# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for y = x^3 - 3x^2 - 9x +15?

Mar 21, 2016

At $x = 3$, we have a relative minima and at $x = - 1$ we have a relative maxima.

#### Explanation:

A function is increasing, when the value of its derivative at that point is positive and is decreasing, when the value of its derivative at that point is negative.

At maxima and minima, the value of derivative is $0$, as at maxima, the slope of the curve stops increasing and starts declining and at minima the slope of the curve stops declining and starts increasing.

Further, at maxima, second derivative is negative and at minima, second derivative is positive.

Derivative of $f \left(x\right) = {x}^{3} - 3 {x}^{2} - 9 x + 15$ is $f ' \left(x\right) = 3 {x}^{2} - 6 x - 9$, which is zero when $3 {x}^{2} - 6 x - 9 = 0$ or ${x}^{2} - 2 x - 3 = 0$. Further second derivative $f ' ' \left(x\right) = 6 x - 6$

Factorizing ${x}^{2} - 2 x - 3 = 0$ becomes $\left(x - 3\right) \left(x + 1\right) = 0$ and hence minima or maxima is at $x = 3$ or $x = - 1$. As at $x = 3$, $f ' ' \left(x\right) = 12$ we have a minima and at $x = - 1$ $f ' ' \left(x\right) = - 12$ we have a maxima.

However, as elsewhere the functions takes lower value than the minima or higher value than the maxima these are relative minima and maxima.

graph{x^3-3x^2-9x+15 [-10, 10, -25, 25]}