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The given function is \( F(s) = \frac{e^{-9s}}{s^2 + 2s - 8} \), and we are tasked with finding its inverse Laplace transform, \( f(t) \), using the provided notation, which includes the Heaviside step function, \( u(t - c) \), for shifts.

### Step-by-Step Solution:

1. **Factor the Denominator**:
   \[
   s^2 + 2s - 8 = (s + 4)(s - 2)
   \]
   This makes the function:
   \[
   F(s) = \frac{e^{-9s}}{(s + 4)(s - 2)}
   \]

2. **Partial Fraction Decomposition**:
   We decompose the fraction \( \frac{1}{(s + 4)(s - 2)} \) as follows:
   \[
   \frac{1}{(s + 4)(s - 2)} = \frac{A}{s + 4} + \frac{B}{s - 2}
   \]
   Multiplying both sides by \( (s + 4)(s - 2) \) gives:
   \[
   1 = A(s - 2) + B(s + 4)
   \]
   Expanding this:
   \[
   1 = A(s) - 2A + B(s) + 4B
   \]
   Group terms:
   \[
   1 = (A + B)s + (-2A + 4B)
   \]
   Setting up the system of equations:
   \[
   A + B = 0
   \]
   \[
   -2A + 4B = 1
   \]
   Solving this system, we get:
   \[
   A = \frac{1}{6}, \quad B = -\frac{1}{6}
   \]
   So, the partial fraction decomposition is:
   \[
   \frac{1}{(s + 4)(s - 2)} = \frac{1/6}{s + 4} - \frac{1/6}{s - 2}
   \]

3. **Inverse Laplace of Individual Terms**:
   The inverse Laplace transform of \( \frac{1}{s + a} \) is \( e^{-at} \). Thus:
   \[
   \mathcal{L}^{-1}\left( \frac{1}{6(s + 4)} \right) = \frac{1}{6} e^{-4t}
   \]
   and
   \[
   \mathcal{L}^{-1}\left( -\frac{1}{6(s - 2)} \right) = -\frac{1}{6} e^{2t}
   \]
   So the inverse Laplace transform of \( \frac{1}{(s + 4)(s - 2)} \) is:
   \[
   \frac{1}{6}e^{-4t} - \frac{1}{6}e^{2t}
   \]

4. **Shifting Using the Exponential Term**:
   The term \( e^{-9s} \) indicates a time shift by 9 units. According to the shift theorem, if \( \mathcal{L}^{-1}[F(s)] = f(t) \), then:
   \[
   \mathcal{L}^{-1}[e^{-cs} F(s)] = u(t - c) f(t - c)
   \]
   Therefore, the inverse Laplace transform of the original function \( F(s) = \frac{e^{-9s}}{(s + 4)(s - 2)} \) is:
   \[
   f(t) = \left( \frac{1}{6}e^{-4(t-9)} - \frac{1}{6}e^{2(t-9)} \right) u(t - 9)
   \]

### Final Answer:
\[
f(t) = \left( \frac{1}{6}e^{-4(t-9)} - \frac{1}{6}e^{2(t-9)} \right) u(t - 9)
\]

Would you like further details on any of these steps?

Here are 5 related questions you can explore:
1. How do you perform partial fraction decomposition in more complex cases?
2. What is the Laplace transform of the Heaviside step function?
3. How does the time-shifting property of Laplace transforms work?
4. Can you apply the inverse Laplace transform to other functions with exponential shifts?
5. What are the conditions for the existence of Laplace transforms?

**Tip:** Always check the denominator for easy factoring before applying partial fraction decomposition—it simplifies the inverse Laplace process.

Find the inverse Laplace transform of F(s) = (e^{-9s}) / (s^2 + 2s - 8). Use the Heaviside step function notation.

Solve the inverse Laplace transform of F(s) = (e^{-9s}) / (s^2 + 2s - 8). This problem explores partial fraction decomposition, exponential shift, and the Heaviside step function. Suitable for advanced calculus students.

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