# How do you use the first and second derivatives to sketch y=x^4-2x?

Mar 30, 2017

minimum (0.79;-1.19)
inflection (0;0)
convex

#### Explanation:

The first derivative is:

$f ' \left(x\right) = 4 {x}^{3} - 2$

Let's study the solution of the inequality

$f ' \left(x\right) \ge 0$

that's

$4 {x}^{3} - 2 \ge 0$

$x \ge \sqrt[3]{\frac{1}{2}} \cong 0.79$

It means that

if $x < \sqrt[3]{\frac{1}{2}}$ the function is decreasing;

if $x > \sqrt[3]{\frac{1}{2}}$ the function is increasing;

then if x=if $x < \sqrt[3]{\frac{1}{2}}$ the function has a minimum and there $f \left(x\right) = {\left(\sqrt[3]{\frac{1}{2}}\right)}^{4} - 2 \cdot \sqrt[3]{\frac{1}{2}} = \left(\sqrt[3]{\frac{1}{16}}\right) - 2 \cdot \sqrt[3]{\frac{1}{2}} \cong - 1.19$;

The second derivative is:

$f ' ' \left(x\right) = 12 {x}^{2}$

Let's study the solution of the inequality

$f ' ' \left(x\right) \ge 0$

that's

$12 {x}^{2} \ge 0$ that is verified $\forall x \in \mathbb{R}$

It tells the function is convex and if $x = 0$ there is a point of inflection and it is the origin O(0;0)

graph{y=x^4-2x [-2, 3, -2, 5]}