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)


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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
	$let x∈R and y=f(x) _((if it exists)) (x−5)g(x)=3x+2 so (y−5)g(y)=3y+2 thus (y−5)(2x−5)=(3y+2)(3x+7) ⇔(7x+26)y=11−16x so, f: { ((R\{((−26)/7)}),→,R),(x,→,((11−16x)/(26+7x))) :} I hope I did not make any calculation mistake but the reasonning is the right one. ;•)$
	$let x \in R$ $and y = f (x)_{(if it exists)}$ $(x - 5) g (x) = 3 x + 2$ $so (y - 5) g (y) = 3 y + 2$ $thus (y - 5) (2 x - 5) = (3 y + 2) (3 x + 7)$ $\Leftrightarrow (7 x + 26) y = 11 - 16 x$ $so, f : {\begin{cases} R ∖ {\frac{- 26}{7}} & \to & R \\ x & \to & \frac{11 - 16 x}{26 + 7 x} \end{cases}$ $I h o p e I d i d n o t m a k e a n y c a l c u l a t i o n m i s t a k e$ $b u t t h e r e a s o n n i n g i s t h e r i g h t o n e .$ $; ∙)$
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