# Examine y = x^3 – 3*x^2 – 9x + 5 for maxima and minima using the sign of the second derivative method.

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The function y = x^3 - 3*x^2 - 9x + 5. The extreme points of the function are the roots of the equation y' = 0. If at a root x = a, y'' is negative it is a point of maxima and if y'' is positive it is a point of minima.

y' = 3x^2 - 6x - 9

3x^2 - 6x - 9 = 0

=> x^2 - 2x - 3 = 0

=> x^2 - 3x + x - 3 = 0

=> x(x - 3) + 1(x - 3) = 0

=> (x - 3)(x + 1) = 0

=> x = -1 and x = 3

y'' = 6x - 6

At x = -1, y'' is negative; a point of maxima lies at x = -1

At x = 3, y'' is positive; a point of minima lies at x = 3

**The given function has a point of maxima at x = -1 and a point of minima at x = 3.**