## Tutorial: Maxima and minima

* Adaptive game version*

(This topic is also in Section 5.1 in *Applied Calculus*or Section 12.1 in

*Finite Mathematics and Applied Calculus*)

### Resources

Function evaluator and grapher | Excel grapher |

Relative and absolute extrema

#[Let us start with a quick review and quiz on using the "Calculation Thought Experiment (CTE)" discussed in the %%prevsectut:][Empecemos con un repaso y concurso rápido sobre el uso del "Experimento mental de cálculo (EMC)" descrito en el %%prevsectut:]#
**Relative extrema: Definition**$f$ has a

**relative maximum**at $x_0$ if there is some interval $(x_0 - h, x_0 + h)$ (even a very small one) for which $f(x_0) \geq f(x)$ for all $x$ in $(x_0 - h, x_0 + h)$ for which $f(x)$ is defined. Similarly, $f$ has a

**relative minimum**at $x_0$ if there is some interval $(x_0 - h, x_0 + h)$ (even a very small one) for which $f(x_0) \leq f(x)$ for all $x$ in $(x_0 - h, x_0 + h)$ for which $f(x)$ is defined. Collectively, maxima and minima are referred to as

**extrema.**If $f$ has a relative extremum at $x_0,$ then the corresponding point $(x_0, f(x_0))$ on the graph of $f$ is also referred to as a relative maximum or relative minimum as the case may be.

**Visualizing relative maximima and minima**

**One for you**If a relative maximum happens to be the highest point on the entire graph of $f$, we refer to it as an

**absolute maximum.**Similarly, if a relative minimum happens to be the lowest point on the entire graph of $f$, we refer to it as an

**absolute minimum.**

**Absolute extrema: Definition**$f$ has an

**absolute maximum**at $x_0$ if $f(x_0) \geq f(x)$ for every $x$ in the domain of $f$. Similarly, $f$ has an

**absolute minimum**at $x_0$ if $f(x_0) \leq f(x)$ for every $x$ in the domain of $f$. The following figure shows several relative and absolute extrema:

**One for you**

Locating maxima and minima

Sometimes it is not a simple matter to tell from the graph Exactly where the local extrema are situated. For instance, try graphing the curve $y = x^3(x^{1/2}-1)$ with $0 \leq x \leq 2$, and see if you can tell exactly where the absolute minimum lies. (This example is discussed in the on-line review exercises).
To help us locate extrema accurately, we first classify them into three types, and use calculus to tell us exactly where they are:
**Types of extrema and how to find them**If $f$ is continuous on its domain and differentiable except at a few isolated points, then its extrema occur among the following types of points:

**Stationary points:**Points $x$ in the interior of the domain where $f'(x) = 0$. To locate stationary points, set $f'(x) = 0$ and solve for $x$.**Singular points:**Points $x$ in the interior of the domain where $f'(x)$ is not defined. To locate singular points, find values $x$ where $f'(x)$*is not*defined, but $f(x)$*is*defined.**Endpoints:**The endpoints of the domain, if any. Recall that closed intervals contain endpoints, but open intervals do not.

**critical points.**(So, we consider two kinds of critical: stationary and singular.) The following figure shows several instances of all three types. The following figure shows several relative and absolute extrema:

**Examples**

**Stationary points:**$f(x) = 2x^{3} - 24x$ has $f'(x)=6x^2-24$. To locate stationary points, set $f'(x) = 0$ and solve for $x$:

$6x^2-24 = 0 \implies x^2 = 4$ \t ${}\implies x = \pm 2^{\ }$

The corresponding points on the graph are
$(-2, f(-2)) = (-2, 32)\ $ \t and $\ (2, f(2)) = (2, -32)$.

**One for you**

**Singular points:**$f(x) = 5(x-2)^{3/5}$ has $f'(x)=\dfrac{15}{(x-2)^{2/5}}$. The only value of $x$ for which $f'(x)$ is not defined is $x = 2$. As $f'(2)$ is not defined but $f(2)$ is defined, there is a singular point at $x = 2$. The corresponding point on the graph is

$(2, f(2)) = (2, 0)$.

**One for you**

First derivative test

The first derivative test gives a systematic way of checking whether a critical point is a maximum, minimum, or neither without having to use a plot of the graph, or to plot sufficiently many points to be able to tell. It is based on the fact that the sign of the derivative tells you the direction of the curve (see the table in the exaple below).
**First derivative test**Assume that $c$ is a critical point of $f$ so that $f$ is continuous near $c$ and the derivative is defined close to and on both sides of $c$. Calculate the sign of the derivative to the left and right of the point $c$.

- If $f'$ is positive to the left of $c$ and negative to the right, then $f$ has a maximum at $c$
- If $f'$ is negative to the left of $c$ and positive to the right, then $f$ has a minimum at $c$
- If $f'$ has the same sign on both sides of $c$, then $f$ has a neither a maximum nor minimum at $c$

**Example**

We saw above that $f(x) = 2x^{3} - 24x$ has stationary points at $x = -2$ and $x = 2$. The following table shows the values of $f'$ at points on either side of the stationary points.

Extreme value theorem

When the domain of a continuous function is a *closed interval*, then the

**Extreme value theorem**tells us that there is always an absolute maximum and also an absolute minimum. Notice that all the examples we have seen of continuous functions were there is no absolute maximum or no absolute minimum had a domain something other than a closed interval.

**Extreme value theorem**If $f$ is continuous on a closed interval $[a, b],$ then it will have an absolute maximum and an absolute minimum value on that interval. Each absolute extremum must occur at either an endpoint or a critical point. Becuase of the theorem, if we make a list of the endpoints of the domain and critical points, any point that gives the largest value of $f$ on the list is an absolute maximum, while any point giving the least value of $f$ on the list is an absolute minimum.

**#[Note][Nota]#**The theorem says nothing about functions whose domain is not a closed interval; for instance, $f(x) = x$ on the interval $[0, 1)$ has an absolute minimum at $0$ but no absolute maximum (it

*wants*to have an absolute maximum at $1$ but $x=1$ is not in the domain!

**Example**

Let's look once again at $f(x) = 2x^{3} - 24x$, but this time with domain $[-3,4]$. By the theorem all we need is a table of values of $f$ at the endpoints and critical points:

Now try the exercises in Section 5.1 in

*Applied Calculus*or Section 12.1 in*Finite Mathematics and Applied Calculus*.*July 2021*

Copyright © 2019 Stefan Waner and Steven R. Costenoble