Answer:
[tex] - \frac{2[1 - x]}{3} = g[f(x)] \\ \\ \frac{3x}{2 - x} = f[g(x)][/tex]
Step-by-step 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:
[tex] \frac{2}{ \frac{3}{x - 1}} [/tex]
Then, you bring x - 1 to the top while changing the expression to its conjugate [same expressions with opposite symbols]:
[tex] - \frac{2[1 - x]}{3}[/tex]
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:
[tex] \frac{3}{ \frac{2}{x} - 1}[/tex]
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:
[tex] \frac{3x}{2 - x} [/tex]
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.
