#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)#.