Let \[g(x) =
\begin{cases}
x^2 + bx - 1 & \text{if } x \leq 1, \\
3x + 2 & \text{if } x > 1.
\end{cases}
\]
Determine the value of \(b\) such that the function \(g(x)\) is continuous at \(x = 1\). Let's think step by step and output the final answer within \boxed{}.
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