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-

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 {\begin{cases} f (2 x - 1) + g (1 - x) = x + 1 \dots (i) \\ f (\frac{x}{x + 1}) + 2 g (\frac{1}{2 x + 2}) = 3 \dots . (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} \dots . (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 \dots \dots (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 \dots (i) \\ f (\frac{x}{x + 1}) + 2 g (\frac{1}{2 x + 2}) = 3 \dots . . (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 \dots . . (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 \dots \dots (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