# What Are Stationary Points of a Function?

It’s often useful to figure out where a graph is at its highest (maxima) or lowest (minima). These points are also called the extrema, or extremes, of the graph. There is also a third type of points called saddle points, where the graph is neither increasing nor decreasing (the first derivative is equal to 0), but it is neither a maximum nor a minimum. The collective name for points where the first derivative equals 0 is stationary points.

The sign chart of ${f}^{\prime }\left(x\right)$ tells us where the graph of $f\left(x\right)$ is increasing or decreasing. It also tells us where we have stationary points. Furthermore, it tells us where the derivative ${f}^{\prime }\left(x\right)$ of the function is above or below the $x$-axis. You find the points in the following way:

Rule

### DeterminingStationaryPoints

• The maxima of a function are located where the graph changes from increasing to decreasing.

• The minima of a function are located where the graph changes from decreasing to increasing.

• The saddle points of a function are located where the graph changes from increasing to leveling off and then increasing again, or where it changes from decreasing to leveling off and then continues to decrease.

Study the figure closely. It shows you the link between $f\left(x\right)$, ${f}^{\prime }\left(x\right)$ and a sign chart for both functions. A sign chart is a very useful tool to find where the function $f\left(x\right)$ has stationary points.

• ${f}^{\prime }\left(x\right)>0⇔f$ increases

• ${f}^{\prime }\left(x\right)<0⇔f$ decreases

• ${f}^{\prime }\left(x\right)=0⇔f$ levels off

The sign chart helps you find different properties of the functions.

Example 1

Find the stationary points of

 $f\left(x\right)={x}^{2}+5x+6$

You find these points by solving the equation ${f}^{\prime }\left(x\right)=0$. Thus, find ${f}^{\prime }\left(x\right)$ and set it equal to 0: $\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)=2x+5& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2x& =-5\phantom{\rule{1em}{0ex}}|÷2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =-\frac{5}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

To find the $y$-value, you insert $x=-\frac{5}{2}$ into the main function $f\left(x\right)$:

 $f\phantom{\rule{-0.17em}{0ex}}\left(-\frac{5}{2}\right)=\phantom{\rule{-0.17em}{0ex}}{\left(-\frac{5}{2}\right)}^{2}+5\phantom{\rule{-0.17em}{0ex}}\left(-\frac{5}{2}\right)+6=-\frac{1}{4}$

Thus, you have a stationary point at $\phantom{\rule{-0.17em}{0ex}}\left(-\frac{5}{2},-\frac{1}{4}\right)$. You now need to find out if it’s a maximum, a minimum or a saddle point. When working with quadratic functions ($a{x}^{2}+bx+c$) the answer can be found easily by looking at the value of $a$ in the function. In $f\left(x\right)={x}^{2}+5x+6$, we have $a=1>0$, which means that the graph is smiling and you have a minimum.

Example 2

Find the zeros and the stationary point of the function $f\left(x\right)={x}^{3}+{x}^{2}-2x$. Also, find the regions where the function is increasing and decreasing.

### Zeros

You find the zeros where $f\left(x\right)=0$:

 ${x}^{3}+{x}^{2}-2x=x\left({x}^{2}+x-2\right)=0$

 $x\left(x+2\right)\left(x-1\right)=0$

You use the zero product property and find that $\begin{array}{llll}\hfill x& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x+2& =0⇒x=-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x-1& =0⇒x=1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You then find that the zeros are $\left(-2,0\right)$, $\left(0,0\right)$ and $\left(1,0\right)$.

### Stationary Points

You find the stationary points by solving the equation

 ${f}^{\prime }\left(x\right)=3{x}^{2}+2x-2=0$

You use the quadratic formula and get: $\begin{array}{llll}\hfill x& =\frac{-2±\sqrt{4-4\cdot 3\cdot \left(-2\right)}}{2\cdot 3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-2±\sqrt{28}}{6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-2±2\sqrt{7}}{6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-1±\sqrt{7}}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Thus, ${x}_{1}=\frac{-1+\sqrt{7}}{3}\approx 0.55$ and $x=\frac{-1-\sqrt{7}}{3}\approx -1.22$.

To find the corresponding $y$-values, you insert the $x$-values into the main function $f\left(x\right)$. You get

$\begin{array}{llll}\hfill f\left(0.55\right)& ={\left(0.55\right)}^{3}+{\left(0.55\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}-2\left(0.55\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-0.63\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill f\left(-1.22\right)& ={\left(-1.22\right)}^{3}+{\left(-1.22\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}-2\left(-1.22\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2.11\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill f\left(0.55\right)& ={\left(0.55\right)}^{3}+{\left(0.55\right)}^{2}-2\left(0.55\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-0.63\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill f\left(-1.22\right)& ={\left(-1.22\right)}^{3}+{\left(-1.22\right)}^{2}-2\left(-1.22\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2.11\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The stationary points are thus $\left(0.55,-0.63\right)$ and $\left(-1.22,2.11\right)$.

To determine which type of stationary points they are, you need to find the value of the derivative before, after and between the stationary points. You can do this by drawing a sign chart for the factorized expression $3\left(x-0.55\right)\left(x+1.22\right)$. It looks like this:

From the figure, you see that:
• $f\left(x\right)$ decreases on the interval $\phantom{\rule{-0.17em}{0ex}}\left(-1.22,0.55\right)$,

• $f\left(x\right)$ increases on the intervals $\phantom{\rule{-0.17em}{0ex}}\left(-\infty ,-1.22\right)\cup \phantom{\rule{-0.17em}{0ex}}\left(0.55,\infty \right)$.

Thus, you know that $\left(-1.22,2.11\right)$ is a maximum and $\left(0.55,-0.63\right)$ is a minimum.

Note! You cannot tell if you have a maximum or a minimum based on your $y$-values alone, just by saying that the largest one is a maximum and the smallest a minimum. This is because the $y$-values do not tell you how the points are aligned in relation to one another. You can have stationary points where the function is neither increasing nor decreasing overall. See the figure below.