how can i solve this question?

Question

Two functions are defined by f(x)=2x-1 and g(x)=x/x-1, x not =1. Obtain expressions in the same form for fg and gf. Find the value of x for which fg=gf.

Answer 1

I assume by fg you mean f(g(x)) and by gf you mean g(f(x))

f(g(x)) = f(x/x - 1) = 2(x/x - 1) - 1
g(f(x)) = g(2x - 1) = (2x - 1)/(2x - 1) - 1 = (2x - 1)/(2x - 2)

Now let f(g(x)) = g(f(x))

2(x/x - 1) - 1 = (2x - 1)/(2x - 2)
2(x/x - 1)(2x - 2) - (2x - 2) = 2x - 1
[2x(2x - 2)]/(x - 1) - (2x - 2) = 2x - 1
2x(2x - 2) - (2x - 2)(x - 1) = (2x - 1)(x - 1)
(4x² - 4x) - (2x² - 4x + 2) = (2x² - 3x + 1)
4x² - 4x - 2x² + 4x - 2 = 2x² - 3x + 1
2x² - 2 = 2x² - 3x + 1
-3 = -3x

x = 1

which, as this would lead to dividing by 0, suggests there is no real number for f(g(x)) = g(f(x))

Answer 2

f(x) = 2x - 1
g(x) = x/(x - 1) for x not equal to 1.

(f o g)(x) is defined to be f(g(x)), so

f(g(x)) = f(x/(x - 1))
= 2[x/(x - 1)] - 1
= 2x/(x - 1) - 1
= 2x/(x - 1) - (x - 1)/(x - 1)

Merging this into a single fraction,

= [2x - (x - 1)] / (x - 1)
f(g(x)) = (x + 1)/(x - 1)

(g of f)(x) is defined to be g(f(x)).

g(f(x)) = g(2x - 1) = [2x - 1] / [(2x - 1) - 1]
= [2x - 1] / [2x - 2]

We want the value of x with which f(g(x)) = g(f(x)), so we equate them.

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

We solve this by cross multiplying.

(x + 1)(2x - 2) = (2x - 1)(x - 1)

Expand both sides

2x^2 - 2 = 2x^2 - 3x + 1

Move everything with an x term to the left hand side; everything else to the right hand side.

2x^2 - 2x^2 + 3x = 2 + 1
3x = 3
x = 1

However, x is not equal to 1.

Therefore, there are no values for x which (f o g)(x) = (g o f)(x)