(i) f(x) = x3 ,x ∈ [- 2,2]
f’(x) = 0 ⇒ 3x2 = 0
⇒ x = 0 for finding the absolute maximum and absolute minimum,
we have to evaluate f (0), f (2), f (-2)
F(0) = (0)3 = 0, F(2) = (2)3 = 8,
F(-2) = (-2)3 = – 8
Absolute maximum = 8 and
Absolute minimum = -8
∴ maximum at 2 is 8 and minimum at -2 is – 8.
(ii)
(iii)
(iv) f (x) = (x – 1)2 + 3, x ∈ [-3,1]
f'(x) = 2(x – 1)
f’(x) = 0 ⇒ (x – 1) = 0
⇒ x = 1 we will evaluate f (-3) and f(1)
f(-3) = (-3 – 1)2 + 3= 16 + 3 = 19
f(1) = (0)2 + 3 = 3
Absolute maximum at x = -3 is 19 and
Absolute minimum x = 1 is 3.