# How do you determine the intervals where #f(x)=3x^2-x^3/3# is concave up or down?

##### 2 Answers

The function is concave up in the interval

#### Explanation:

The function is

This a polynomial function continous and derivable on

Calculate the first and second derivatives

Let

There is an inflection point at

Build a variation chart to determine the concavities

The function is concave up in the interval

graph{3x^2-x^3/3 [-9.73, 18.75, -0.11, 14.13]}

Determine where the second derivative is positive or negative. If negative, then it is concave down. If positive, concave up. If zero, then it's an inflection point.

#### Explanation:

While the first derivative tells us the slope of a function at a given point, the second derivative gives us the concavity of the function.

When the second derivative is positive, the function has a 'concave up' concavity to it. When negative, it's concave down. The point where this changes is the point of inflection. The point of inflection is equal to when the second derivative is equal to zero.

Let's work with the function for a bit to determine the second derivative:

Now that we know the second derivative, we can calculate the points of inflection to determine the intervals for concavity:

We only have one inflection point, so we just need to determine if the function is concave up or down on either side of the function:

The concavity for our function can be described as such:

We can see that when graphing the function as well:

graph{3x^2-(x^3)/3 [-5, 10, -2, 38]}