# How do use the first derivative test to determine the local extrema f(x) = x^3 - x^2 - x + 3?

Sep 26, 2015

See the explanation.

#### Explanation:

$f ' = 3 {x}^{2} - 2 x - 1$
$f ' = 0 \iff 3 {x}^{2} - 2 x - 1 = 0$
$3 {x}^{2} - 3 x + x - 1 = 0$
$3 x \left(x - 1\right) + x - 1 = 0$
$\left(x - 1\right) \left(3 x + 1\right) = 0$
$x = 1 \vee x = - \frac{1}{3}$

$a = 3 > 0$ function is concave.

graph{3x^2-2x-1 [-10, 10, -5, 5]}

$\forall x \in \left(- \infty , - \frac{1}{3}\right) \cup \left(1 , + \infty\right) f ' \left(x\right) > 0$ function is increasing

$\forall x \in \left(- \frac{1}{3} , 1\right) f ' \left(x\right) < 0$ function is decreasing

$f ' \left(x\right)$ changes sign in $x = - \frac{1}{3}$ and $f \left(x\right)$ has a maximim value ${f}_{\max} = f \left(- \frac{1}{3}\right) = \frac{86}{27}$

$f ' \left(x\right)$ changes sign in $x = 1$ and $f \left(x\right)$ has a minimum value ${f}_{\min} = f \left(1\right) = 2$