If-the-function-f-and-g-are-defined-on-the-set-of-real-numbers-are-such-that-gof-x-2x-5-3x-7-and-g-x-3x-2-x-5-find-an-expression-for-f-x-

Question Number 143650 by Rankut last updated on 16/Jun/21

I f t h e f u n c t i o n f a n d g a r e d e f i n e d

o n t h e s e t o f r e a l n u m b e r s, a r e s u c h

t h a t g o f (x) = \frac{2 x - 5}{3 x + 7} a n d

g (x) = \frac{3 x + 2}{x - 5} .

find an expression for f (x)

Answered by ajfour last updated on 16/Jun/21

\frac{3 f + 2}{f - 5} = \frac{2 x - 5}{3 x + 7}

\Rightarrow 9 x f + 21 f + 6 x + 14

= 2 x f - 5 f - 10 x + 25

\Rightarrow f (x) = \frac{11 - 16 x}{7 x + 26}

Answered by Olaf_Thorendsen last updated on 16/Jun/21

g (x) = \frac{3 x + 2}{x - 5}

(x - 5) g (x) = 3 x + 2

x (g (x) - 3) = 5 g (x) + 2

x = \frac{5 g (x) + 2}{g (x) - 3}

\Rightarrow g^{- 1} (x) = \frac{5 x + 2}{x - 3}

f (x) = g^{- 1} o g o f (x) = \frac{5 (\frac{2 x - 5}{3 x + 7}) + 2}{(\frac{2 x - 5}{3 x + 7}) - 3}

f (x) = g^{- 1} o g o f (x) = \frac{16 x - 11}{- 7 x - 26}

f (x) = g^{- 1} o g o f (x) = - \frac{16 x - 11}{7 x + 26}

Answered by TheHoneyCat last updated on 16/Jun/21

let x \in R

and y = f (x)_{(if it exists)}

(x - 5) g (x) = 3 x + 2

so (y - 5) g (y) = 3 y + 2

thus (y - 5) (2 x - 5) = (3 y + 2) (3 x + 7)

\Leftrightarrow (7 x + 26) y = 11 - 16 x

so, f : {\begin{cases} R ∖ {\frac{- 26}{7}} & \to & R \\ x & \to & \frac{11 - 16 x}{26 + 7 x} \end{cases}

I h o p e I d i d n o t m a k e a n y c a l c u l a t i o n m i s t a k e

b u t t h e r e a s o n n i n g i s t h e r i g h t o n e .

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