# What are the inflection points for y = 6x^3 - 3x^4?

Feb 20, 2018

$\left(0 , 0\right) , \left(1 , 3\right)$

#### Explanation:

Take the first derivative:

$y ' = 18 {x}^{2} - 12 {x}^{3}$

Take the second derivative:

$y ' ' = 36 x - 36 {x}^{2}$

Let's factor $y ' '$ a bit to simplify it:

$y ' ' = 36 x \left(1 - x\right)$

Set $y ' '$ equal to zero and solve for $x$:

$36 x \left(1 - x\right) = 0$

$36 x = 0$

So $x = 0$ is our first potential inflection point's $x$-value.

$1 - x = 0$

So $x = 1$ is our second potential inflection point's $x$-value.

We must test values of $y ' '$ in the following intervals (but not at the endpoints, hence the parentheses):

(-∞, 0)
$\left(0 , 1\right)$
(1, ∞)

(-∞,0)

$y ' ' \left(- 1\right) = - 36 \left(2\right) < 0$

In the interval (-∞, 0), $y ' ' < 0$ and so the graph of y is concave down.

$\left(0 , 1\right)$

$y ' ' \left(\frac{1}{2}\right) = 18 \left(\frac{1}{2}\right) > 0$

In the interval $\left(0 , 1\right)$, $y ' ' > 0$ and so the graph of y is concave up. We've switched concavity. This means we have an inflection point at $x = 0$.

Let's plug $x = 0$ back into our original function to get the inflection point's coordinates:

$y \left(0\right) = 6 {\left(0\right)}^{3} - 3 {\left(0\right)}^{4} = 0$

$\left(0 , 0\right)$ is an inflection point.

(1,∞)

$y ' ' \left(2\right) = 72 \left(- 1\right) < 0$

In the interval (1,∞), $y ' ' < 0$ and so the graph of y is concave down. Again, we've switched concavity. We also have an inflection point at $x = 1$.

Plug $x = 1$ into the original function to get the second inflection point's coordinates:

$y \left(1\right) = 6 - 3 = 3$

$\left(1 , 3\right)$ is an inflection point.