Question

Find the local extrema for the following functions using second derivative test:

f(x) = x² e^–2x 

Answer 1

f'(x) = x² [– 2e^–2x] + e^–2x (2x)

= 2e^–2x (x – x²)

f”(x) = 2e^–2x (1 – 2x) + (x – ²) (–4e^–2x)

= 2e^–2x [(1 – 2x) + (x – x²)(– 2)]

= 2e^–2x [2x² – 4x + 1]

f'(x) = 0

⇒ 2e^–2x (x – x²) = 0

⇒ x(1 – x) = 0

⇒ x = 0 or x = 1

At x = 0, f”(x) = 2 × 1[0 – 0 + 1] = + ve

⇒ x = 0 is a local minimum point and the minimum value is f(0) = 0 at x = 1,

f”(x) = 2e^–2 [2 – 4 + 1] = – ve

⇒ x = 1 is a local maximum point and the maximum value is f(1) = `1/"e"^2`

Local maxima `1/"e"^2` and local minima = 0