Question

Show that (f o g)(x) = (g o f)(x) = x

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

Answer 1

`- (2[1 - x])/(3) = g[f(x)] \\ \\ (3x)/(2 - x) = f[g(x)]`

Explanation:

They are not.

For the g[f(x)] function, you substitute ³/ₓ ₋ ₁ from the f(x) function in for x in the g(x) function to get this:

`(2)/( (3)/(x - 1))`

Then, you bring x - 1 to the top while changing the expression to its conjugate [same expressions with opposite symbols]:

`- (2[1 - x])/(3)`

You could also do this [attaching another negative would make that positive].

For the f[g(x)] function, ²/ₓ from the g(x) function for x in the f(x) function to get this:

`(3)/( (2)/(x) - 1)`

Now, if you look closely, ²/ₓ is written as 2x⁻¹, and according to the Negative Exponential Rule, you bring the denominator to the numerator while ALTERING THE INTEGER SYMBOL FROM NEGATIVE TO POSITIVE:

`(3x)/(2 - x)`

When this happens, x leaves the two and gets attached to the three, and 1 gets an x attached to it.

I am joyous to assist you anytime.