# What are the extrema of f(x)=2x^3 + 5x^2 - 4x - 3 ?

Dec 10, 2016

#### Answer:

${x}_{1} = - 2$ is a maximum

${x}_{2} = \frac{1}{3}$ is a minimum.

#### Explanation:

First we identify the critical points by equating the first derivative to zero:

$f ' \left(x\right) = 6 {x}^{2} + 10 x - 4 = 0$

giving us:

$x = \frac{- 5 \pm \sqrt{25 + 24}}{6} = \frac{- 5 \pm 7}{6}$

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

Now we study the sign of the second derivative around the critical points:

$f ' ' \left(x\right) = 12 x + 10$

so that:

$f ' ' \left(- 2\right) < 0$ that is ${x}_{1} = - 2$ is a maximum

$f ' ' \left(\frac{1}{3}\right) > 0$ that is ${x}_{2} = \frac{1}{3}$ is a minimum.

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