Question

if f is a real function defined by f(x) = x-1/x+1, then prove that f(2x) = 3f(x)+1/f(x)+3

Answer 1

and plug in f(x) = (x-1)/(x+1) and simplify 


f(2x)=(3f(x)+1)/(f(x)+3) 


f(2x)=(3*(x-1)/(x+1)+1)/((x-1)/(x+1)+3) 


f(2x)=(3*(x-1)+1(x+1))/(x-1+3(x+1)) ... multiply every term by the inner LCD x+1 


f(2x)=(3x-3+x+1)/(x-1+3x+3) 


f(2x)=(4x-2)/(4x+2) 


f(2x)=(2(2x-1))/(2(2x+1)) 


f(2x)=(2x-1)/(2x+1) 

Answer 2

Answer:

$f(2x)=\frac{3f(x)+1}{f(x)+3}$

Step-by-step explanation:

Given: $f(x)=\frac{x-1}{x+1}$

We need to prove

$f(2x)=\frac{3f(x)+1}{f(x)+3}$              .... (1)

The given function is

$f(x)=\frac{x-1}{x+1}$

Substitute x=2x in the above function.

$f(2x)=\frac{(2x)-1}{(2x)+1}$

$f(2x)=\frac{2x-1}{2x+1}$           .... (2)

Taking RHS of equation (1).

$RHS=\frac{3f(x)+1}{f(x)+3}$

Substitute the value of f(x).

$RHS=\frac{3(\frac{x-1}{x+1})+1}{(\frac{x-1}{x+1})+3}$

$RHS=\frac{\frac{3(x-1)+(x+1)}{x+1}}{\frac{(x-1)+3(x+1)}{x+1}}$

Cancel out common factors.

$RHS=\frac{3x-3+x+1}{x-1+3x+3}$

$RHS=\frac{4x-2}{4x+2}$

$RHS=\frac{2(2x-1)}{2(2x+1)}$

Cancel out common factors.

$RHS=\frac{2x-1}{2x+1}$

Using equation (2), we get

$RHS=f(2x)$

$RHS=LHS$

Hence proved.