Question

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

Answer 1

Answer:

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.

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Answer 2

Answer:

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