The given derivative is:
$$f'(x) = e^{-x}(x + 2)^2(x - 1)(x - 3)^3$$
Finding Critical Points :
Critical points occur where the derivative is equal to zero or is undefined. Since $f'(x)$ is a product of exponential and polynomial functions, it is defined everywhere. We set $f'(x) = 0$ to find the critical points:
$$e^{-x}(x + 2)^2(x - 1)(x - 3)^3 = 0$$
Since the exponential function $e^{-x}$ is strictly positive for all real values of $x$, it can never be zero. Therefore, we only need to set the polynomial factors to zero:
$(x + 2)^2 = 0 \implies x = -2$
$x - 1 = 0 \implies x = 1$
$(x - 3)^3 = 0 \implies x = 3$
The critical points are $x = -2$, $x = 1$, and $x = 3$.
Applying the First Derivative Test :
To classify each critical point as a local maximum, local minimum, or neither, we evaluate the sign of $f'(x)$ on the intervals between the critical points.
The sign of $f'(x)$ depends on the factors. Let's analyze them:
$e^{-x}$ is always positive $(+)$.
$(x + 2)^2$ is always non-negative $(+)$, except at $x = -2$ where it is $0$. It will not cause a sign change.
The sign of $f'(x)$ will entirely depend on the product of the remaining factors with odd powers $:(x - 1)(x - 3)^3$.
Let's test the intervals:
Interval $(-\infty, -2):$ Let $x = -3$.
$f'(-3) = (+) \cdot (+) \cdot (-4) \cdot (-6)^3 = (+) \cdot (+) \cdot (-) \cdot (-) = (+)$
The function is increasing.
Interval $(-2, 1):$ Let $x = 0$.
$f'(0) = (+) \cdot (+) \cdot (-1) \cdot (-3)^3 = (+) \cdot (+) \cdot (-) \cdot (-) = (+)$
The function is increasing.
Interval $(1, 3):$ Let $x = 2$.
$f'(2) = (+) \cdot (+) \cdot (1) \cdot (-1)^3 = (+) \cdot (+) \cdot (+) \cdot (-) = (-)$
The function is decreasing.
Interval $(3, \infty):$ Let $x = 4$.
$f'(4) = (+) \cdot (+) \cdot (3) \cdot (1)^3 = (+) \cdot (+) \cdot (+) \cdot (+) = (+)$
The function is increasing.
Now let's check the given options based on our sign analysis:
At $x = -2:$ The derivative $f'(x)$ is positive on both the left and the right sides. Since the sign does not change, there is no local extremum here. It is just a stationary point. This makes statement A false and statement D true.
At $x = 1:$ The derivative changes sign from positive to negative. This means the function changes from increasing to decreasing, creating a local maximum. This makes statement B true.
At $x = 3:$ The derivative changes sign from negative to positive. This means the function changes from decreasing to increasing, creating a local minimum. This makes statement C true.
