For the function \(f(x) = xe^{-x}\), \(x \in R\):
(A) \(f'(x) = (1 - x)e^{-x}\)
(B) f is strictly increasing in \((-\infty, 1)\).
(C) f is strictly decreasing in \((0, \infty)\).
(D) f is strictly decreasing in \((2, \infty)\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (D) only
By the product rule:
\( f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) \)
\( = (1 - x)e^{-x} \)
Since \(e^{-x} \gt 0\), f′(x) has the sign of 1 − x:
\( f'(x) \gt 0 \) for \( x \lt 1 \)
\( f'(x) \lt 0 \) for \( x \gt 1 \)
So \(f'(x) = (1 - x)e^{-x}\), f is strictly increasing in \((-\infty, 1)\) and strictly decreasing in \((1, \infty)\).
Checking the statements
(A) \(f'(x) = (1 - x)e^{-x}\) — Correct. Product rule: \(e^{-x} - xe^{-x}\).
(B) f is strictly increasing in \((-\infty, 1)\). — Correct. 1 − x > 0 there.
(C) f is strictly decreasing in \((0, \infty)\). — Incorrect. f increases on (0, 1), for example \(f'\left(\frac{1}{2}\right) = \frac{1}{2}e^{-1/2} \gt 0\).
(D) f is strictly decreasing in \((2, \infty)\). — Correct. This lies inside \((1, \infty)\), where 1 − x < 0.