# How do you find concavity, inflection points, and min/max points for the function: f(x)=x(x^2+1) defined on the interval [–5,4]?

Aug 23, 2015

Take the first and second derivative and their zeros.

#### Explanation:

First, look at the graph so you can get an idea of what you are working with.
graph{x(x^2+1) [-10, 10, -5, 5]}
At first glance, there don't appear to be any max or min on the function. The graph looks concave down to the left and up on the right. Just to be sure, lets do the math. We need to take the first derivative, and that will be easier once we multiply the $x$ through.

$f \left(x\right) = {x}^{3} + x$
$f ' \left(x\right) = 3 {x}^{2} + 1$
${x}^{2} = - \frac{1}{3}$

Since ${x}^{2}$ would need to be negative, there are no real zeros. This means the min an max will be the endpoints, $x = - 5$ and $x = 4$. To get inflection points and concavity, we need to take the second derivative.

$f ' ' \left(x\right) = 6 x$

This function is linear, so our work becomes pretty easy. The inflection point is $x = 0$ because that is where our expression $= 0$. The expression is negative for $x < 0$ so its concave down and positive for $x > 0$ so its concave up. To find your points, plug your $x$ values into $f \left(x\right)$ and solve.

min: $\left(- 5 , - 130\right)$
max: $\left(4 , 68\right)$
inflection: $\left(0 , 0\right)$
concave down: $x < 0$
concave up: $x > 0$