Maximums and minimums

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We use derivatives to help locate extrema.

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

We’ll start with some definitions.

(a): A function $f$ has an global maximum at $x=a$ , if $f(a)\ge f(x)$ for every $x$ in the domain of the function.
(b): A function $f$ has an global minimum at $x=a$ , if $f(a)\le f(x)$ for every $x$ in the domain of the function.

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

A local extremum of a function $f$ is a point $(a,f(a))$ on the graph of $f$ where the $y$ -coordinate is larger (or smaller) than all other $y$ -coordinates of points on the graph whose $x$ -coordinates are “close to” $a$ .

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

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

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.

In the figure above the function $f$ has a local minimum at, $a=\answer [given]{1},$ because were able to find an open interval $I$ (marked in the figure) that contains the number $a=\answer [given]{1},$ and for all 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 points are quite distinctive on the graph of a function, and are therefore useful in understanding the shape of the graph. In many applied problems we want to find the largest or smallest value that a function achieves (for example, we might want to find the minimum cost at which some task can be performed) and so identifying maximum and minimum points will be useful for applied problems as well.

Critical points

Consider the graph of the function $f$ .

The function $f$ has four local extremums: at $x=-4$ , $x=-2$ , $x=0$ and $x=4$ . Notice that the function $f$ is not differentiable at $x=-4$ and $x=-2$ . 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 $x=0$ .

Fermat’s Theorem says that the only points at which a function can have a local maximum or minimum are points 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 point $(1,f(1))$ , 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 point $(1.5,f(1.5))$ , 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 point $(2,f(2))$ , 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 point $(-2,f(-2))$ , 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 point $(0,0)$ , 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.

Assume that a function $f$ is defined on an open interval I that contains a point $a$ . The function $f$ has a critical point at $x=a$ if $f'(a) = 0\qquad \text {or}\qquad \text {$f'(a)$ does not exist.}$

When looking for local maximum and minimum points, you are likely to make two sorts of mistakes:

You may forget that a maximum or minimum can occur where the derivative does not exist, and so forget to check whether the derivative exists everywhere.
You might assume that any place that the derivative is zero is a local maximum or minimum point, 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 at $(0,f(0))$ .

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

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

Write $\ddx f(x)=\answer [given]{3x^2-1}.$ We can easily express $f'(x)$ as a product of its factors $\ddx f(x)=3(x+\answer [given]{\frac {\sqrt {3}}{3}})(x-\answer [given]{\frac {\sqrt {3}}{3}}),$ which implies that the function $f$ has only two critical points, $x=-\frac {\sqrt {3}}{3}$ 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 derivative $f'(x)$ does not change the sign on the intervals $(-\infty ,-\frac {\sqrt {3}}{3})$ , $(-\frac {\sqrt {3}}{3},\frac {\sqrt {3}}{3})$ , and $(\frac {\sqrt {3}}{3},\infty )$ , because these intervals do not contain any zeros of $f'(x)$ .

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

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

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 point where $x=-\frac {\sqrt {3}}{3}$ .

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

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

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

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

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 point where $x=\frac {\sqrt {3}}{3}$ .

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

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

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

The first derivative test

We will further explore and refine the method of the previous section for deciding whether there is a local maximum or minimum at a critical point. Recall that

If $f'(x) >0$ on an interval, then $f$ is increasing on that interval.
If $f'(x) <0$ on an interval, then $f$ is decreasing on that interval.

So how exactly does the derivative tell us whether there is a maximum, minimum, or neither at a point? Use the first derivative test.

First Derivative Test Suppose that $f$ is continuous on an interval, and that $f'(a)=0$ for some value of $a$ in that interval.

If $f'(x)>0$ to the left of $a$ and $f'(x)<0$ to the right of $a$ , then $f(a)$ is a local maximum.
If $f'(x)<0$ to the left of $a$ and $f'(x)>0$ to the right of $a$ , then $f(a)$ is a local minimum.
If $f'(x)$ has the same sign to the left and right of $a$ , then $f(a)$ is not a local extremum.

Consider the function $f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$ Find the intervals on which $f$ is increasing and decreasing and identify the local extrema of $f$ .

Start by computing $\ddx f(x) = \answer [given]{x^3+x^2-2x}.$ Now we need to find when this function is positive and when it is negative. To do this, solve $f'(x) = \answer [given]{x^3+x^2-2x} =0.$ Factor $f'(x)$ $\begin{align*} f'(x) &= \answer[given]{x^3+x^2-2x} \\ &=x(\answer[given]{x^2+x-2})\\ &=x(x+2)\answer[given]{(x-1)}. \end{align*}$

So the critical points (when $f'(x)=0$ ) are when $x=-2$ , $x=0$ , and $x=1$ . Now we can check points between the critical points to find when $f'(x)$ is increasing and decreasing:

$\begin{align*} f'(-3)&=\answer[given]{-12},\\ f'(.5)&=\answer[given]{-0.625},\\ f'(-1)&=\answer[given]{2},\\ f'(2)&=\answer[given]{8}. \end{align*}$

From this we can make a sign-chart:

Hence $f$ is increasing on $(-2,0)$ and $(1,\infty )$ and $f$ is decreasing on $(-\infty ,-2)$ and $(0,1)$ . Moreover, from the first derivative test, the local maximum is at $x=0$ while the local minima are at $x=-2$ and $x=1$ , see the graphs of $f(x) =x^4/4 + x^3/3 -x^2$ and $f'(x) = x^3 + x^2 -2x$ .

Hence we have seen that if $f'$ is zero at a point and increasing on an interval containing that point, then $f$ has a local minimum at the point. If $f'$ is zero at a point and decreasing on an interval containing that point, then $f$ has a local maximum at the point. Thus, we see that we can gain information about $f$ by studying how $f'$ changes. This leads us to our next section.

Inflection points

If we are trying to understand the shape of the graph of a function, knowing where it is concave up and concave down helps us to get a more accurate picture. It is worth summarizing what we have seen already in to a single theorem.

Test for Concavity Suppose that $f''(x)$ exists on an interval.

(a): If $f''(x)>0$ on an interval, then $f$ is concave up on that interval.
(b): If $f''(x)<0$ on an interval, then $f$ is concave down on that interval.

Of particular interest are points at which the concavity changes from up to down or down to up.

If $f$ is continuous at $x=a$ and its concavity changes either from up to down or down to up at $x=a$ , then $f$ has an inflection point at $x=a$ .

It is instructive to see some examples of inflection points:

It is also instructive to see some nonexamples of inflection points:

We identify inflection points by first finding $x$ such that $f''(x)$ is zero or undefined and then checking to see whether $f''(x)$ does in fact go from positive to negative or negative to positive at these points.

Even if $f''(a) = 0$ , the point determined by $x=a$ might not be an inflection point.

Describe the concavity of $f(x)=x^3-x$ .

To start, compute the first and second derivative of $f(x)$ with respect to $x$ , $f'(x)=\answer [given]{3x^2-1}\qquad \text {and}\qquad f''(x)=\answer [given]{6x}.$ Since $f''(0)=0$ , there is potentially an inflection point at $x=0$ . Using test points, we note the concavity does change from down to up, hence there is an inflection point at $x=0$ . The curve is concave down for all $x<0$ and concave up for all $x>0$ , see the graphs of $f(x) = x^3-x$ and $f''(x) = 6x$ .

Note that we need to compute and analyze the second derivative to understand concavity, so we may as well try to use the second derivative test for maxima and minima. If for some reason this fails we can then try one of the other tests.

The second derivative test

Recall the first derivative test:

If $f'(x)>0$ to the left of $a$ and $f'(x)<0$ to the right of $a$ , then $f(a)$ is a local maximum.
If $f'(x)<0$ to the left of $a$ and $f'(x)>0$ to the right of $a$ , then $f(a)$ is a local minimum.

If $f'$ changes from positive to negative it is decreasing. In this case, $f''$ might be negative, and if in fact $f''$ is negative then $f'$ is definitely decreasing, so there is a local maximum at the point in question. On the other hand, if $f'$ changes from negative to positive it is increasing. Again, this means that $f''$ might be positive, and if in fact $f''$ is positive then $f'$ is definitely increasing, so there is a local minimum at the point in question. We summarize this as the second derivative test.

Second Derivative Test Suppose that $f''(x)$ is continuous on an open interval and that $f'(a)=0$ for some value of $a$ in that interval.

If $f''(a) <0$ , then $f$ has a local maximum at $a$ .
If $f''(a) >0$ , then $f$ has a local minimum at $a$ .
If $f''(a) =0$ , then the test is inconclusive. In this case, $f$ may or may not have a local extremum at $x=a$ .

The second derivative test is often the easiest way to identify local maximum and minimum points. Sometimes the test fails and sometimes the second derivative is quite difficult to evaluate. In such cases we must fall back on one of the previous tests.

Once again, consider the function $f(x) = \frac {x^4}{4}+\frac {x^3}{3}-x^2$ Use the second derivative test, to locate the local extrema of $f$ .

Start by computing $f'(x) = \answer [given]{x^3 + x^2 -2x} \qquad \text {and}\qquad f''(x) = \answer [given]{3x^2 + 2x-2}.$ Using the same technique as we used before, we find that $f'(-2) = \answer [given]{0},\qquad f'(0) = \answer [given]{0}, \qquad f'(1) = \answer [given]{0}.$ Now we’ll attempt to use the second derivative test, $f''(-2) = \answer [given]{6}, \qquad f''(0) =\answer [given]{ -2}, \qquad f''(1) = \answer [given]{3}.$ Hence we see that $f$ has a local minimum at $x=-2$ , a local maximum at $x=0$ , and a local minimum at $x=1$ , see below for a plot of $f(x) =x^4/4 + x^3/3 -x^2$ and $f''(x) = 3x^2 + 2x -2$ :

If $f''(a)=0$ , what does the second derivative test tell us?

The function has a local extrema at $x=a$ . The function does not have a local extrema at $x=a$ . It gives no information on whether $x=a$ is a local extremum.