At what point, the slope of the curve y = – x3 + 3x2 + 9x –

1.	At what point, the slope of the curve y = – x3 + 3x2 + 9x – 27 is maximum? Also find the maximum slope.
Answer» Given, curve y = – x³ + 3x² + 9x – 27 Differentiating both sides w.r.t. x, we get dy/dx = -3x² + 6x + 9 Let slope of the curve dy/dx = z So, 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 d²z/dx² = -6 < 0 Maxima Putting 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 Therefore, (1, -16) is the point at which the slope of the given curve is maximum and maximum slope = 12.

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

Answer»

Given, curve y = – x³ + 3x² + 9x – 27

Differentiating both sides w.r.t. x, we get

dy/dx = -3x² + 6x + 9

Let slope of the curve dy/dx = z

So, 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

d²z/dx² = -6 < 0 Maxima

Putting 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

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