Answer is: option1
\( f \) has a relative maximum.Solution:
Using the Product Rule
\( f(x) = x^2 e^{-x} \)
Let \( u = x^2 \) and \( v = e^{-x} \), then:
\( u' = 2x, \quad v' = -e^{-x} \)
Applying the product rule:
\[ f'(x) = u'v + uv' \]
\[ f'(x) = (2x e^{-x}) + (x^2 (-e^{-x})) \]
\[ = 2x e^{-x} - x^2 e^{-x} \]
\[ = e^{-x} (2x - x^2) \]
Setting \( f'(x) = 0 \):
\[ e^{-x} (2x - x^2) = 0 \]
Since \( e^{-x} \neq 0 \) for all \( x \), we solve:
\[ 2x - x^2 = 0 \]
\[ x (2 - x) = 0 \]
Thus, \( x = 0 \) or \( x = 2 \).
Checking \( x = 2 \) for maximum or minimum:
- Pick \( x = 1 \):
\[ f'(1) = e^{-1} (2(1) - 1^2) = e^{-1} (2 - 1) = e^{-1} > 0 \]
\( f \) is increasing before \( x = 2 \).
- Pick \( x = 3 \):
\[ f'(3) = e^{-3} (2(3) - 3^2) = e^{-3} (6 - 9) = e^{-3} (-3) < 0 \]
\( f \) is decreasing after \( x = 2 \).
Since \( f'(x) \) changes from positive to negative at \( x = 2 \), \( f(x) \) has a relative maximum at \( x = 2 \).
\( f \) has a relative maximum.