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.