Find the absolute maximum and absolute minimum values of f on the give

1.	Find the absolute maximum and absolute minimum values of f on the given interval.f(x) = x3 – 12x + 1, [-3, 5]
Answer» f(x) = x³ – 12x + 1 f'(x) = 3x² – 12 f'(x) = 0 ⇒ 3x² – 12 = 0 3x² = 12 ⇒ x² = 4 ⇒ x = ±2 The x values are -3, -2, 2, 5 f(x) (at x = -3) = (-3)³ – 12 (-3) + 1 = -27 + 36 + 1 = 10 f(x) (at x = -2) = (-2)³ – (12)(-2) + 1 = -8 + 24 + 1 = 17 f(x) (at x = 2) = 2³ – 12 (2) + 1 = 8 – 24 + 1 = -15 f(x) (at x = 5) = 53 – 12 (5) + 1 = 125 – 60 + 1 = 66 From the above four values 10,17,-15 and 66, we see that absolute maximum is 66 and the absolute minimum is -15.

Find the absolute maximum and absolute minimum values of f on the given interval.f(x) = x3 – 12x + 1, [-3, 5]

Answer»

f(x) = x³ – 12x + 1

f'(x) = 3x² – 12

f'(x) = 0 ⇒ 3x² – 12 = 0

3x² = 12 ⇒ x² = 4 ⇒ x = ±2

The x values are -3, -2, 2, 5

f(x) (at x = -3) = (-3)³ – 12 (-3) + 1 = -27 + 36 + 1 = 10

f(x) (at x = -2) = (-2)³ – (12)(-2) + 1 = -8 + 24 + 1 = 17

f(x) (at x = 2) = 2³ – 12 (2) + 1 = 8 – 24 + 1 = -15

f(x) (at x = 5) = 53 – 12 (5) + 1 = 125 – 60 + 1 = 66

From the above four values 10,17,-15 and 66, we see that absolute maximum is 66 and the absolute minimum is -15.