Question

Find both the maximum value and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3].

Answer 1

f (x) = 3x4 – 8x3 + 12x2 – 48 x + 25

f’(x) = 12x3 – 24x2 + 24x – 48

f‘(x) = 0 ⇒ 12 (x3 – 2x2 + 2x - 4) = 0

⇒ 12 (x2 (x – 2) + 2 (x – 2))

⇒ 12 ( (x – 2) (x2 + 2) ) = 0 (x – 2)(x2 + 2) = 0

⇒ x = 2 but x2 + 2 ≠ 0

The points are f(0), f(2), f(3)

f(x) = 3x4 – 8x3 + 12x2 – 48x + 25

f(0) = 25

f(2) = 3(16) – 8(8) + 12(4) – 48(2) + 25 = 48 – 64 + 48 – 96 + 25 = -39

f(3) = 3(34) – 8 (33) + 12(32) – 48(3) + 25 = 243 – 216 + 108 – 144 + 25

376 – 360 = 16

∴ maximum of f (x) at x = 0 is 25

minimum of f (x) at x = 2 is – 39.