Let's go through the given problem step by step.

Given:

We are asked to analyze the function whose derivative is:

$f'(x) = (x + 4) e^{-x}$

Part (a): Critical Points of $f$

Critical points occur where the derivative is either zero or undefined. Since the derivative $f'(x) = (x + 4) e^{-x}$ is defined for all $x$, we only need to find where $f'(x) = 0$.

$f'(x) = (x + 4) e^{-x} = 0$

Since $e^{-x}$ is never zero, we set $x + 4 = 0$.

$x + 4 = 0 \implies x = -4$

Thus, the critical point is at $x = -4$.

Part (b): Intervals where $f$ is Increasing or Decreasing

To determine whether the function is increasing or decreasing, we analyze the sign of $f'(x)$ on either side of the critical point $x = -4$.

• When $x < -4$, $(x + 4) < 0$ and $e^{-x} > 0$, so $f'(x) < 0$. Hence, $f(x)$ is decreasing on $(-\infty, -4)$.
• When $x > -4$, $(x + 4) > 0$ and $e^{-x} > 0$, so $f'(x) > 0$. Hence, $f(x)$ is increasing on $(-4, \infty)$.

Thus, $f$ is decreasing on $(-\infty, -4)$ and increasing on $(-4, \infty)$.

Part (c): Local Maximum or Minimum Values

At $x = -4$, the derivative changes sign from negative to positive. This indicates that $f(x)$ has a local minimum at $x = -4$.

Summary:

• (a) The critical point is at $x = -4$.
• (b) $f(x)$ is decreasing on $(-\infty, -4)$ and increasing on $(-4, \infty)$.
• (c) There is a local minimum at $x = -4$.

Would you like more detailed steps or have any other questions? Here are some related questions for deeper understanding:

1. How would you determine the concavity of the function $f(x)$?
2. Can the second derivative test be applied to this function to confirm the local minimum?
3. How would a graph of $f(x)$ look, based on the information we've derived?
4. What happens to the behavior of the function as $x \to \infty$?
5. How would the analysis change if the derivative were $f'(x) = (x + a)e^{-x}$ for a general constant $a$?

Tip: For finding critical points, always consider both where the derivative equals zero and where it is undefined!