Stationary points

We are often interested in the location of the highest and lowest points on the graph of a function $f$ (see for example the section about optimisation), or other special points.

Definition 1

Consider the graph below. We call point $A$ a local minimum (or a high point), and point $C$ a local maximum (or a low point). We also call $A$ or $B$ a local extremum (or turning point). The plural forms are local maxima, local minima, and local extrema. Note that $A$ and $B$ are the highest and lowest points only in a local environment, that is, $A$ forms the top of a hill, and $B$ is at the bottom of a valley. Or to be more formal, $C$ is a local maximum because if we move away from this point (to the left or right) we go downwards. And $A$ is a local minimum because moving away from this point we go upwards.

Points on the graph where the tangent is horizontal are called stationary points. Local maxima and local minima are clearly stationary points. Their is also a third type of stationary point, called a saddle point, in the figure it is point $B$ . Moving to the left from such a point, we go downwards on the graph, moving to the right we go upwards (or vice versa).

Let's summarise:

Theorem 1

Consider a function $f$ , and a point $P(x|y)$ on the graph of $f$ . A point $P$ on the graph with $x$ -coordinate $x$ is a stationary point, if the tangent at $P$ is horizotnal, that is, we have

Equation 1

\text{stationary point at $x$}\leftrightarrow f'(x)=0

Stationary point at point $P$ with $x$ -coordinate $x$

A stationary point is either a local extremum (local maximum or minimum), or a saddle point.

Example 1

Find the stationary points of the function $f(x)=x^3-2x^2$ by calculation. Then draw the graph in order to classify the stationary points as local maximum, local minimum or saddle point.

Solution

To find the $x$ -coordinate of the stationary points, determine all values of $x$ with

f'(x)=0

Because $f'(x)=3x^2-4x$ , we have to find all values of $x$ with

3x^2-4x=0

Use the midnight formula to solve the equation or alternatively, observe that I can write

x(3x-4)=0

from which follows that $x_1=0$ and $x_2=\frac{4}{3}$ . With $f(0)=0$ and $f(\frac{4}{3})=-\frac{32}{27}$ , we get the stationary points $\underline{P_1(0|0)}$ and $\underline{P_2(\frac{4}{3} | -\frac{32}{27})}$ . From the drawing (not shown) follows that $P_1$ is a local maximum and $P_2$ is a local minimum.

Can we somehow determine the type of stationary point (maximum, minimum, saddle point)? This might be convenient if we cannot draw the graph, or if there are so many stationary points that it is hard to identify and classify all of them (e.g. plot the graph of the function $f(x)=\sin(1/x)$ ).

Fortunately, there are some simple criteria to classify stationary points.

Theorem 2

Assume that $P(x|y)$ is a point on the graph of the function $f$ . Then we have the following is true:

Equation 2

\begin{array}{lll} f^{\prime}(x)=0 \text{ and}&&\\ \quad\dots\, f^{\prime\prime}(x)<0 & \rightarrow & \text{local max at $x$}\\ \quad\dots\, f^{\prime\prime}(x)>0 & \rightarrow & \text{local min at $x$}\\ \quad\dots\, f^{\prime\prime}(x)=0 \text{ and } f^{\prime\prime\prime}(x)\neq 0 & \rightarrow & \text{saddle pt at $x$} \\ \quad\dots\, f^{\prime\prime}(x)=0 \text{ and } f^{\prime\prime\prime}(x)= 0 & \rightarrow & \text{draw $f$ to decide}\end{array}

Classifying stationary points.

Proof

In order to see this, let's first have a closer look at the "hills" and "valleys" of a graph $f$ , and how their derivatives look like (see below). You can see that in the case of a valley, the graph of $f'$ has a positive slope at the deepest point $x$ , that is, $f^{\prime\prime}(x)>0$ , and in the case of a hill, the graph of $f'$ has a negative slope at the highest point $x$ , that is, $f^{\prime\prime}(x)<0$ .

Looking at a saddle and inspecting the derivative, we see that $f'$ has a horizontal tangent at the saddle point $x$ , and thus $f^{\prime\prime}(x)=0$ , but clearly $f^{\prime\prime\prime}(x)\neq 0$ .

Finally, it $f^{\prime\prime}(x)=f^{\prime\prime\prime}(x)=0$ if could either be a flat hill, flat valley, or flat saddle point (see figure below). In this case we have to draw the graph to decide if $P$ is a local maximum, local minimum, or saddle point.

Exercise 1

Determine by calculation the stationary points, and classify them using the criteria given above. Draw the graph with the calculator to verify your results.
1. $f(x)=x^2$
2. $g(x)=x^3$
3. $h(x)=5-x^4$
4. $k(x)=x^4-\frac{1}{2}x^2$
Consider the function $f(x)=x^3-u\cdot x^2$ , where $u$ is a fixed, but unknown number (a so called parameter).
1. Show that $f$ has a stationary point at $x=0$ for all values of $u$ .
2. For what values of $u$ does $f$ have a local maximum, a local minimum, or a saddle point at $x=0$ ?

Solution

It is
1. Find stationary point $P$ : $f^\prime(x)=2x=0 \rightarrow x=0$ , thus $\underline{P(0|0)}$ . Type? Because $f^{\prime\prime}(x)=2 \rightarrow f^{\prime\prime}(0)=2>0 \rightarrow$ local minimum.
2. Find stationary point $P$ : $g^\prime(x)=3x^2=0 \rightarrow x=0$ , thus $\underline{P(0|0)}$ .Because $g^{\prime\prime}(x)=6x \rightarrow g^{\prime\prime}(0)=0$ and $g^{\prime\prime\prime}(x)=6\neq 0$ it is a saddle point.
3. Find stationary point $P$ : $h^\prime(x)=-4x^3=0 \rightarrow x=\underline{0}$ , thus $\underline{P(0|5)}$ .Type? Because $h^{\prime\prime}(x)=-12x^2 \rightarrow h^{\prime\prime}(0)=0$ and $h^{\prime\prime\prime}(x)=-24x \rightarrow h^{\prime\prime\prime}(0)=0$ . Draw graph $\rightarrow$ "flat" local maximum.
4. Find stationary point $P$ : $k^\prime(x)=4x^3-x=0 \rightarrow x(4x^2-1)=0 \rightarrow x_1=\underline{0}, x_2=-0.5, x_3=0.5$ , thus $\underline{P_1(0|0)}, \underline{P_2(-0.5|-0.0625)}, \underline{P_3(0.5|-0.0625)}$ .Type? Because $k^{\prime\prime}(x)=12x^2-1 \rightarrow k^{\prime\prime}(0)=-1<0 \rightarrow$ local maximum, $k^{\prime\prime}(-0.5)=2>0 \rightarrow$ local minimum, $k^{\prime\prime}(0.5)=2>0 \rightarrow$ local minimum.
$f^\prime(x)=3x^2-2ux$ , $f^{\prime\prime}(x)=6x-2u$ , $f^{\prime\prime\prime}(x)=6$
1. For every fixed value of $u$ we have $f^\prime(0)=3\cdot 0^2-2u\cdot 0=0 \rightarrow x=0 \rightarrow$ is stationary point for all values $u$ .
2. $f^{\prime\prime}(0)=6\cdot 0-2u=-2u<0 \rightarrow$ at $x=0$ there is a local maximum for $\underline{u>0}$ , and a local minimum for $\underline{u<0}$ . For $\underline{u=0}$ it is $f^{\prime\prime}(0)=0$ and $f^{\prime\prime\prime}(0)=6\neq 0$ , so at $x=0$ there is a saddle point.