To solve the problem, we need to find the inverse of the composition $(f \circ g)^{-1}$. Given the information:

1. $f(3x) = x$
2. $g(x) = \frac{x-1}{x-2} = 2x$ (this statement seems off, let's assume it means $g(x) = \frac{x-1}{x-2}$)

Step 1: Finding the inverse of $f(x)$

Given $f(3x) = x$. To find $f^{-1}(x)$, let’s rewrite it in terms of $x$:

$f(3x) = x \implies y = 3x \implies x = \frac{y}{3} \implies f(x) = \frac{x}{3}.$

Now, finding the inverse:

$f(x) = \frac{x}{3} \implies f^{-1}(x) = 3x.$

Step 2: Finding the inverse of $g(x)$

Given $g(x) = \frac{x-1}{x-2}$. To find $g^{-1}(x)$:

Set $y = \frac{x-1}{x-2}$. Cross-multiply to solve for $x$:

$y(x-2) = x-1 \implies yx - 2y = x - 1 \implies yx - x = 2y - 1 \implies x(y-1) = 2y - 1 \implies x = \frac{2y-1}{y-1}.$

So,

$g^{-1}(x) = \frac{2x-1}{x-1}.$

Step 3: Finding $(f \circ g)^{-1}(5)$

First, find $f \circ g(x)$:

$f(g(x)) = f\left(\frac{x-1}{x-2}\right).$

Substituting $g(x)$ into $f$:

$f\left(\frac{x-1}{x-2}\right) = \frac{\frac{x-1}{x-2}}{3} = \frac{x-1}{3(x-2)}.$

The inverse of $f \circ g$ is needed at $5$:

Let $y = \frac{x-1}{3(x-2)}$. To find the inverse, solve for $x$:

$5 = \frac{x-1}{3(x-2)}.$

Cross-multiplying gives:

$5 \cdot 3(x-2) = x-1 \implies 15x - 30 = x - 1 \implies 14x = 29 \implies x = \frac{29}{14}.$

Thus, $(f \circ g)^{-1}(5) = \frac{29}{14}$.

Would you like further clarification on any step, or explore related questions?

Related Questions:

1. How do you find the composition of functions in general?