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Find the points of local maxima, local minima and the points of inflection of the function f(x) = x5 – 5x4 + 5x3 – 1. Also find the corresponding local maximum and local minimum values.

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प्रश्न

Find the points of local maxima, local minima and the points of inflection of the function f(x) = x5 – 5x4 + 5x3 – 1. Also find the corresponding local maximum and local minimum values.

योग
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उत्तर

We have f(x) = x5 – 5x4 + 5x3 – 1

⇒ f '(x) = 5x4 – 20x3 + 15x2

For local maxima and local minima, f '(x) = 0

⇒ 5x4 – 20x3 + 15x2 = 0

⇒ 5x2(x2 – 4x + 3) = 0

⇒ 5x2(x2 – 3x – x + 3) = 0

⇒ x2(x – 3)(x – 1) = 0

∴ x = 0, x = 1 and x = 3

Now f '(x) = 20x3 – 60x2 + 30x

⇒ `"f''"(x)_("at" x = 0)` = 20(0)3 – 60(0)2 + 30(0) = 0

Which is neither maxima nor minima.

∴ f (x) has the point of inflection at x = 0

`"f''"(x)_("at" x = 1)` = 20(1)3 – 60(1)2 + 30(1)

= 20 – 60 + 30

= –10 < 0 Maxima

`"f''"(x)_("at" x = 2)` = 20(3)3 – 60(3)2 + 30(3)

= 540 – 540 + 90

= 90 > 0 Minima

The maximum value of the function at x = 1

f (x) = (1)5 – 5(1)4 + 5(1)3 – 1

= 1 – 5 + 5 – 1

= 0

The minimum value at x = 3 is

f (x) = (3)5 – 5(3)4 + 5(3)3 – 1

= 243 – 405 + 135 – 1

= 378 – 406

= – 28

Hence, the function has its maxima at x = 1 and the maximum value = 0 and it has minimum value at x = 3 and its minimum value is – 28.

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अध्याय 6: Application Of Derivatives - Exercise [पृष्ठ १३७]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 6 Application Of Derivatives
Exercise | Q 26 | पृष्ठ १३७

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