# How many points of inflection does the function f(x)=x^7-x^2 have?

Aug 19, 2015

$f \left(x\right) = {x}^{7} - {x}^{2}$ has $\textcolor{red}{\text{one inflection point}}$.

#### Explanation:

Inflection points are where the function changes concavity.

The second derivative must equal zero when the function changes concavity.

But we must check points on either side to make sure that the concavity really does change.

$f \left(x\right) = {x}^{7} - {x}^{2}$

$f ' \left(x\right) = 7 {x}^{6} - 2 x$

$f ' ' \left(x\right) = 42 {x}^{5} - 2 = 0$

$42 {x}^{5} = 2$

${x}^{5} = \frac{2}{42} = \frac{1}{21}$

x=1/root(5)(21)≈0.54

So, $x = \frac{1}{\sqrt[5]{21}}$ is a possible inflection point.

To make sure that the concavity actually changes, we pick a number on either side of $x = \frac{1}{\sqrt[5]{21}}$ and check what the concavity is at those locations.

Let's use $x = - 1$ and $x = 1$.

$f ' ' \left(- 1\right) = 42 {\left(- 1\right)}^{5} - 2 = - 42 - 2 = - 44$ (concave down).

$f ' ' \left(1\right) = 42 {\left(1\right)}^{5} - 2 = 42 - 2 = 40$ (concave up).

The concavity changes, so $x = \frac{1}{\sqrt[5]{21}}$ is the only inflection point.

The brown square represents the inflection point.