The function \(f(x) = (x^{2} - 3)e^{x}\), \(x \in R\), is strictly decreasing in the interval
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-3, 1)\)
By the product rule:
\( f'(x) = 2x \cdot e^{x} + (x^{2} - 3)e^{x} \)
\( = e^{x}(x^{2} + 2x - 3) = e^{x}(x + 3)(x - 1) \)
Since \(e^{x} \gt 0\) for every x, the sign of f′(x) is the sign of (x + 3)(x − 1):
\( f'(x) \lt 0 \) for \( -3 \lt x \lt 1 \)
\( f'(x) \gt 0 \) for \( x \lt -3 \) and for \( x \gt 1 \)
So f is strictly decreasing in \((-3, 1)\).
Why the other options are incorrect
Option (1) \((-\infty, -3)\) — Incorrect. For x < −3 both factors x + 3 and x − 1 are negative, so f′(x) > 0 and f is increasing.
Option (2) \((1, \infty)\) — Incorrect. For x > 1 both factors are positive, so f is increasing.
Option (4) \((-\sqrt{3}, \sqrt{3})\) — Incorrect. This is where f(x) itself is negative, not where f′(x) is negative; f increases on \((1, \sqrt{3})\).