1. **State the problem:** We have a piecewise function $ f(x) = \begin{cases} e^{3x}, & x \leq 3 \\ 2.5x + b, & x > 3 \end{cases} $. We want to find the value of $ b $ such that $ f(x) $ is continuous at $ x = 3 $. 2. **Recall the continuity condition:** For $ f(x) $ to be continuous at $ x = 3 $, the left-hand limit and right-hand limit at $ x=3 $ must be equal, and equal to $ f(3) $. That is, $$\lim_{x \to 3^-} f(x) = f(3) = \lim_{x \to 3^+} f(x)$$ 3. **Evaluate the left-hand limit:** Since $ x \leq 3 $ uses $ e^{3x} $, $$\lim_{x \to 3^-} f(x) = e^{3 \cdot 3} = e^9$$ 4. **Evaluate the right-hand limit:** For $ x > 3 $, $ f(x) = 2.5x + b $, so $$\lim_{x \to 3^+} f(x) = 2.5 \cdot 3 + b = 7.5 + b$$ 5. **Set the limits equal for continuity:** $$e^9 = 7.5 + b$$ 6. **Solve for $ b $:** $$b = e^9 - 7.5$$ 7. **Calculate the numerical value:** $$e^9 \approx 8103.084\quad \Rightarrow \quad b \approx 8103.084 - 7.5 = 8095.584$$ **Final answer:** $ b \approx 8095.584 $ to the nearest thousandth. This ensures the function is continuous at $ x=3 $.