Let's go through the given problem step by step:

a) Maximal Definitionsmängd for $g(x) = \ln(x^2)$

The function $g(x) = \ln(x^2)$ is a logarithmic function, and for a logarithmic function to be defined, its argument must be strictly greater than zero:

$x^2 > 0$

Since $x^2 > 0$ for all $x \neq 0$, the domain of $g(x)$ is:

$x \in (-\infty, 0) \cup (0, \infty)$

In other words, the maximal domain (or definitionsmängd) of $g(x)$ is all real numbers except $x = 0$.

Maximal Definitionsmängd: $x \in \mathbb{R}, x \neq 0$

---

b) $g \circ f(x)$ and $f^{-1}(x)$

1. Compute $g \circ f(x)$

The composition of functions $g \circ f(x)$ means applying $f(x)$ first, then applying $g(x)$ to the result. We are given:

$f(x) = 3x + 1 \quad \text{and} \quad g(x) = \ln(x^2)$

Thus, $g \circ f(x) = g(f(x))$:

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

So:

$g \circ f(x) = \ln((3x + 1)^2)$

---

2. Compute $f^{-1}(x)$ (the inverse of $f(x)$)

To find the inverse of $f(x) = 3x + 1$, we solve for $x$ in terms of $y$ where $y = f(x)$:

$y = 3x + 1$

Solve for $x$:

$y - 1 = 3x$

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

Thus, the inverse function is:

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

---

Final Results:

a) The maximal definitionsmängd for $g(x)$ is $x \in \mathbb{R}, x \neq 0$.

b)

• $g \circ f(x) = \ln((3x + 1)^2)$
• $f^{-1}(x) = \frac{x - 1}{3}$

Would you like more details on any of these steps? Here are some related questions to deepen understanding:

1. What is the definition of the composition of functions?
2. How do you determine the domain of a composite function like $g \circ f(x)$?
3. Why does $x^2$ remain positive for all $x \neq 0$?
4. How would you graph $f(x) = 3x + 1$ and its inverse $f^{-1}(x)$?
5. What are the properties of logarithmic functions that help solve $g(x) = \ln(x^2)$?

Tip: When finding the domain of a logarithmic function, always ensure the argument inside the logarithm is positive.