# How do you ise interval notation indicate where f(x) is concave up and concave down for f(x)=x^(4)-6x^(3)?

Jun 16, 2017

$f \left(x\right)$ is concave up from $\left(- \infty , 0\right) \cup \left(3 , \infty\right)$
$f \left(x\right)$ is concave down from $\left(0 , 3\right)$

#### Explanation:

What you want to do is find the second derivative using the power rule:

$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$

The first derivative is:

$\frac{d}{\mathrm{dx}} = 4 {x}^{3} - 18 {x}^{2}$

The second derivative is:

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

What you want to do now is factor it and set it equal to zero:

$12 x \left(x - 3\right) = 0$

$12 x = 0$ & $x - 3 = 0$

$x = 0 , 3$

Now you make a test interval from:
$\left(- \infty , 0\right) \cup \left(0 , 3\right) \cup \left(3 , \infty\right)$

You test values from the left and right into the second derivative but not the exact values of $x$.
If you get a negative number then it means that at that interval the function is concave down and if it's positive its concave up.

If done so correctly you should get that:

$f \left(x\right)$ is concave up from $\left(- \infty , 0\right) \cup \left(3 , \infty\right)$ and that
$f \left(x\right)$ is concave down from $\left(0 , 3\right)$

You should also note that the points $f \left(0\right)$ and $f \left(3\right)$ are inflection points. 