# How do you find all local maximum and minimum points given y=3x^4-4x^3?

Jan 3, 2017

$x = 1$ is a local minimum.
$x = 0$ is an inflection point.

#### Explanation:

First we find the critical values for $f \left(x\right)$ equating the first derivative to zero:

$f ' \left(x\right) = 12 {x}^{3} - 12 {x}^{2} = 0$

$12 {x}^{2} \left(x - 1\right) = 0 \implies {x}_{1} = 0 , {x}_{2} = 1$ are the critical points.

Now we calculate the second derivative:

$f ' ' \left(x\right) = 36 {x}^{2} - 24 x = 12 x \left(3 x - 2\right)$

So we have $f ' ' \left(1\right) = 12 > 0$ and ${x}_{2} = 1$ is a local minimum.
Instaed we have $f ' ' \left(0\right) = 0$, so to determine the nature of ${x}_{1} = 0$ we can look at the sign of the first derivative and we can see that:

$f ' \left(x\right) = 12 {x}^{2} \left(x - 1\right) < 0$ around $x = 0$, so that ${x}_{1}$cannot be a local maximum or minimum and is instead an inflection point.

graph{3x^4-4x^3 [-2.813, 2.812, -1.406, 1.407]}