# What are the global and local extrema of f(x)=x^3-x^2-x ?

Aug 31, 2016

There is a maxima at $x = - \frac{1}{3}$
There is a minima at $x = 1$

#### Explanation:

Given -

$y = {x}^{3} - {x}^{2} - x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} - 2 x - 1$
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 x - 2$

Set the first derivative equal to zero

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 3 {x}^{2} - 2 x - 1 = 0$

$3 {x}^{2} - 3 x + x - 1 = 0$

$3 x \left(x - 1\right) + 1 \left(x - 1\right) = 0$
$\left(3 x + 1\right) \left(x - 1\right) = 0$
$3 x = - 1$
$x = - \frac{1}{3}$
$x - 1 = 0$
$x = 1$

At $x = - \frac{1}{3}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 \left(- \frac{1}{3}\right) - 2 = - 2 - 2 = - 4 < 0$

There is a maxima at $x = - \frac{1}{3}$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 6 \left(1\right) - 2 = 6 - 2 = 4 > 0$

There is a minima at $x = 1$