To find the critical points of the function \( f(x) = (7x + 8)e^{-4x} \), we need to find the values of \( x \) where the derivative \( f'(x) \) is equal to zero or undefined. Here's the process:

1. **Differentiate \( f(x) \):**

   \[
   f(x) = (7x + 8)e^{-4x}
   \]

   Using the product rule, \( (uv)' = u'v + uv' \), with \( u = 7x + 8 \) and \( v = e^{-4x} \), we get:

   \[
   f'(x) = (7)(e^{-4x}) + (7x + 8)(-4e^{-4x})
   \]

2. **Simplify \( f'(x) \):**

   Factor out \( e^{-4x} \):

   \[
   f'(x) = e^{-4x} (7 - 4(7x + 8))
   \]

3. **Set \( f'(x) = 0 \):**

   Since \( e^{-4x} \) is never zero, we only need to solve:

   \[
   7 - 4(7x + 8) = 0
   \]

4. **Solve for \( x \):**

   Expanding and simplifying:

   \[
   7 - 28x - 32 = 0
   \]
   \[
   -28x - 25 = 0
   \]
   \[
   x = -\frac{25}{28}
   \]

So, the critical point occurs at:

\[
x = -\frac{25}{28}
\]

Let me know if you'd like a more detailed explanation or have any questions!

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### Related Questions
1. How do you determine if a critical point is a maximum, minimum, or inflection point?
2. What role does the exponential function \( e^{-4x} \) play in the behavior of this function?
3. How would the critical points change if we had \( f(x) = (7x + 8)e^{4x} \) instead?
4. Can we find critical points by looking only at the sign changes in \( f'(x) \)?
5. How does the product rule help in finding the derivative of a product of functions?

### Tip:
Always check for both zero and undefined points when finding critical points of a function.

Find the critical points of the function (7x+8)e^-4x

Learn how to find the critical points of the function (7x + 8)e^-4x using calculus techniques. This solution includes applying the product rule, simplifying derivatives, and solving for where the derivative equals zero. Suitable for advanced high school or college-level calculus students.

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