हिंदी

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

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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].

योग
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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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अध्याय 6: Application of Derivatives - Exercise 6.5 [पृष्ठ २३२]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 6 Application of Derivatives
Exercise 6.5 | Q 7 | पृष्ठ २३२

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