Which Interval for the Graphed Function Contains the Local Maximum

August 30, 2022 admin 120 Views

Which Interval for the Graphed Function Contains the Local Maximum

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Section 4-3 : Minimum and Maximum Values

Many of our applications in this chapter will revolve around minimum and maximum values of a function. While nosotros can all visualize the minimum and maximum values of a function we want to be a fiddling more specific in our work hither. In particular, we want to differentiate between two types of minimum or maximum values. The following definition gives the types of minimums and/or maximums values that nosotros’ll be looking at.

Definition

We say that $f\left( ten \right)$ has an
accented (or global) maximum
at $x = c$ if$f\left( x \correct) \le f\left( c \right)$ for every $10$ in the domain we are working on.
We say that $f\left( x \right)$ has a
relative (or local) maximum
at $x = c$ if $f\left( ten \right) \le f\left( c \right)$ for every $x$ in some open interval around $ten = c$.
We say that $f\left( x \right)$ has an
absolute (or global) minimum
at $10 = c$ if $f\left( x \right) \ge f\left( c \right)$ for every $ten$ in the domain nosotros are working on.
We say that $f\left( 10 \right)$ has a
relative (or local) minimum
at $x = c$ if$f\left( 10 \right) \ge f\left( c \right)$ for every $x$ in some open interval effectually $x = c$.

Note that when we say an “open interval around$x = c$” we mean that we can notice some interval $\left( {a,b} \right)$, non including the endpoints, such that $a < c < b$. Or, in other words, $c$ volition be contained somewhere inside the interval and volition non be either of the endpoints.

Too, nosotros will collectively call the minimum and maximum points of a function the
extrema
of the function. And then, relative extrema will refer to the relative minimums and maximums while accented extrema refer to the absolute minimums and maximums.

Now, let’southward talk a little bit nigh the subtle difference betwixt the absolute and relative in the definition to a higher place.

We will have an absolute maximum (or minimum) at $10 = c$ provided $f\left( c \correct)$ is the largest (or smallest) value that the function volition ever take on the domain that we are working on. Too, when we say the “domain we are working on” this just means the range of $x$’southward that we take called to work with for a given problem. There may be other values of $ten$ that we can actually plug into the function but have excluded them for some reason.

A relative maximum or minimum is slightly different. All that’south required for a bespeak to be a relative maximum or minimum is for that point to be a maximum or minimum in some interval of $x$’south around $10 = c$. There may exist larger or smaller values of the office at another place, but relative to $ten = c$, or local to $x = c$, $f\left( c \right)$ is larger or smaller than all the other function values that are near it.

Note besides that in order for a betoken to exist a relative extrema nosotros must be able to await at role values on both sides of $x = c$ to see if it really is a maximum or minimum at that indicate. This means that relative extrema do not occur at the stop points of a domain. They can only occur interior to the domain.

There is actually some contend on the preceding betoken. Some folks do feel that relative extrema tin occur on the end points of a domain. Withal, in this class we will be using the definition that says that they can’t occur at the terminate points of a domain. This will be discussed in a picayune more particular at the end of the section once we have a relevant fact taken care of.

Information technology’s unremarkably easier to go a feel for the definitions by taking a quick look at a graph.

For the function shown in this graph we accept relative maximums at $x = b$ and $x = d$. Both of these points are relative maximums since they are interior to the domain shown and are the largest betoken on the graph in some interval effectually the bespeak. We likewise have a relative minimum at $10 = c$ since this betoken is interior to the domain and is the lowest point on the graph in an interval around it. The far-correct cease point, $x = due east$, volition non be a relative minimum since it is an cease point.

The role volition have an absolute maximum at $x = d$ and an absolute minimum at $x = a$. These ii points are the largest and smallest that the function volition ever exist. We tin also detect that the absolute extrema for a function will occur at either the endpoints of the domain or at relative extrema. Nosotros will utilise this thought in after sections so it’s more than important than it might seem at the present fourth dimension.

Permit’southward take a quick look at some examples to make sure that we have the definitions of accented extrema and relative extrema straight.

Case one
Place the absolute extrema and relative extrema for the post-obit function. \[f\left( x \right) = {x^two}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ { – 1,2} \right]\]

Show Solution

Since this function is easy plenty to graph let’s do that. However, we only want the graph on the interval $\left[ { – one,2} \correct]$. Here is the graph,

$Graph of $f\left( x \right)={{x}^{2}}$ on the domain $-1 \le x \le 2$ with dots at the (-1,1) and (2,4) to indicate that the graph does not extend past these two points.$

Annotation that we used dots at the terminate of the graph to remind us that the graph ends at these points.

We can now identify the extrema from the graph. It looks like we’ve got a relative and absolute minimum of zip at $x = 0$ and an accented maximum of four at $x = two$. Notation that $x = – ane$ is non a relative maximum since information technology is at the end point of the interval.

This function doesn’t have any relative maximums.

As we saw in the previous example functions do not have to take relative extrema. Information technology is completely possible for a function to not have a relative maximum and/or a relative minimum.

Case 2
Identify the absolute extrema and relative extrema for the post-obit function. \[f\left( x \right) = {x^2}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ { – 2,2} \right]\]

Testify Solution

Here is the graph for this function.

$Graph of $f\left( x \right)={{x}^{2}}$ on the domain $-2 \le x \le 2$ with dots at the (-2,4) and (2,4) to indicate that the graph does not extend past these two points.$

In this example we even so have a relative and accented minimum of nix at $ten = 0$. Nosotros too still have an accented maximum of four. However, different the offset case this will occur at 2 points, $10 = – 2$ and $10 = ii$.

Again, the function doesn’t have whatsoever relative maximums.

Every bit this example has shown in that location can only be a single absolute maximum or absolute minimum value, merely they can occur at more one place in the domain.

Example 3
Identify the absolute extrema and relative extrema for the following function. \[f\left( ten \correct) = {x^2}\]

Show Solution

In this instance we’ve given no domain and so the assumption is that nosotros will take the largest possible domain. For this role that means all the real numbers. Here is the graph.

$Graph of $f\left( x \right)={{x}^{2}}$ with no domain specified and arrows at the ends pointing upwards to indicate the graph continues.$

In this case the graph doesn’t finish increasing at either finish and so there are no maximums of any kind for this function. No thing which point we pick on the graph at that place will be points both larger and smaller than it on either side and then we can’t take whatsoever maximums (of any kind, relative or accented) in a graph.

We however have a relative and absolute minimum value of zilch at $10 = 0$.

And so, some graphs can have minimums but not maximums. As well, a graph could accept maximums but not minimums.

Example 4
Place the accented extrema and relative extrema for the following function. \[f\left( x \correct) = {x^three}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ { – 2,two} \right]\]

Show Solution

Here is the graph for this function.

$Graph of $f\left( x \right)={{x}^{3}}$ on the domain $-2 \le x \le 2$ with dots at the (-2,-8) and (2,8) to indicate that the graph does not extend past these two points.$

This function has an absolute maximum of eight at $10 = 2$ and an absolute minimum of negative eight at $ten = – ii$. This function has no relative extrema.

Then, a part doesn’t have to have relative extrema as this example has shown.

Case five
Identify the absolute extrema and relative extrema for the following function. \[f\left( x \right) = {x^3}\]

Show Solution

Again, we aren’t restricting the domain this fourth dimension so here’s the graph.

$Graph of $f\left( x \right)={{x}^{3}}$ with no domain specified and arrows at the ends to indicate the graph continues.$

In this instance the function has no relative extrema and no accented extrema.

As we’ve seen in the previous example functions don’t accept to have any kind of extrema, relative or absolute.

Example half-dozen
Identify the accented extrema and relative extrema for the following role. \[f\left( 10 \right) = \cos \left( x \right)\]

Show Solution

Nosotros’ve non restricted the domain for this function. Here is the graph.

$Graph of $f\left( x \right)=\cos \left( x \right)$ on the domain $\left[ -3\pi ,3\pi \right]$ and arrows at the ends to indicate the graph continues.$

Cosine has extrema (relative and absolute) that occur at many points. Cosine has both relative and absolute maximums of one at

\[10 = \ldots – 4\pi ,\, – 2\pi ,\,\,0,\,\,two\pi ,\,\,4\pi , \ldots \]

Cosine likewise has both relative and absolute minimums of -one at

\[10 = \ldots – three\pi ,\, – \pi ,\,\,\pi ,\,\,3\pi , \ldots \]

Every bit this instance has shown a graph tin in fact accept extrema occurring at a large number (space in this instance) of points.

We’ve now worked quite a few examples and we can use these examples to see a overnice fact about absolute extrema. First let’s find that all the functions above were continuous functions. Next notice that every time we restricted the domain to a airtight interval (i.e.
the interval contains its cease points) we got absolute maximums and absolute minimums. Finally, in only 1 of the three examples in which we did non restrict the domain did we get both an absolute maximum and an accented minimum.

These observations atomic number 82 us the following theorem.

Farthermost Value Theorem

Suppose that $f\left( x \correct)$ is continuous on the interval $\left[ {a,b} \right]$ and then there are two numbers $a \le c,d \le b$ and so that $f\left( c \correct)$ is an accented maximum for the part and $f\left( d \correct)$ is an absolute minimum for the part.

Then, if we have a continuous part on an interval $\left[ {a,b} \right]$ then we are guaranteed to have both an absolute maximum and an absolute minimum for the office somewhere in the interval. The theorem doesn’t tell u.s.a. where they will occur or if they will occur more than than once, but at least it tells us that they do exist somewhere. Sometimes, all that we need to know is that they practice be.

This theorem doesn’t say anything about absolute extrema if we aren’t working on an interval. We saw examples of functions above that had both absolute extrema, one accented extrema, and no absolute extrema when we didn’t restrict ourselves down to an interval.

The requirement that a role be continuous is likewise required in order for us to use the theorem. Consider the example of

\[f\left( 10 \right) = \frac{ane}{{{x^2}}}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}[ – 1,one]\]

Here’s the graph.

$Graph of $f\left( x \right)=\frac{1}{{{x}^{2}}}$ on the domain $-1 \le x \le 1$ with dots at the (-1,1) and (1,1) to indicate that the graph does not extend past these two points. There is also a gap as the graph approaches the y-axis to indicate that the graph will continue to move up the y-axis as it approaches.$

This function is not continuous at $x = 0$ as we motion in towards nil the function is approaching infinity. Then, the function does not have an absolute maximum. Annotation that it does have an accented minimum withal. In fact the absolute minimum occurs twice at both $ten = – ane$ and $x = ane$.

If we changed the interval a petty to say,

\[f\left( x \right) = \frac{i}{{{ten^ii}}}\hspace{0.25in}{\mbox{on}}\hspace{0.25in}\left[ {\frac{1}{2},i} \right]\]

the role would now have both accented extrema. We may only encounter problems if the interval contains the bespeak of discontinuity. If it doesn’t then the theorem will concord.

We should also betoken out that just because a role is not continuous at a point that doesn’t hateful that it won’t accept both absolute extrema in an interval that contains that signal. Below is the graph of a part that is not continuous at a indicate in the given interval and even so has both absolute extrema.

$This is the graph of some unknown function. The left most point is in the third quadrant and labeled “a”. The graph increases to a peak (but not the highest point on the graph) in the second quadrant and then decreases into the third quadrant, passing through the y-axis into the fourth quadrant to a point labeled “c” that is the lowest point on the graph that is marked as the “Abs. Min.”. At this point the graph jumps up to a higher point that is still in the fourth quadrant and still at the same x value. The graph increases into the first quadrant to a peak (not highest point), decreases back into the fourth quadrant to a valley (not lowest point) and finally increases back into the first quadrant to the right most point that is labeled as “b” that is the highest point on the graph and marked as the “Abs. Max.”.$

This graph is not continuous at $10 = c$, yet it does have both an absolute maximum ($x = b$) and an accented minimum ($ten = c$). Also note that, in this example ane of the absolute extrema occurred at the point of aperture, but it doesn’t need to. The absolute minimum could just accept hands been at the other end indicate or at some other betoken interior to the region. The point here is that this graph is not continuous and yet does have both absolute extrema

The point of all this is that we need to be careful to just use the Extreme Value Theorem when the conditions of the theorem are met and non misinterpret the results if the weather condition aren’t met.

In order to use the Extreme Value Theorem nosotros must have an interval that includes its endpoints, often called a closed interval, and the function must be continuous on that interval. If we don’t accept a closed interval and/or the function isn’t continuous on the interval and then the function may or may not have absolute extrema.

We need to discuss one terminal topic in this section earlier moving on to the outset major application of the derivative that we’re going to be looking at in this chapter.

Fermat’s Theorem

If $f\left( ten \right)$ has a relative extrema at $10 = c$ and $f’\left( c \correct)$ exists then $x = c$ is a critical point of $f\left( x \right)$. In fact, it will be a critical betoken such that $f’\left( c \right) = 0$.

To come across the proof of this theorem see the Proofs From Derivative Applications department of the Extras chapter.

Too annotation that nosotros can say that $f’\left( c \right) = 0$ because nosotros are also assuming that $f’\left( c \right)$ exists.

This theorem tells us that there is a nice human relationship between relative extrema and critical points. In fact, it volition let u.s.a. to get a list of all possible relative extrema. Since a relative extrema must be a critical point the listing of all critical points will give the states a list of all possible relative extrema.

Consider the instance of $f\left( x \right) = {x^2}$. We saw that this function had a relative minimum at $10 = 0$ in several before examples. And so according to Fermat’s theorem $x = 0$ should exist a critical point. The derivative of the function is,

\[f’\left( x \correct) = 2x\]

Sure plenty $x = 0$ is a critical point.

Exist careful not to misuse this theorem. Information technology doesn’t say that a critical point will be a relative extrema. To see this, consider the following case.

\[f\left( x \right) = {x^3}\hspace{0.25in}\hspace{0.25in}f’\left( x \right) = three{x^2}\]

Clearly $x = 0$ is a critical bespeak. However, we saw in an earlier example this function has no relative extrema of any kind. So, critical points do not take to be relative extrema.

Also note that this theorem says nothing virtually absolute extrema. An absolute extrema may or may non be a critical point.

Before we leave this section we need to discuss a couple of issues.

Kickoff, Fermat’s Theorem simply works for disquisitional points in which $f’\left( c \right) = 0$. This does not, however, hateful that relative extrema won’t occur at critical points where the derivative does non exist. To encounter this consider $f\left( x \correct) = \left| x \correct|$. This function clearly has a relative minimum at $x = 0$ and withal in a previous section we showed in an example that $f’\left( 0 \right)$ does not exist.

What this all ways is that if nosotros want to locate relative extrema all we really need to do is look at the critical points as those are the places where relative extrema may exist.

Finally, recall that at that start of the section nosotros stated that relative extrema will non exist at endpoints of the interval we are looking at. The reason for this is that if we allowed relative extrema to occur there it may well (and in fact most of the time) violate Fermat’s Theorem. At that place is no reason to look end points of intervals to be critical points of whatsoever kind. Therefore, nosotros do not allow relative extrema to exist at the endpoints of intervals.