# How do you graph y = 2x^3 - 4x + 1 using the first and second derivative?

Feb 21, 2015

You can start by using the first derivative to find points of maximum and minimum:
$y ' = 6 {x}^{2} - 4$
You set the first derivative equal to zero:
$6 {x}^{2} - 4 = 0$ which gives you:
$x = \pm \sqrt{\frac{2}{3}}$

So:
maximum: $x = - \sqrt{\frac{2}{3}}$ and $y = 3.17$
minimum: $x = \sqrt{\frac{2}{3}}$ and $y = - 1.17$

You use the second derivative to find inflection points:
$y ' ' = 12 x$
Setting it equal to zero you get the point of inflection:
$x = 0$ $y = 1$

And finally the graph: