For the function \(f(x) = e^{-2x} (2-x)^2\), the point of local maxima is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(x=3\)
By the product rule:
\(f'(x) = -2e^{-2x}(2 - x)^2 + e^{-2x} \cdot 2(2 - x)(-1)\)
\(= -2e^{-2x}(2 - x)\left[(2 - x) + 1\right]\)
\(= -2e^{-2x}(2 - x)(3 - x)\)
Since \(e^{-2x} \gt 0\), the critical points are x = 2 and x = 3. The sign of f’(x) on each interval:
\(x \lt 2\): \((2 - x) \gt 0,\ (3 - x) \gt 0\), so \(f'(x) \lt 0\)
\(2 \lt x \lt 3\): \((2 - x) \lt 0,\ (3 - x) \gt 0\), so \(f'(x) \gt 0\)
\(x \gt 3\): \((2 - x) \lt 0,\ (3 - x) \lt 0\), so \(f'(x) \lt 0\)
At x = 3, f’(x) changes from positive to negative, so by the first derivative test x = 3 is the point of local maximum.
Why the other options are incorrect
Option (1) \(x=1\) — Incorrect. \(f'(1) = -2e^{-2}(1)(2) \ne 0\), so x = 1 is not a critical point.
Option (2) \(x=2\) — Incorrect. f’(x) changes from negative to positive at x = 2, so it is the point of local minimum (f(2) = 0).
Option (4) \(x = 5/2\) — Incorrect. \(f'(5/2) = -2e^{-5}\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) \ne 0\), so it is not a critical point.