1. **State the problem:** Find the inverse Laplace transform of $$F(s) = \frac{s + \frac{1}{2}}{s^2 + 3s + \frac{5}{4}}$$ using the first shift theorem. 2. **Recall the first shift theorem:** If $$\mathcal{L}\{f(t)\} = F(s)$$, then $$\mathcal{L}\{e^{at}f(t)\} = F(s - a)$$. This means shifting the variable $$s$$ in the Laplace domain corresponds to multiplying by an exponential in the time domain. 3. **Rewrite the denominator to complete the square:** $$s^2 + 3s + \frac{5}{4} = \left(s^2 + 3s + \frac{9}{4}\right) - \frac{9}{4} + \frac{5}{4} = \left(s + \frac{3}{2}\right)^2 - 1$$ 4. **Rewrite the function:** $$F(s) = \frac{s + \frac{1}{2}}{\left(s + \frac{3}{2}\right)^2 - 1}$$ 5. **Make the substitution $$u = s + \frac{3}{2}$$:** $$F(s) = \frac{u - 1}{u^2 - 1} = \frac{u - 1}{(u - 1)(u + 1)}$$ 6. **Simplify the fraction by canceling common factors:** $$F(s) = \frac{\cancel{u - 1}}{\cancel{u - 1}(u + 1)} = \frac{1}{u + 1}$$ 7. **Rewrite back in terms of $$s$$:** $$F(s) = \frac{1}{s + \frac{3}{2} + 1} = \frac{1}{s + \frac{5}{2}}$$ 8. **Apply the first shift theorem:** Since $$F(s) = \frac{1}{s + \frac{5}{2}} = \mathcal{L}\{e^{-\frac{5}{2}t}f(t)\}$$ where $$f(t) = 1$$ because $$\mathcal{L}\{1\} = \frac{1}{s}$$. 9. **Find the inverse Laplace transform:** $$\mathcal{L}^{-1}\{F(s)\} = e^{-\frac{5}{2}t} \cdot 1 = e^{-\frac{5}{2}t}$$ **Final answer:** $$\boxed{f(t) = e^{-\frac{5}{2}t}}$$