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Let's go through this problem step by step.

### Part (a): Represent \( f(t) \) using Heaviside step functions

The function \( f(t) \) is defined in pieces:
1. \( f(t) = 3 \) for \( t < 6 \)
2. \( f(t) = t - 6 \) for \( 6 \leq t < 7 \)
3. \( f(t) = 2 \) for \( t \geq 7 \)

To express \( f(t) \) using Heaviside functions, \( h(t - c) \), we can set it up by noting the following:
- Start with the base value for each interval.
- Use Heaviside functions to "switch" the function at each discontinuity.

From the given solution:
\[
f(t) = 3 - 3h(t - 6) + (t - 6)(h(t - 6) - h(t - 7)) + 2h(t - 7)
\]

Here's the breakdown:
1. \( 3 \) is the value of \( f(t) \) before \( t = 6 \).
2. \( -3h(t - 6) \): At \( t = 6 \), subtracting \( 3 \) sets up the change to the next interval.
3. \( (t - 6)(h(t - 6) - h(t - 7)) \): Adds \( t - 6 \) for \( 6 \leq t < 7 \), due to the difference between \( h(t - 6) \) and \( h(t - 7) \), which is 1 in this range.
4. \( 2h(t - 7) \): Finally, adding 2 at \( t = 7 \) captures the constant value for \( t \geq 7 \).

### Part (b): Finding the Laplace transform \( F(s) = \mathcal{L}\{f(t)\} \)

To find the Laplace transform of \( f(t) \), we'll take the Laplace transform of each term separately.

1. **Transform of \( 3 \)**:
   \[
   \mathcal{L}\{3\} = \frac{3}{s}
   \]

2. **Transform of \( -3h(t - 6) \)**:
   \[
   -3 \cdot \frac{e^{-6s}}{s}
   \]

3. **Transform of \( (t - 6)(h(t - 6) - h(t - 7)) \)**:
   - For \( (t - 6) h(t - 6) \): Shifted by 6, so
     \[
     \mathcal{L}\{(t - 6)h(t - 6)\} = \frac{e^{-6s}}{s^2}
     \]
   - For \( (t - 6) h(t - 7) \): Shifted by 7, so
     \[
     \mathcal{L}\{(t - 6)h(t - 7)\} = \frac{e^{-7s}}{s^2}
     \]
   - Therefore, the term becomes:
     \[
     \frac{e^{-6s}}{s^2} - \frac{e^{-7s}}{s^2}
     \]

4. **Transform of \( 2h(t - 7) \)**:
   \[
   2 \cdot \frac{e^{-7s}}{s}
   \]

Combining these results gives:
\[
F(s) = \frac{3}{s} - \frac{3e^{-6s}}{s} + \left( \frac{e^{-6s}}{s^2} - \frac{e^{-7s}}{s^2} \right) + \frac{2e^{-7s}}{s}
\]

Would you like further details or have any questions on specific parts of this solution?

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Here are some related questions:
1. How do Heaviside step functions work in piecewise-defined functions?
2. What is the Laplace transform of a constant?
3. How can we apply shifting in the Laplace domain using Heaviside functions?
4. What is the significance of using \( h(t - c) \) for modeling discontinuities?
5. How does the Laplace transform help in solving differential equations involving piecewise functions?

**Tip**: In Laplace transforms, the shift theorem \( \mathcal{L}\{f(t - a)h(t - a)\} = e^{-as}F(s) \) is crucial for handling piecewise functions effectively.

Consider the function: f(t) = { 3, for t < 6; t - 6, for 6 <= t < 7; 2, for t >= 7 }

Represent f(t) using a combination of Heaviside step functions. Use h(t - c) for the Heaviside function h_c(t).

Find the Laplace transform F(s) = L{f(t)}.

Learn how to represent a piecewise function using Heaviside step functions and find its Laplace transform. This solution covers key concepts in Laplace transforms and Heaviside functions, ideal for undergraduate-level students.

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