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We are given the derivative of a function \( f'(x) = 7 \cos x - 3 \sin x \) and the initial condition \( f(0) = -3 \). We need to find the function \( f(x) \).

### Step-by-step solution:

1. **Integrate \( f'(x) \)**:
   To find \( f(x) \), we need to integrate \( f'(x) \).

   \[
   f'(x) = 7 \cos x - 3 \sin x
   \]
   Now, integrate each term:
   \[
   \int 7 \cos x \, dx = 7 \sin x
   \]
   \[
   \int -3 \sin x \, dx = 3 \cos x
   \]
   Therefore, the general form of \( f(x) \) is:
   \[
   f(x) = 7 \sin x + 3 \cos x + C
   \]
   where \( C \) is the constant of integration.

2. **Use the initial condition \( f(0) = -3 \)**:
   Substitute \( x = 0 \) and \( f(0) = -3 \) into the equation \( f(x) = 7 \sin x + 3 \cos x + C \):
   \[
   f(0) = 7 \sin(0) + 3 \cos(0) + C = -3
   \]
   Since \( \sin(0) = 0 \) and \( \cos(0) = 1 \), this simplifies to:
   \[
   3 + C = -3
   \]
   Solving for \( C \):
   \[
   C = -6
   \]

3. **Final solution**:
   Substitute \( C = -6 \) back into the general solution:
   \[
   f(x) = 7 \sin x + 3 \cos x - 6
   \]

Thus, the function is:
\[
f(x) = 7 \sin x + 3 \cos x - 6
\]

Let me know if you need further clarification or more details!

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### Follow-up questions:
1. How do you differentiate trigonometric functions like sine and cosine?
2. Can you explain the process of integrating sine and cosine functions?
3. What are initial conditions and how do they affect the solution of an integral?
4. How can we use definite integrals for solving similar problems?
5. How do constants of integration affect the shape of a function?

### Tip:
When solving integrals, always remember to add the constant of integration, especially if no limits are given. The initial conditions are crucial for determining this constant!

Given f'(x) = 7 cos x - 3 sin x and f(0) = -3, find f(x)

Learn how to find the function f(x) given its derivative f'(x) = 7 cos x - 3 sin x and an initial condition f(0) = -3. This problem involves key concepts of calculus, including integration and trigonometric functions. Suitable for students in Grades 11-12.

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