find function f(x) and g(x) such that { ((f(2x−1)+g(1−x)=x+1)),((f((x/(x+1)))+2g((1/(2x+2)))=3)) :}


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Question Number 187988 by cortano12 last updated on 24/Feb/23

$find function f (x) and g (x)$ $such that {\begin{cases} f (2 x - 1) + g (1 - x) = x + 1 \\ f (\frac{x}{x + 1}) + 2 g (\frac{1}{2 x + 2}) = 3 \end{cases}$
	Answered by MathGuy last updated on 24/Feb/23

	$A n s w e r : - d o x \to x + 1 i n e q . 1 & x \to x - 1 i n e q . 2$ $y o u w i l l g e t$ $f (2 x + 1) + g (- x) = x + 2 a s n e w e q . 1 . c a l l i t e q . 3$ $&$ $f (\frac{x - 1}{x}) + 2 g (\frac{1}{2 x}) = 3 a s n e w e q . 2 . c a l l i t e q . 4$ $d o x \to - x i n e q . 3 & x \to \frac{1}{2 x} i n e q . 4$ $y o u w i l l g e t$ $f (1 - 2 x) + g (x) = 2 - x; c a l l i t e q . 5$ $&$ $f (1 - 2 x) + 2 g (x) = 3; c a l l i t e q . 6$ $d o i n g [e q . 6 - e q . 5] g i v e s$ $g (x) = x + 1$ $p u t v a l u e o f g (x) i n e q . 5$ $y o u w i l l g e t$ $f (1 - 2 x) = 1 - 2 x$ $n o w p u t t i n g x \to \frac{1 - x}{2}, g i v e s$ $f (x) = x$ $s o w e h a v e f (x) = x & g (x) = x + 1$
	Answered by Rasheed.Sindhi last updated on 24/Feb/23
	$such that { ((f(2x−1)+g(1−x)=x+1...(i))),((f((x/(x+1)))+2g((1/(2x+2)))=3....(ii))) :} (i):_(−) 2x−1=y⇒x=((y+1)/2) (i)⇒f(y)+g(1−((y+1)/2))=((y+1)/2)+1 f(y)+g(((1−y)/2))=((y+3)/2)....(iii) (ii):_(−) (x/(x+1))=y x−yx=y x=(y/(1−y)) (ii)⇒f(y)+2g((1/(2((y/(1−y)))+2)))=3 f(y)+2g((1/((2y+2−2y)/(1−y))))=3 f(y)+2g(((1−y)/2))=3......(iv) (iv)−(iii):g(((1−y)/2))=3−((y+3)/2)=((3−y)/2) Let ((1−y)/2)=x g(x)=((1−y+2)/2)=((1−y)/2)+1=x+1 2(iii)−(iv): (vi): f(y)+2g(((1−y)/2))=3 2(iii): 2f(y)+2g(((1−y)/2))=y+3 2(iii)−(iv): f(y)=y+3−3 f(y)=y f(x)=x$
	$such that {\begin{cases} f (2 x - 1) + g (1 - x) = x + 1 . . . (i) \\ f (\frac{x}{x + 1}) + 2 g (\frac{1}{2 x + 2}) = 3 . . . . (i i) \end{cases}$ $\underline{(i) :}$ $2 x - 1 = y \Rightarrow x = \frac{y + 1}{2}$ $(i) \Rightarrow f (y) + g (1 - \frac{y + 1}{2}) = \frac{y + 1}{2} + 1$ $f (y) + g (\frac{1 - y}{2}) = \frac{y + 3}{2} . . . . (i i i)$ $\underline{(i i) :}$ $\frac{x}{x + 1} = y$ $x - y x = y$ $x = \frac{y}{1 - y}$ $(i i) \Rightarrow f (y) + 2 g (\frac{1}{2 (\frac{y}{1 - y}) + 2}) = 3$ $f (y) + 2 g (\frac{1}{\frac{2 y + 2 - 2 y}{1 - y}}) = 3$ $f (y) + 2 g (\frac{1 - y}{2}) = 3 . . . . . . (i v)$ $(i v) - (i i i) : g (\frac{1 - y}{2}) = 3 - \frac{y + 3}{2} = \frac{3 - y}{2}$ $L e t \frac{1 - y}{2} = x$ $g (x) = \frac{1 - y + 2}{2} = \frac{1 - y}{2} + 1 = x + 1$ $2 (i i i) - (i v) :$ $(v i) : f (y) + 2 g (\frac{1 - y}{2}) = 3$ $2 (i i i) : 2 f (y) + 2 g (\frac{1 - y}{2}) = y + 3$ $2 (i i i) - (i v) : f (y) = y + 3 - 3$ $f (y) = y$ $f (x) = x$
	Answered by Rasheed.Sindhi last updated on 25/Feb/23

	$find function f (x) and g (x)$ $such that {\begin{cases} f (2 x - 1) + g (1 - x) = x + 1 . . . (i) \\ f (\frac{x}{x + 1}) + 2 g (\frac{1}{2 x + 2}) = 3 . . . . . (i i) \end{cases}$ $(i) : 1 - x = y \Rightarrow x = 1 - y$ $(i) \Rightarrow f (2 (1 - y) - 1) + g (y) = 1 - y + 1$ $f (1 - 2 y) + g (y) = 2 - y . . . . . (iii)$ $(ii) : \frac{1}{2 x + 2} = y \Rightarrow 2 xy + 2 y = 1 \Rightarrow x = \frac{1 - 2 y}{2 y}$ $(ii) \Rightarrow f (\frac{\frac{1 - 2 y}{2 y}}{\frac{1 - 2 y}{2 y} + 1}) + 2 g (y) = 3$ $\frac{\frac{1 - 2 y}{2 y}}{\frac{1 - 2 y}{2 y} + 1} = 1 - 2 y$ $f (1 - 2 y) + 2 g (y) = 3 . . . . . . (iv)$ $(iv) - (iii) : g (y) = 3 - (2 - y) = y + 1$ $g (x) = x + 1$ $2 (iii) - (iv) :$ $2 (iii) \Rightarrow 2 f (1 - 2 y) + 2 g (y) = 4 - 2 y$ $(iv) \Rightarrow f (1 - 2 y) + 2 g (y) = 3$ $f (1 - 2 y) = 4 - 2 y - 3$ $f (1 - 2 y) = 1 - 2 y$ $f (x) = x$
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