1. Given;
f(x) = 2x3 – 15x2 + 36x + 1, x ∈ [1, 5]
⇒ f'(x) = 6x2 – 30x + 36
For turning point f'(x) = 0 ⇒ 6x2 – 30x + 36 = 0
⇒ x2 – 5x + 6 = 0 ⇒ (x -3)(x – 2) = 0
⇒ x = 3, 2
f(1) = 2(1)3 – 15(1)2 + 36(1) + 1 = 24
f(2) = 2(2)3 – 15(2)2 + 36(2) + 1 = 29
f(3) = 2(3)3 – 15(3)2 + 36(3) + 1 = 28
f(5) = 2(5)3 – 15(5)2 + 36(5) + 1 = 56
Absolute maximum = max {24, 29, 28, 56} = 56
Absolute minimum = min {24, 29, 28, 56} = 24
2. Given;
f'(x) = 0 at x = \(\frac{1}{8}\) 18 and f'(x) is not defined at x = 0.
Therefore;
Absolute maximum = max {18, 0, 6, −\(\frac{9}{4}\)} = 18
Absolute minimum = min {18, 0, 6, −\(\frac{9}{4}\)} = −\(\frac{9}{4}\).