Question

f(x)=3x⁴-8x³+12x²-48x+125 x ∈ [0,3],Find the maximum and minimum value of the given function.

Answer 1

Let f (x) = 3x⁴ – 8x³ + 12x² – 48x + 25, x ∈[0,3]

⇒ f’(x) = 12x³ - 24x² + 24x – 48

=12(x³ - 2x² + 2x – 4)

=12[x² (x – 2)+ 2(x – 2)]

=12(x -2)( x²+ 2)

Now, f’(x) = 0

⇒ x =2 or (x2+ 2) ≠ 0
Therefore, we will only consider 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) – 96 + 26
=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 absolute maximum value of f on [0,3] is 25 occurring at x =0.
And, the absolute minimum value of f on [0,3] is -39 occurring at x = 2.