Question

if f(x) = x-1/x+1 then f(ax) in terms of f(x) is given by

Answer 1

Answers
∵ ƒ(x) = ( x - 1 ) / ( x + 1 ) ............ (1)

∴ ( x + 1 ). ƒ(x) = x - 1

∴ x. ƒ(x) + ƒ(x) = x - 1

∴ x. ƒ(x) - x = - 1 - ƒ(x)

∴ x. [ ƒ(x) - 1 ] = - [ 1 + ƒ(x) ]

∴ x = [ 1 + ƒ(x) ] / [ 1 - ƒ(x) ] .................. (2)
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∴ from (1),

ƒ(2x) = [ (2x) - 1 ] / [ (2x) + 1 ]

. . . . .= { 2( [1+ƒ(x)] / [1-ƒ(x)] ) - 1 } / { 2( [1+ƒ(x)] / [1-ƒ(x)] + 1 }

. . . . .= { 2[ 1 + ƒ(x) ] - [ 1 - ƒ(x) ] } / { 2[ 1 + ƒ(x) ] + [ 1 - ƒ(x) ] }

. . . . .= { 2 + 2.ƒ(x) - 1 + ƒ(x) } / { 2 + 2.ƒ(x) + 1 - ƒ(x) }

. . . . .= [ 1 + 3.ƒ(x) ] / [ 3 + ƒ(x) ]

Answer 2

Answer:

Given that ƒ(x) = ( x - 1 ) / ( x + 1 ) ............ (1)

( x + 1 ). ƒ(x) = x - 1

x. ƒ(x) + ƒ(x) = x - 1

x. ƒ(x) - x = - 1 - ƒ(x)

x. [ ƒ(x) - 1 ] = - [ 1 + ƒ(x) ]

x = [ 1 + ƒ(x) ] / [ 1 - ƒ(x) ] .................. (2)

put x = ax in equation (1),

ƒ(ax) = [ (ax) - 1 ] / [ (ax) + 1 ]............... (3)

put the value of x from equation (2) in equation (3)...

ƒ(ax) = [ (a[ 1 + ƒ(x) ] / [ 1 - ƒ(x) ]) - 1 ] / [ (a[ 1 + ƒ(x) ] / [ 1 - ƒ(x) ]) + 1 ]

simplify the above equation.