Question

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Answer 1

Answer:

12

Step-by-step explanation:

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Answer 2

Given the equation of the curve , $y = -x^3 + 3x^{2} + 9x - 27$

let us find the points where the curve attains its maxima/minima by equating its first derivative to zero.

$\frac{dy}{dx} = -3x^{2} + 6x + 9 = 0$

⇒ $-x^{2} +2x+3 = 0$

⇒ $x^{2} -2x-3 = 0$

⇒$(x+1)(x-3) = 0$

so, for values of x = -1 , x = +3 the curve has maxima or minima

to find where the curve attains maxima we need to check the nature of its second derivative at these points.

   $\frac{d^2y}{dx^2} = -6x + 6$

at x = -1 , $\frac{d^2y}{dx^2} = 12 > 0$ , so the curve attains its minima at x = -1

at x = 3 , $\frac{d^2y}{dx^2} = -12 < 0$ , so the curve attains its maxima at x = 3

so the maximum value the curve = y(3)

y(3) = -3³ + 3(3²) + 9(3) -27  = 0

∴The local maximum value of the curve is 0.

HOPE THIS HELPS YOU !!