To find the critical points of the function $f'(x) = (x+2)e^{-x}$, we need to find the values of $x$ where the derivative $f'(x)$ is equal to zero or undefined.

Since $f'(x) = (x+2)e^{-x}$, we set the expression equal to zero and solve for $x$:

$(x+2)e^{-x} = 0$

The exponential function $e^{-x}$ is never zero, so the equation simplifies to:

$x + 2 = 0$

Solving for $x$:

$x = -2$

So, $x = -2$ is a critical point of $f'(x)$.

Next, we need to check the second derivative to determine the nature of this critical point. The second derivative $f''(x)$ can be found using the product rule:

$f''(x) = \frac{d}{dx} \left[ (x+2)e^{-x} \right]$

Applying the product rule $(uv)' = u'v + uv'$, we have $u = x+2$ and $v = e^{-x}$:

$f''(x) = \frac{d}{dx}(x+2)e^{-x} + (x+2)\frac{d}{dx}(e^{-x})$

$f''(x) = e^{-x} + (x+2)(-e^{-x})$

$f''(x) = e^{-x} - (x+2)e^{-x}$

$f''(x) = e^{-x}(1 - (x+2))$

$f''(x) = e^{-x}(1 - x - 2)$

$f''(x) = e^{-x}(-x - 1)$

Evaluating $f''(x)$ at $x = -2$:

$f''(-2) = e^{2}(-(-2) - 1)$

$f''(-2) = e^{2}(2 - 1)$

$f''(-2) = e^{2}(1)$

$f''(-2) = e^{2}$

Since $e^{2} > 0$, the critical point $x = -2$ is a local minimum.

Therefore, the critical point of $f'(x) = (x+2)e^{-x}$ is $x = -2$, and it is a local minimum.

Would you like further details or have any questions?

Here are 5 related questions you might ask:

1. How do you find critical points for more complex functions?
2. What are the steps to use the product rule in differentiation?
3. How can you determine the nature of critical points?
4. What role does the second derivative play in identifying critical points?
5. Can you provide examples of critical points in real-world applications?

Tip: When solving for critical points, always check if the derivative can be zero and then verify the nature of the critical points using the second derivative or other methods.