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प्रश्न
Find both the maximum value and the minimum value of 3x4 − 8x3 + 12x2 − 48x + 25 on the interval [0, 3].
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उत्तर
Let, f (x) = 3x4 - 8x3 + 12x2 - 48x + 25
f‘(x) = 12x3 - 24x2 + 24x - 48
= 12 [x3 - 2x2 + 2x - 4]
= 12 [x2 (x - 2) + 2 (x - 2)]
= 12 (x - 2) (x2 + 2)
If f'(x) = 0, then x - 2 = 0 ⇒ x = 2
x2 + 2 = 0 is impossible.
We now find the value of f at x = 2 and the endpoints of the interval [0, 3].
At, x = 0 f(0) = 25
At, x = 2 f(2) = 3(2)4 - 8(2)3 + 12(2)2 - 48(2) + 25
= 3 × 16 - 8 × 8 + 12 × 4 - 48 × 2 + 25
= 48 - 64 + 48 - 96 + 25
= - 39
At, x = 3 f(3) = 3(3)4 - 8(3)3 + 12(3)2 - 48(3) + 25
= 3 × 81 - 8 × 27 + 12 × 9 - 48 × 3 + 25
= 243 - 216 + 108 - 144 + 25
= 16
∴ Maximum value off (x) = 25 at x = 0 and minimum value of f(x ) = -39 at x = 2.
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