To solve this problem, we need to find an explicit formula for the function $f \circ g^{-1}$, where:

1. $f(x) = x^2$
2. $g(x) = 3x + 1$

Let's break down the steps:

Step 1: Find $g^{-1}(x)$

To find $g^{-1}(x)$, we need to determine the inverse of $g(x) = 3x + 1$.

1. Set $y = g(x) = 3x + 1$.

2. Solve for $x$ in terms of $y$:

   $y - 1 = 3x$ $x = \frac{y - 1}{3}$

So, the inverse function $g^{-1}(x)$ is:

$g^{-1}(x) = \frac{x - 1}{3}$

Step 2: Find $f \circ g^{-1}(x)$

Now, we need to compose $f$ with $g^{-1}$. This means substituting $g^{-1}(x)$ into $f$.

Since $f(x) = x^2$, we have:

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

Step 3: Simplify the Expression

Now, square the expression:

$f(g^{-1}(x)) = \frac{(x - 1)^2}{9}$

So, the explicit formula for $f \circ g^{-1}$ is:

$f \circ g^{-1}(x) = \frac{(x - 1)^2}{9}$

Would you like a detailed explanation of any part of this solution?

Here are some related questions that might further your understanding:

1. How do we find the inverse of a function in general?
2. Why is it important to check that a function has an inverse before solving problems like this?
3. What are some properties of compositions of functions?
4. Can $f \circ g$ ever be the same as $g \circ f$?
5. How does squaring affect the domain and range of a function?

Tip: When dealing with compositions involving inverses, always ensure that the inverse function exists and is unique before proceeding.