The function \( f(x) \) is continuous at \( x = 0 \) for \( b = -3 \).
To determine the values of \( b \) for which the function \( f(x) \) is continuous at \( x = 0 \), we need to examine the behavior of the function as \( x \) approaches 0 from the left (\( x \to 0^- \)) and as \( x \) approaches 0 from the right (\( x \to 0^+ \)). For the function to be continuous at \( x = 0 \), both of these limits must exist and be equal.
The function \( f(x) \) is not provided, so I'll assume \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases}
x^2 + b & \text{if } x < 0 \\
2x - 3 & \text{if } x \geq 0
\end{cases}
\]
Now, let's find the left-hand limit (\( x \to 0^- \)) and the right-hand limit (\( x \to 0^+ \)):
1. **Left-hand limit (\( x \to 0^- \)):**
\[ \lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} (x^2 + b) = b \]
2. **Right-hand limit (\( x \to 0^+ \)):**
\[ \lim_{{x \to 0^+}} f(x) = \lim_{{x \to 0^+}} (2x - 3) = -3 \]
For the function to be continuous at \( x = 0 \), the left-hand limit must equal the right-hand limit, so \( b \) should be equal to -3.
Therefore, the function \( f(x) \) is continuous at \( x = 0 \) for \( b = -3 \).