The function f(x) = (x - 1)e^x is strictly decreasing in the interval

The correct answer is Option (3) → \((-\infty, 0)\)

By the product rule:

    \( f'(x) = 1 \cdot e^{x} + (x - 1)e^{x} \)

    \( = xe^{x} \)

Since \(e^{x} \gt 0\) for every x, the sign of f′(x) is the sign of x:

    \( f'(x) \lt 0 \) for \( x \lt 0 \)
    \( f'(x) \gt 0 \) for \( x \gt 0 \)

So f is strictly decreasing in \((-\infty, 0)\) and strictly increasing in \((0, \infty)\).

Why the other options are incorrect

• Option (1) \((0, \infty)\) — Incorrect. For x > 0, \(f'(x) = xe^{x} \gt 0\), so f is increasing there.

• Option (2) \((-\infty, 1)\) — Incorrect. The change of sign is at x = 0, the zero of \(xe^{x}\), not at x = 1, the zero of f; f increases on (0, 1).

• Option (4) \((1, \infty)\) — Incorrect. For x > 1, \(f'(x) = xe^{x} \gt 0\), so f is increasing there.