# How do you find the local maximum and minimum values of f(x) = 2x^3 - 5x +1 in the the interval is (-3,3)?

Nov 15, 2016

$x = - \sqrt{\frac{5}{6}}$ is a local maximum
$x = \sqrt{\frac{5}{6}}$ is a local minimum
graph{2x^3-5x+1 [-10, 10, -5, 5]}

#### Explanation:

Find local extrema on the interval by finding where $f ' \left(x\right)$ is equal to zero. First find $f ' \left(x\right)$

$f \left(x\right) = 2 {x}^{3} - 5 x + 1$

$f ' \left(x\right) = 6 {x}^{2} - 5$
$0 = 6 {x}^{2} - 5$
$5 = 6 {x}^{2}$
${x}^{2} = \frac{5}{6}$
$x = \pm \sqrt{\frac{5}{6}}$
$x \approx \pm .9129$

Find whether each is a local max or local min by checking values around $\pm \sqrt{\frac{5}{6}}$;
$x = - \sqrt{\frac{5}{6}}$ is a local maximum
$x = \sqrt{\frac{5}{6}}$ is a local minimum