghisom last updated on 28/Oct/24

\frac{u - 3}{u + 1} = v \Leftrightarrow u = \frac{v + 3}{1 - v}

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

f (x) = \frac{x (x^{2} + 7)}{2 (1 - x^{2})}

Answered by ajfour last updated on 28/Oct/24

f (u) + f (v) = x

u = h (x) = \frac{x - 3}{1 + x}

\Rightarrow x = \frac{u + 3}{1 - u}

v = - h (- x) = \frac{x + 3}{1 - x}

x = \frac{v - 3}{1 + v} = \frac{u + 3}{1 - u}

v - 3 - u v + 3 u = u + 3 + u v + 3 v

(v - u) + u v + 3 = 0

o r v = \frac{u - 3}{1 + u} & u = \frac{v + 3}{1 - v}

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

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

\Rightarrow f (x) - f (\frac{x + 3}{1 - x}) = \frac{x^{2} + 3}{x (1 - x)} . . (i)

& f (x) + f (\frac{x + 3}{1 - x}) = \frac{x - 3}{1 - x} . . (i i)

a d d i n g (i) & (i i)

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

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

Commented by Ghisom last updated on 30/Oct/24

test your solution!

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