Question

Find both the maximum value and the minimum value of 3x ^4 − 8x^ 3 + 12x^ 2 − 48x + 25 on the interval [0, 3]

Answer 1

given, f(x) = 3x⁴ - 8x³ + 12x² - 48x + 25
differentiate with respect to x,
f'(x) = 12x³ - 24x² + 24x - 48
= 12(x³ - 2x² + 2x - 4)
= 12(x² + 2)(x -2)

now, f'(x) = 0 = 12(x² + 2)(x - 2)
(x² + 2) ≠ 0 for all real value of x
so, x = 2

Now, we evaluate the value of f at critical point x = 2 and at end points of the interval [0,3].
f(2) = 3(2)⁴ - 8(2)³ + 12(2)² - 48(2) + 25
= 3(16) - 8(8) + 12(4) +25
=48 - 64 +48 – 96 + 25
= -39
f(0) = 3(0)⁴ - 8(0)³ + 12(0)² - 48(0) + 25
= 0+ 0 + 0 +25
= 25
f(3) = 3(3)⁴ - 8(3)³ + 12(3)² - 48(3) + 25
= 3(81) – 8(27) + 12(9) +25
=243 – 216 +108 – 144 + 25
= 16
Therefore, we have the maximum value of f(x) on [0,3] is 25 occurring at x =0.
And, the minimum value of f(x) on [0,3] is -39 occurring at x = 2.