yes no
Consider , , but does not have a local maximum or minimum at .
We use derivatives to help locate extrema.
Extrema
We’ll start with some definitions.
- (a)
- A function has an global maximum at , if for every in the domain of the function.
- (b)
- A function has an global minimum at , if for every in the domain of the function.
A global extremum is either a global maximum or a global minimum.
Local extrema on a function are points on the graph where the -coordinate is larger (or smaller) than all other -coordinates on the graph at points “close to” .
- (a)
- A function has a local maximum at , if for every in some open interval I containing .
- (b)
- A function has a local minimum at , if for every in some open interval I containing .
A local extremum is either a local maximum or a local minimum.
In our next example, we clarify the definition of a local minimum.
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 .
After this example, the following theorem should not come as a surprise.
Does Fermat’s Theorem say that if , then has a local extrema at ?
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 and .
Make a correct choice that completes the sentence below.
At the point , the function has
a local maximum a local minimum no local extremum
Make a correct choice that completes the sentence below.
At the point , the function has
a local maximum a local minimum no local extremum
Make a correct choice that completes the sentence below.
As an illustration of the second scenario, consider the plots of and : At the point , the function has
a local maximum a local minimum no local extremum
Make a correct choice that completes the sentence below.
At the point , the function has
a local maximum a local minimum no local extremum
Make a correct choice that completes the sentence below.
At the point , the function has
a local maximum a local minimum no local extremum
This brings us to our next definition.
Assume that a function is defined on an open interval I that contains a point . The
function has a critical point at if
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 and . While , there is neither a maximum nor minimum at .
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 .
Write
We can easily express as a product of its factors which implies that the
function has only two critical points, and . Notice that the derivative
is a polynomial, and polynomials do not change signs except possibly at
their zeros. This implies that derivative does not change the sign on
the intervals , , and , because these intervals do not contain any zeros of
.
Do your answers agree with the graphs of and given in the picture below?
Select the correct statement about the sign of on the intervals and .
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 .
on and on . on in and on . on and on . on and on
At the critical point where , the function has
no local extremum, because is increasing on and increasing on . a local
maximum, because is increasing on and decreasing on . a local minimum,
because is decreasing on and increasing on . no local extremum, because is
decreasing on and decreasing on .
Select the correct statement about the sign of on the intervals and .
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 . on and on . on and on . on and on . on and on .
At the critical point where , the function has
no local extremum, because is increasing on and increasing on . a local
maximum, because is increasing on and decreasing on . a local minimum,
because is decreasing on and increasing on . no local extremum, because is
decreasing on and decreasing on .
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 on an interval, then is increasing on that interval.
- If on an interval, then 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 is continuous on an interval, and that for some
value of in that interval.
- If to the left of and to the right of , then is a local maximum.
- If to the left of and to the right of , then is a local minimum.
- If has the same sign to the left and right of , then is not a local extremum.
Consider the function Find the intervals on which is increasing and decreasing and
identify the local extrema of .
Start by computing Now we need to find when this function is positive and when it
is negative. To do this, solve Factor
So the critical points (when ) are when , , and . Now we can check points between the critical points to find when is increasing and decreasing:
From this we can make a sign-chart:
Hence is increasing on and and is decreasing on and . Moreover, from the first derivative test, the local maximum is at while the local minima are at and , see the graphs of and .
Hence we have seen that if is zero at a point and increasing on an interval containing that point, then has a local minimum at the point. If is zero at a point and decreasing on an interval containing that point, then has a local maximum at the point. Thus, we see that we can gain information about by studying how 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 exists on an interval.
- (a)
- If on an interval, then is concave up on that interval.
- (b)
- If on an interval, then 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 is continuous at and its concavity changes either from up to down or down to up
at , then has an inflection point at .
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 such that is zero or undefined and then checking to see whether does in fact go from positive to negative or negative to positive at these points.
Describe the concavity of .
To start, compute the first and second derivative of with respect to , Since , there is
potentially an inflection point at . Using test points, we note the concavity does
change from down to up, hence there is an inflection point at . The curve
is concave down for all and concave up for all , see the graphs of and .
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 to the left of and to the right of , then is a local maximum.
- If to the left of and to the right of , then is a local minimum.
If changes from positive to negative it is decreasing. In this case, might be negative, and if in fact is negative then is definitely decreasing, so there is a local maximum at the point in question. On the other hand, if changes from negative to positive it is increasing. Again, this means that might be positive, and if in fact is positive then 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 is continuous on an open interval and that for
some value of in that interval.
- If , then has a local maximum at .
- If , then has a local minimum at .
- If , then the test is inconclusive. In this case, may or may not have a local extremum at .
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.