$f(x)= (x+1)/(2x-1) =>f(2x)= ?$ result $f(2x)=(4f(x)+1)/(2f(x)+5)$ but how find it ?

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$f(x)= (x+1)/(2x-1) =>f(2x)= ?$

$f(2x)=(4f(x)+1)/(2f(x)+5)$
but how find it ?

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Answer 1 · 2018-03-07T11:42:43

$f(x)=(x+1)/(2x-1)$ .

To find, $f(2x)$ , let, $2x=t$ . Then,

$"Reqd. Value="f(2x)=f(t)$ ,

$=(t+1)/(2t-1)$ ,

$=(2x+1)/{2(2x)-1}............[because, t=2x]$ .

$rArr "Reqd. Value="f(2x)=(2x+1)/(4x-1)............(star)$ .

Next, $f(x)=(x+1)/(2x-1)$ ,

$rArr 4f(x)+1=4{(x+1)/(2x-1)}+1$ ,

$={(4x+4)+(2x-1)}/(4x-1)$ ,

$:. 4f(x)+1=(6x+3)/(2x-1).............(star1)$ .

Also, $2f(x)+5=2{(x+1)/(2x-1)}+5$ ,

$={(2x+2)+5(2x-1)}/(2x-1)$ ,

$rArr 2f(x)+5=(12x-3)/(2x-1).............(star2)$ .

$"Therefore, "(star1) and (star2) rArr (4f(x)+1)/(2f(x)+5)$ ,

$=(6x+3)/(12x-3)={3(2x+1)}/{3(4x-1)}$ .

$:. f(2x)=(2x+1)/(4x-1)=(4f(x)+1)/(2f(x)+5)$ .