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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.
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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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