Answer:
![- \frac{2[1 - x]}{3} = g[f(x)] \\ \\ \frac{3x}{2 - x} = f[g(x)]](https://tex.z-dn.net/?f=%20-%20%5Cfrac%7B2%5B1%20-%20x%5D%7D%7B3%7D%20%20%3D%20g%5Bf%28x%29%5D%20%5C%5C%20%20%5C%5C%20%20%5Cfrac%7B3x%7D%7B2%20-%20x%7D%20%20%3D%20f%5Bg%28x%29%5D)
Step-by-step explanation:
They are not.
For the <em>g[f(x)]</em> function, you substitute ³/ₓ ₋ ₁ from the <em>f</em><em>(</em><em>x</em><em>)</em><em> </em>function in for <em>x</em><em> </em>in the <em>g</em><em>(</em><em>x</em><em>)</em><em> </em>function to get this:

Then, you bring <em>x</em><em> </em><em>-</em><em> </em><em>1</em><em> </em>to the top while changing the expression to its conjugate [same expressions with opposite symbols]:
![- \frac{2[1 - x]}{3}](https://tex.z-dn.net/?f=%20-%20%5Cfrac%7B2%5B1%20-%20x%5D%7D%7B3%7D)
You could also do this [attaching another negative would make that positive].
For the <em>f[g(x)]</em> function, ²/ₓ from the <em>g(x)</em> function for <em>x</em><em> </em>in the <em>f(x)</em> function to get this:

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:

When this happens, <em>x</em><em> </em>leaves the <em>two</em> and gets attached to the <em>three</em><em>,</em><em> </em>and 1 gets an <em>x</em><em> </em>attached to it.
I am joyous to assist you anytime.