Extrema

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maximums and minimums

Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function looks like. We can obtain a good picture of the graph using certain crucial information provided by derivatives of the function.

Extrema

A function has a value at each domain number. The set of all function values might have a greatest (maximum) or least(minimum) value. Together these are know as extreme values or extrema. These correspond to highest and lowest points on the graph.

A local extremum of a function $f$ is a number, a function value, $f(a)$. It occurs at a domain number, $a$. These correspond to a point $(a,f(a))$ on the graph of $f$ where the $vertical$-coordinate is larger (or smaller) than all other $vertical$-coordinates of points on the graph whose $horizontal$-coordinates are “close to” $a$.

Local Extrema

(a): A function $f(x)$ has a local maximum at $x=a$, if $f(a)\ge f(x)$ for every domain number, $x$, in some open interval I containing $a$.
(b): A function $f(x)$ has a local minimum at $x=a$, if $f(a)\le f(x)$ for every domain number, $x$, in some open interval I containing $a$.

A local extremum is either a local maximum or a local minimum.

Local extremum are numbers and they correspond to highest and lowest points on the graph.

In our next example, we clarify the definition of a local minimum.

Consider the graph of a function $f$:

Identify the local extrema of $f$ and give an explanation.

explanation

In the figure above the function $f$ has a local minimum at,

\[ a=\answer [given]{1}, \]

because we can find an open interval $I$ (marked in the figure) that contains the number

\[ a=\answer [given]{1}, \]

and for all domain numbers $x$ in $I$ the following statement is true:

\[ f\left (\answer [given]{1}\right )\le f\left (\answer [given]{x}\right ). \]

Local maximum and minimum values of a function correspond to quite distinctive points on the graph of a function, and are, therefore, useful in understanding the shape of the graph. Many problems in real world and in different scientific fields turn out to be about finding the smallest (or largest) value that a function achieves (for example, we might want to find the minimum cost at which some task can be performed).

Critical Numbers

Consider the graph of the function $f$.

The function $f$ has five local extrema: $5$, $6$, $-1$, and $4.5$ These occur at $x=-4$, $x=-2$, $x=0$, and $x=4$.

Differentiable

Let $f$ be a function and $a$ a member of its domain.

We say that $f$ is differentiable at $a$ if there is a tangent line to graph at $(a, f(a))$ and this tangent line has a slope.

i.e. differentiable means has a derivative value.

Can a tangent line not have a slope? Yes. We’ll see an example later.

Notice that the function $f$ above is not differentiable at $x=-4$ and $x=-2$.

$f$ is not differentiable at $x=-4$, because the graph doesn’t have a tangent line at $(-4, 5)$.

How do we know this?

If there was a tangent line, then it would need to be close to the graph on the left side of $(-4, 5)$. Well, the left side is a line and a line is its own tangent line. That tells us that the tangent line must be $y = x + 9$. Unfortunately, the same reasoning on the right side tells us that the tangent line must be $y = -x +1$.

There is no line that does both of those. There is no tangent line.

$f$ is not differentiable at $x=-2$, because the graph doesn’t have a tangent line at $(-2, 6)$.

How do we know this?

Well, the tangent line must go through the tangent point, which is $(-2, 6)$. On the right side of $(-2, 6)$ the tangent line would need to have a tremendously large negative slope to immediate get down near $1$. But, no line like this could get down near $3$ on the left sife of $(-2, 6)$.

There is no tangent line.

This example function illustrates that extreme function values can occur where the derivative does not exist.

Extreme function values can occur at other places as well.

Notice that $f'(0)=0$ and $f'(4)=0$.

After this example, the following theorem should not come as a surprise.

Fermat’s Theorem If $f$ has a local extremum at $x=a$ and $f$ is differentiable at $a$, then $f'(a)=0$.

Does Fermat’s Theorem say that if $f'(a) = 0$, then $f$ has a local extrema at $x=a$?

yes no

Consider $f(x) = x^3$, $f'(0) = 0$, but $f$ does not have a local maximum or minimum at $x=0$.

Following the example above and Fermat’s Theorem, it is believable that the only numbers at which a function can have a local maximum or minimum are numbers at which the derivative is zero or the derivative is undefined. As an illustration of the first scenario, consider the plots of $f(x) = x^3-4.5x^2+6x$ and $f'(x) = 3x^2-9x+6$.

Make a correct choice that completes the sentence below.

At the number $1$, the function $f$ has

a local maximum a local minimum no local extremum

Select the correct statement.

$f'(1)$ is undefined $f'(1)>0$ $f'(1)=0$ $f'(1)<0$

Make a correct choice that completes the sentence below.

At the number $1.5$, the function $f$ has

a local maximum a local minimum no local extremum

Select the correct statement.

$f'(1.5)$ is undefined $f'(1.5)>0$ $f'(1.5)=0$ $f'(1.5)<0$

Make a correct choice that completes the sentence below.

At the number $2$, the function $f$ has

a local maximum a local minimum no local extremum

Select the correct statement.

$f'(2)$ is undefined $f'(2)>0$ $f'(2)=0$ $f'(2)<0$

As an illustration of the second scenario, consider the plots of $f(x) = x^{2/3}$ and $f'(x) = \frac {2}{3x^{1/3}}$:

Make a correct choice that completes the sentence below.

At the number $-2$, the function $f$ has

a local maximum a local minimum no local extremum

Select the correct statement.

$f'(-2)$ is undefined $f'(-2)>0$ $f'(-2)=0$ $f'(-2)<0$

Make a correct choice that completes the sentence below.

At the number $0$, the function $f$ has

a local maximum a local minimum no local extremum

Select the correct statement.

$f'(0)$ is undefined $f'(0)>0$ $f'(0)=0$ $f'(0)<0$

This brings us to our next definition.

(from earlier) Critical Number

Let $f$ be a function defined on an open interval, I, that contains a number $a$. Then we say that the function $f$ has a critical number at $x=a$ if

\[ f'(a) = 0\qquad \text {or}\qquad \text {$f'(a)$ does not exist.} \]

Notice, if a function $f$ has a critical number at $x=a$, then the number $a$ is inside some open interval $I$, and $I$ is in the domain of $f$.

When looking for local maximum and minimum numbers, be careful not to make two common mistakes:

You may forget that a maximum or minimum can occur where the derivative does not exist, and only check where the derivative is $0$.
You might assume that any place that the derivative is zero is a local maximum or minimum number, but this is not true. Consider the plots of $f(x) = x^3$ and $f'(x) = 3x^2$.

While $f'(0)=0$, there is neither a maximum nor minimum of $f$ at $x=0$.

Since both local maximum and local minimum occur at a critical number, when we locate a critical number, we need to determine which, if either, actually occurs.

Critical numbers are candidates for locations of extrema.

Find all local maximum and minimum values for the function $f(x)=x^3-x$.

explanation

The derivative of $f$ is given by

\[ 3x^2 - 1 \]

We can easily express $f'(x)$ as a product of its factors

\[ \frac {d}{dx} f(x)=3\left (x+\answer [given]{\frac {\sqrt {3}}{3}}\right )\left (x-\answer [given]{\frac {\sqrt {3}}{3}}\right ), \]

which implies that the function $f$ has only two critical numbers, $x=-\frac {\sqrt {3}}{3}$ and $x=\frac {\sqrt {3}}{3}$. Notice that the derivative $f'(x)$ is a polynomial, and polynomials do not change signs except possibly at their zeros. This implies that the derivative $f'(x)$ does not change the sign on the intervals $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$, $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$, and $\left (\frac {\sqrt {3}}{3},\infty \right )$, because these intervals do not contain any zeros of $f'(x)$.

Select the correct statement about the sign of $f'(x)$ on the intervals $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$.

$f'(x)>0$ on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. $f'(x)>0$ on $x$ in $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. $f'(x)<0$ on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. $f'(x)<0$ on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$

If we know the sign of the derivative on an interval, we also know whether the function is increasing or decreasing on that interval. This will help us determine whether the function has a local extremum at the critical number where $x=-\frac {\sqrt {3}}{3}$.

At the critical number where $x=-\frac {\sqrt {3}}{3}$, the function $f$ has

no local extremum, because $f$ is increasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. a local maximum, because $f$ is increasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. a local minimum, because $f$ is decreasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$. no local extremum, because $f$ is decreasing on $\left (-\infty ,-\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$.

Select the correct statement about the sign of $f'(x)$ on the intervals $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $\left (\frac {\sqrt {3}}{3},\infty \right )$.

$f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$. $f'(x)>0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$. $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)>0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$. $f'(x)<0$ on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and $f'(x)<0$ on $\left (\frac {\sqrt {3}}{3},\infty \right )$.

Again, the sign of the derivative on an interval determines whether the function is increasing or decreasing on that interval. This will help us determine whether the function has a local extremum at the critical number where $x=\frac {\sqrt {3}}{3}$.

At the critical number where $x=\frac {\sqrt {3}}{3}$, the function $f$ has

no local extremum, because $f$ is increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$. a local maximum, because $f$ is increasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$. a local minimum, because $f$ is decreasing on $\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and increasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$. no local extremum, because $f$ is decreasing on$\left (-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3}\right )$ and decreasing on $\left (\frac {\sqrt {3}}{3},\infty \right )$.

Do your answers agree with the graphs of $f$ and $f'$ given in the picture below?

Differentiable means the graph has a tangent line and this tangent line has a slope. (No. Not all tangent lines have slopes.)

Consider the function $f(x) = 6 (x-2)^{\frac {2}{3}}$.

The graph has a “cusp”. It isn’t a corner.

The two pieces of the graph do, in fact, become parallel as you approach the cusp. And, there is a tangent line. It is a vertical tangent line.

So, we have a tangent line.

However, the tangent line has no slope, because it is vertical.

Hence, differentiable means there is a tangent line and that tangent line has a slope, a number.

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more examples can be found by following this link
More Examples of Optimization

2026-01-29 19:22:58