मराठी

At what point, the slope of the curve y = – x3 + 3x2 + 9x – 27 is maximum? Also find the maximum slope.

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प्रश्न

At what point, the slope of the curve y = – x3 + 3x2 + 9x – 27 is maximum? Also find the maximum slope.

बेरीज
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उत्तर

Given that: y = – x3 + 3x2 + 9x – 27

Differentiating both sides w.r.t. x,

We get `"dy"/'dx"` = – 3x2 + 6x + 9

Let slope of the cuve `"dy"/"dx"` = Z

∴ z = – 3x2 + 6x + 9

Differentiating both sides w.r.t. x,

We get `"dz"/"dx"` = – 6x + 6

For local maxima and local minima,

`"dz"/"dx"` = 0

∴ – 6x + 6 = 0

⇒ x = 1

⇒ `("d"^2z)/("d"x^2)` = – 6 < 0 Maxima

Put x = 1 in equation of the curve y = (– 1)3 + 3(1)2 + 9(1) – 27

= – 1 + 3 + 9 – 27

= – 16

Maximum slope = – 3(1)2 + 6(1) + 9 = 12

Hence, (1, – 16) is the point at which the slope of the given curve is maximum and maximum slope = 12.

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

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 6 Application Of Derivatives
Exercise | Q 23 | पृष्ठ १३७

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