# How do you find the local extrema for f(x) = 2x^3 - x^2 - 4x +3?

Dec 29, 2016

$f \left(x\right)$ has a local maximum for $x = - \frac{2}{3}$ and a local minimum for $x = 1$

#### Explanation:

You find the critical values for $f \left(x\right)$ by identifying where the first derivative equals zero:

$f ' \left(x\right) = 6 {x}^{2} - 2 x - 4 = 0$

$x = \frac{1 \pm \sqrt{1 + 24}}{6} = \frac{1 \pm 5}{6}$,

so: ${x}_{1} = - \frac{2}{3}$ and ${x}_{2} = 1$

Now, as $f \left(x\right)$ is a second degree polynomial with leading positive coefficient, we know that:

$f ' \left(x\right) > 0$ for $x \in \left(- \infty , {x}_{1}\right)$ and $x \in \left({x}_{2} , + \infty\right)$
$f ' \left(x\right) < 0$ for $x \in \left({x}_{1} , {x}_{2}\right)$

so ${x}_{1}$ is a maximum and ${x}_{2}$ is a minimum.

graph{2x^3-x^2-4x+3 [-10, 10, -5, 5]}