# Inflection points review

Review your knowledge of inflection points and how we use differential calculus to find them.

## What are inflection points?

Inflection points (or points of inflection) are points where the graph of a function changes concavity (from $\cup$ to $\cap$ or vice versa).

*Want to learn more about inflection points and differential calculus? Check out this video.*

## Practice set 1: Analyzing inflection points graphically

*Want to try more problems like this? Check out this exercise.*

## Practice set 2: Analyzing inflection points algebraically

Inflection points are found in a way similar to how we find extremum points. However, instead of looking for points where the derivative changes its sign, we are looking for points where the

*second derivative*changes its sign.Let's find, for example, the inflection points of $f(x)=\dfrac{1}{2}x^4+x^3-6x^2$.

The second derivative of $f$ is $f''(x)=6(x-1)(x+2)$.

$f''(x)=0$ for $x=-2,1$, and it's defined everywhere. $x=-2$ and $x=1$ divide the number line into three intervals:

Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval.

Interval | $x$-value | $f''(x)$ | Verdict |
---|---|---|---|

$x<-2$ | $x=-3$ | $f''(-3)=24>0$ | $f$ is concave up $\cup$ |

$-2<x<1$ | $x=0$ | $f''(0)=-12<0$ | $f$ is concave down $\cap$ |

$x>1$ | $x=2$ | $f''(2)=24>0$ | $f$ is concave up $\cup$ |

We can see that the graph of $f$ changes concavity at both $x=-2$ and $x=1$, so $f$ has inflection points at both of those $x$-values.

*Want to try more problems like this? Check out this exercise.*