(i) f(x) = – 3x5 + 5x3
f'(x) = 0, f”(x) = -ve at x = a
⇒ x = a is a maximum point
f'(x) = 0, f”(x) = +ve at x = 6
⇒ x = b is a minimum point
f(x) = – 3x5 + 5x3
f’(x) = -15x4 + 15x2
f”(x) = -60x3 + 30x
f'(x) = 0 ⇒ – 15x2(x2 – 1) = 0
⇒ x = 0, +1, -1
at x = 0, f”(x) = 0
at x = 1, f”(x) = -60 + 30 = – ve
at x = -1, f”(x) = 60 – 30 = + ve
So at x = 1, f'(x) = 0 and f”(x) = -ve
⇒ x = 1 is a local maximum point and f(1) = 2
So the local maximum is (1, 2)
at x = -1, f'(x) = 0 and f”(x) = +ve
⇒ x = -1 is a local maximum point and f(-1) = -2.
So the local minimum point is (-1, -2)
∴ local minimum is -2 and local maximum is 2.
(ii) f(x) = x log x
(iii) f(x) = x2 e-2x
f'(x) = x2 [-2e-2x] + e-2x(2x)
= 2e-2x(x – x2)
f”(x) = 2e-2x(1 – 2x) + (x – x2) (-4e-2x)
= 2e-2x [(1 – 2x) + (x – x2) (- 2)]
= 2e-2x [2x – 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/e2
Local maxima 1/e2 and local minima = 0