Let's analyze the problem:

We are given two functions:

• $f(x) = \log(x^2)$
• $g(x) = 2x + 1$

Analysis of $f \circ g(x)$ (composition $f(g(x))$)

1. Substituting $g(x) = 2x + 1$ into $f(x)$, we have: $f(g(x)) = \log((2x + 1)^2)$ Since logarithms are defined only for positive arguments, $2x + 1 \neq 0$, which means: $x \neq -\frac{1}{2}$ Hence, the domain of $f \circ g(x)$ is $\mathbb{R} \setminus \{-\frac{1}{2}\}$.

Analysis of $g \circ f(x)$ (composition $g(f(x))$)

2. Substituting $f(x) = \log(x^2)$ into $g(x)$, we have: $g(f(x)) = 2\log(x^2) + 1$ Since $\log(x^2)$ requires $x^2 > 0$, $x \neq 0$. Thus: $x \neq 0$ Hence, the domain of $g \circ f(x)$ is $\mathbb{R} \setminus \{0\}$.

Validation of options

Option 1:

"The domain of $f \circ g(x)$ is $\mathbb{R} \setminus \{-\frac{1}{2}\}$" – Correct, as derived above.

Option 2:

$f \circ g(x) = \log(2x + 1)$ – Incorrect, because $f(g(x)) = \log((2x + 1)^2)$, not $\log(2x + 1)$.

Option 3:

"The domain of $g \circ f(x)$ is $\mathbb{R} \setminus \{-\frac{1}{2}\}$" – Incorrect, because the domain is $\mathbb{R} \setminus \{0\}$.

Option 4:

$g \circ f(x) = 2\log(x^2) + 1$ – Correct, as derived above.

Option 5:

"The domain of $g \circ f(x)$ is $\mathbb{R} \setminus \{0\}$" – Correct, as derived above.

Final Answer:

Correct options are: 1, 4, and 5.

Would you like further clarifications or steps? Here are five related questions for deeper understanding:

1. What happens to the domain if the logarithm base is changed in $f(x)$?
2. How can the general composition $(f \circ g)(x)$ be calculated for other functions?
3. What are the key properties of logarithms that influence domain restrictions?
4. How do you check if a composition is well-defined for any $f(x)$ and $g(x)$?
5. Why does $\log(x^2)$ require $x \neq 0$, even though $x^2 > 0$ always?

**Tip