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Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is ______.

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प्रश्न

Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is ______.

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उत्तर

Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is 12.

Explanation:

Given that y = –x3 + 3x2 + 9x – 27

`"dy"/'dx"` = – 3x2 + 6x + 9

∴ Slope of the given curve,

m = – 3x2 + 6x + 9   ....`("dy"/"dx" = "m")`

`"dm"/"dx"` = –6x + 6

For local maxima and local minima, `"dm"/"dx"` = 0

∴ – 6x + 6 = 0

⇒ x = 1

Now `("d"^2"m")/("dx"^2)` = = – 6 < 0 maxima

∴ Maximum value of the slope at x = 1 is

`"m"_(x = 1)` = – 3(1)2 + 6(1) + 9

= – 3 + 6 + 9

= 12

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अध्याय 6: Application Of Derivatives - Exercise [पृष्ठ १४१]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 6 Application Of Derivatives
Exercise | Q 57 | पृष्ठ १४१

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