The given function is: \[
f(x) =
\begin{cases}
x^2 + 3, & x \neq 0, \\
1, & x = 0?
\end{cases}
\]
Step 1: Check for continuity at \( x = 0 \)
For \( f(x) \) to be continuous at \( x = 0 \), the following condition must hold: \[ \lim_{x \to 0} f(x) = f(0). \] Here, \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 + 3) = 3, \quad f(0) = 1. \] Since \( \lim_{x \to 0} f(x) \neq f(0) \), the function is discontinuous at \( x = 0 \).
Step 2: Check for differentiability at \( x \neq 0 \)
For \( x \neq 0 \), \( f(x) = x^2 + 3 \), which is a polynomial function. Polynomial functions are differentiable everywhere on \( \mathbb{R} \), so \( f(x) \) is differentiable for all \( x \neq 0 \).
Step 3: Verify other points of continuity and differentiability
Since \( f(x) = x^2 + 3 \) for \( x \neq 0 \), it is both continuous and differentiable for \( x \in \mathbb{R} \setminus \{0\} \). Hence, the function \( f(x) \) is continuous and differentiable \( \forall x \in \mathbb{R} \setminus \{0\} \).
Let \( f(x) = \log x \) and \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \] Then the domain of \( f \circ g \) is: