Lesson Explainer: Absolute Extrema Mathematics • Higher Education

In this explainer, we will learn how to find the absolute maximum and minimum values of a function over a given interval using derivatives.

We can use the derivative of a function at a point to help us determine what happens to that function locally. For example, we know that the local extrema of a function must occur at its critical points. Remember, we say ๐‘ฅ=๐‘ is a critical point of a function ๐‘“ if ๐‘“โ€ฒ(๐‘)=0 or does not exist.

However, this only gives a local bound for our function. What if we wanted to bound our function over a larger interval?

For example, consider the following curve.

We can see it has two minimum values, one at ๐‘ฅ=0 and another at ๐‘ฅ=โˆ’3. We can refer to the minimum at ๐‘ฅ=0 as a local minimum since it is the lowest output of our function near ๐‘ฅ=0; however, the minimum at ๐‘ฅ=โˆ’3 is the lowest output of the function entirely, so we will want to give this a different name.

Definition: Absolute Extrema

For a function ๐‘“(๐‘ฅ), we say the function has

  • an absolute maximum at ๐‘ฅ=๐‘, if ๐‘“(๐‘ฅ)โ‰ค๐‘“(๐‘) for all values of ๐‘ฅ in the domain of ๐‘“;
  • an absolute minimum at ๐‘ฅ=๐‘‘, if ๐‘“(๐‘‘)โ‰ค๐‘“(๐‘ฅ) for all values of ๐‘ฅ in the domain of ๐‘“.

We call ๐‘“(๐‘) the absolute maximum of ๐‘“(๐‘ฅ) and ๐‘“(๐‘‘) the absolute minimum of ๐‘“(๐‘ฅ).

There are a few problems with finding absolute extrema in general. Firstly, not all functions have absolute extrema. For example, we know the function ๐‘“(๐‘ฅ)=๐‘ฅ is unbounded; we can always choose higher values of ๐‘ฅ to get larger unbounded outputs (and the same is true for a lower bound). Secondly, sometimes functions are bounded but do not actually reach their absolute extrema. For example, ๐‘“(๐‘ฅ)=1๐‘ฅ๏Šจ is always positive, so we know it is bounded below by 0, but we can always find an input value of ๐‘ฅ to make the output lower.

To circumvent these problems, we will introduce one restriction on our functions; instead of looking for absolute extrema of our functions on their entire domain, we will look for absolute extrema on a subset of their domain.

Definition: Absolute Extrema of a Function on a Closed Interval

For a function ๐‘“(๐‘ฅ) on an interval [๐‘Ž,๐‘],

  • we say the function has an absolute maximum at ๐‘ฅ=๐‘ on the interval [๐‘Ž,๐‘], if ๐‘“(๐‘ฅ)โ‰ค๐‘“(๐‘) for all values of ๐‘ฅโˆˆ[๐‘Ž,๐‘];
  • we say the function ๐‘“ has an absolute minimum at ๐‘ฅ=๐‘‘ on the interval [๐‘Ž,๐‘], if ๐‘“(๐‘‘)โ‰ค๐‘“(๐‘ฅ) for all values of ๐‘ฅโˆˆ[๐‘Ž,๐‘].

We call ๐‘“(๐‘) the absolute maximum and ๐‘“(๐‘‘) the absolute minimum of ๐‘“ on the interval [๐‘Ž,๐‘].

Essentially, these are just the maximum and minimum values a function can obtain for inputs in the closed interval.

There is still one more problem: if we are given a function that is not continuous on an interval, then the function is not necessarily bounded on this interval. For example, ๐‘“(๐‘ฅ)=1๐‘ฅ๏Šจ is not bounded on the interval [โˆ’1,1]. We can see this around ๐‘ฅ=0 in our graph above.

However, if we are working with a continuous function, we can use a very useful result called the extreme value theorem.

Theorem: Extreme Value Theorem

If ๐‘“(๐‘ฅ) is continuous on the interval [๐‘Ž,๐‘], then there exists ๐‘,๐‘‘โˆˆ[๐‘Ž,๐‘] such that ๐‘“(๐‘) is the absolute maximum and ๐‘“(๐‘‘) is the absolute minimum of ๐‘“ on the interval [๐‘Ž,๐‘].

It can be useful to see this in an example. Consider the following diagram of ๐‘ฆ=(๐‘ฅโˆ’1)(๐‘ฅ+3) on the interval [โˆ’3,2].

We can see an example of the extreme value theorem in action in our diagram; the absolute minimum of our function on this interval is at the turning point ๐‘ฅ=โˆ’1 and the absolute maximum of our function is at the endpoint ๐‘ฅ=2.

While it is beyond the scope of this explainer to prove this theorem, this does give us some really useful information to work with. First, if we are asked to find the absolute extrema of a continuous function on a closed interval, then the extreme value theorem guarantees that these exist. Second, if we are given a piecewise continuous function, we can apply the extreme value theorem to each of the pieces and then separately deal with the points of discontinuity.

Finally, to find these extrema, we need to recall some information about local extrema and increasing/decreasing intervals. We know local extrema will occur at critical points, and while we might be tempted to assume that absolute extrema must also be local extrema, this is not necessarily true at the endpoints of our interval [๐‘Ž,๐‘] . Since our function could be increasing/decreasing beyond these endpoints, we will have to check the endpoints of our interval separately.

This gives us a method for finding the absolute extrema of a continuous function on a closed interval.

How To: Finding the Absolute Extrema of a Continuous Function on a Closed Interval

If ๐‘“(๐‘ฅ) is a continuous function, then we can find the absolute extrema of ๐‘“ on the interval [๐‘Ž,๐‘] by

  1. finding all of the critical points of ๐‘“(๐‘ฅ) on [๐‘Ž,๐‘],
  2. evaluating ๐‘“ at all of the critical points in [๐‘Ž,๐‘],
  3. checking the value of the function at the endpoints of [๐‘Ž,๐‘].

The largest of these outputs is the absolute maximum of ๐‘“ on [๐‘Ž,๐‘] and the lowest is the absolute minimum of ๐‘“ on [๐‘Ž,๐‘].

We can use this method on any continuous function on a closed interval. For piecewise continuous functions, we can analyze each continuous part by using this method and then check the points of discontinuity separately.

In our first example, we will look at how to apply this process to find the absolute extrema of a polynomial function.

Example 1: Finding the Absolute Maximum and Minimum Values of a Polynomial Function on a Closed Interval

Determine the absolute maximum and minimum values of the function ๐‘“(๐‘ฅ)=2๐‘ฅโˆ’8๐‘ฅโˆ’13๏Šช๏Šจ in the interval [โˆ’1,2].

Answer

We are tasked with finding the absolute maximum and minimum of a function on a closed interval. Remember, the local extrema of our function occur at critical points, while absolute extrema can also occur at the endpoints of the interval, so we will need to check all of these to find our absolute extrema.

The first thing we should check is the continuity of our function; in this case, ๐‘“(๐‘ฅ) is a polynomial, so ๐‘“ is continuous for all real numbers. This means that the absolute extrema of ๐‘“ on [โˆ’1,2] must occur at the critical points or the endpoints.

Next, we will start by finding the critical points of ๐‘“ on this interval; remember, these are the values of ๐‘ฅ where the derivative is equal to 0 or does not exist. Of course, the derivative of a polynomial will exist for any real value of ๐‘ฅ, so we only need to look for the derivative equal to 0.

We can do this by finding the derivative of ๐‘“: ๐‘“โ€ฒ(๐‘ฅ)=4ร—2๐‘ฅโˆ’2ร—8๐‘ฅ=8๐‘ฅโˆ’16๐‘ฅ.๏Šช๏Šฑ๏Šง๏Šจ๏Šฑ๏Šง๏Šฉ

Then, we set this expression equal to zero and solve by factoring: 8๐‘ฅโˆ’16๐‘ฅ=08๐‘ฅ๏€ป๐‘ฅโˆ’โˆš2๏‡๏€ป๐‘ฅ+โˆš2๏‡=0.๏Šฉ

This has solutions ๐‘ฅ=โˆ’โˆš2,0,โˆš2. Remember, we are only interested in critical points in the interval [โˆ’1,2], so only two of these critical points lie in our interval. These are located at ๐‘ฅ=0 and ๐‘ฅ=โˆš2.

These are the possible locations of the absolute extrema for our function, so we should evaluate our function at these locations to compare to the endpoints of our curve: ๐‘“(0)=2(0)โˆ’8(0)โˆ’13=โˆ’13,๐‘“๏€ปโˆš2๏‡=2๏€ปโˆš2๏‡โˆ’8๏€ปโˆš2๏‡โˆ’13=โˆ’21.๏Šช๏Šจ๏Šช๏Šจ

Since the endpoints of [โˆ’1,2] could still give us absolute extrema, we must check the value of the function at these points: ๐‘“(โˆ’1)=2(โˆ’1)โˆ’8(โˆ’1)โˆ’13=โˆ’19,๐‘“(2)=2(2)โˆ’8(2)โˆ’13=โˆ’13.๏Šช๏Šจ๏Šช๏Šจ

We have now checked all of the possible points where extrema can occur in the interval [โˆ’1,2]. The greatest of these four outputs is โˆ’13, which will be the absolute maximum of ๐‘“ on [โˆ’1,2], and the lowest is โˆ’21, which will be the absolute minimum of ๐‘“ on [โˆ’1,2]. We can see this in the graph of this function, restricted to [โˆ’1,2].

This gives us our final answer. On the interval [โˆ’1,2], the function ๐‘“ will have an absolute maximum value of โˆ’13 at ๐‘ฅ=0 and ๐‘ฅ=2 and an absolute minimum value of โˆ’21 at ๐‘ฅ=โˆš2.

Example 2: Finding the Absolute Maximum and Minimum Values of a Rational Function on a Given Interval

Determine the absolute maximum and minimum values of the function ๐‘ฆ=๐‘ฅ2๐‘ฅ+8 on the interval [2,6].

Answer

We are asked to find the absolute maximum and minimum values of a function on a closed interval. To do this, we want to find the local extrema and also check the value of the function at the endpoints of our interval.

So, the first thing we should do is check the continuity of our function; this will then allow us to find extrema on the continuous parts separately. In this case, our function is rational, and we know all rational functions are continuous everywhere except where their denominator is equal to zero. For this function, that is when ๐‘ฅ=โˆ’4, which is not in the interval [2,6].

This means our function is continuous on the interval [2,6]; so by the extreme value theorem there must exist both an absolute maximum and an absolute minimum in this interval. To find the absolute extrema of this function, we need to compare the value of the function at its critical points with the value at its endpoints.

Let us start by finding the critical points (i.e., where the derivative is equal to 0 or does not exist). We will do this by using the quotient rule, which we recall tells us that, for differentiable functions ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)and, dd๐‘ฅ๏€ฝ๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)๏‰=๐‘”(๐‘ฅ)(๐‘“(๐‘ฅ))โˆ’๐‘“(๐‘ฅ)(๐‘”(๐‘ฅ))(๐‘”(๐‘ฅ)).dddd๏—๏—๏Šจ

Applying this to our function, with ๐‘“(๐‘ฅ)=๐‘ฅ and ๐‘”(๐‘ฅ)=2๐‘ฅ+8, we get: dd๐‘ฆ๐‘ฅ=(2๐‘ฅ+8)(๐‘ฅ)โˆ’๐‘ฅ(2๐‘ฅ+8)(2๐‘ฅ+8)=(2๐‘ฅ+8)โˆ’2๐‘ฅ(2๐‘ฅ+8)=8(2๐‘ฅ+8).dddd๏—๏—๏Šจ๏Šจ๏Šจ

We see the derivative exists everywhere except when ๐‘ฅ=โˆ’4, which is not in the domain of our function, so we only need to check for the values of ๐‘ฅ where the derivative is equal to 0. For this to be equal to 0, the numerator must be equal to 0; however, the numerator is always equal to 8. Therefore, there are no critical points for this function in our interval. That means we only need to check the endpoints.

When ๐‘ฅ=2, ๐‘“(2)=22(2)+8=16.

When ๐‘ฅ=6, ๐‘“(6)=62(6)+8=310.

This tells us that 310 must be the maximum of the function on this interval and 16 must be the minimum. We can see this on the following graph of our function on the interval [2,6].

We were able to show that the maximum value of ๐‘ฅ2๐‘ฅ+8 on the interval [2,6] is 310 at ๐‘ฅ=6 and the minimum value on the same interval is 16 at ๐‘ฅ=2.

So far, the functions in all of our examples have been continuous on the interval we are working on. However, we can also treat discontinuities in a function separately. Let us see an example of how to apply this process to a piecewise-defined function.

Example 3: Finding the Absolute Maximum and Minimum Values of a Piecewise Function in a Given Interval

Determine the absolute maximum and minimum values of the function ๐‘“(๐‘ฅ)=๏ฎ(6๐‘ฅโˆ’3),๐‘ฅโ‰ค2,2โˆ’9๐‘ฅ,๐‘ฅ>2,๏Šจ in the interval [1,6].

Answer

We are tasked with finding the absolute maximum and minimum values of a function on a closed interval; we can do this by comparing the value of the function at its critical points and the endpoints of our interval, provided our function is continuous. So, we should start by checking the continuity of our function.

We can see that our function ๐‘“(๐‘ฅ) is piecewise defined and both pieces are polynomials. This guarantees that ๐‘“(๐‘ฅ) will be continuous everywhere except where we join the pieces, in this case, when ๐‘ฅ=2. To see if ๐‘“(๐‘ฅ) is continuous when ๐‘ฅ=2, we check what happens when ๐‘ฅ=2 in both pieces: (6(2)โˆ’3)=81,2โˆ’9(2)=โˆ’16.๏Šจ

These are not equal, so the two pieces of ๐‘“(๐‘ฅ) do not join when ๐‘ฅ=2. In other words, we have a jump discontinuity here.

We might be worried that this means we cannot find the absolute maximum and minimum values of the function on the interval [1,6]; however, we do know that ๐‘“(๐‘ฅ) is continuous and equal to (6๐‘ฅโˆ’3)๏Šจ when ๐‘ฅโ‰ค2 and ๐‘“(๐‘ฅ) is continuous and equal to 2โˆ’9๐‘ฅ when ๐‘ฅ>2. So, we can treat these two cases separately.

First, when ๐‘ฅโ‰ค2, ๐‘“(๐‘ฅ)=(6๐‘ฅโˆ’3)๏Šจ. We are looking for extrema in the interval [1,6], so our values of ๐‘ฅ must be in the interval [1,2]. To find the extrema of this piece, we want to find all of the critical points and then check the endpoints of our new interval [1,2].

We will start by finding the critical points, that is when the derivative is equal to zero or does not exist: ๐‘“โ€ฒ(๐‘ฅ)=6(2)(6๐‘ฅโˆ’3)=36(2๐‘ฅโˆ’1).

We can see this is defined for all of our values of ๐‘ฅ and is zero when ๐‘ฅ=12, so this is the only critical point of the first function; however, we are only interested in values of ๐‘ฅ in the interval [1,2]. So, we only need to evaluate ๐‘“ at the endpoints of this first interval: ๐‘“(1)=(6(1)โˆ’3)=9,๐‘“(2)=(6(2)โˆ’3)=81.๏Šจ๏Šจ

We are not done yet; we still need to see what happens when ๐‘ฅ>2.

In this case, we know that ๐‘“(๐‘ฅ)=2โˆ’9๐‘ฅ, and we want to find extrema on the interval ]2,6]. Now, this is a problem since this interval is not closed. So, to get around this, we will look for extrema on the interval [2,6], and if we find an extremum at ๐‘ฅ=2, we will need to sketch a graph to see what happens here.

To avoid confusion, let us call ๐‘”(๐‘ฅ)=2โˆ’9๐‘ฅ. We want to find the extrema of this function on the interval [2,6], so we need to find the critical points: ๐‘”โ€ฒ(๐‘ฅ)=โˆ’9.๐‘”โ€ฒ(๐‘ฅ) is therefore defined for all ๐‘ฅ and cannot be equal to zero, so ๐‘”(๐‘ฅ) has no critical points. This means we just need to look at the endpoints of our interval: ๐‘”(2)=2โˆ’9(2)=โˆ’16,๐‘”(6)=2โˆ’9(6)=โˆ’52.

We have now found all of the possible extrema for our original piecewise function ๐‘“(๐‘ฅ) on the interval [1,6], so all that is left to do is compare these values.

๐‘ฅ12๏Šฑ2๏Šฐ6
๐‘“(๐‘ฅ)981โˆ’16โˆ’52

We see the absolute maximum is 81 at ๐‘ฅ=2, and the absolute minimum is โˆ’52 at ๐‘ฅ=6. We can also see this information in the graph of the function on [1,6].

The next few examples show that, sometimes, we will need to use some of our advanced differentiation techniques to find the critical points of the function.

Example 4: Finding the Absolute Maximum and Minimum Values of a Function in a Given Interval Involving Using the Product Rule

Find the absolute maximum and minimum values, rounded to two decimal places, of the function ๐‘“(๐‘ฅ)=5๐‘ฅ๐‘’,๐‘ฅโˆˆ[0,4]๏Šฑ๏—where.

Answer

We want to find the absolute maximum and minimum values of this function and we can immediately notice that ๐‘“(๐‘ฅ) is the product of two continuous functions, so ๐‘“(๐‘ฅ) itself is continuous. To find the absolute extrema of the function on this closed interval, we need to find all of the critical points and check the endpoints of our function.

Let us start with finding all of the critical points; these occur when the derivative is equal to zero or does not exist, so we will begin by differentiating ๐‘“(๐‘ฅ) by using the product rule.

Recall that the product rule tells us if ๐‘ข(๐‘ฅ) and ๐‘ฃ(๐‘ฅ) are differentiable; then, (๐‘ข(๐‘ฅ)๐‘ฃ(๐‘ฅ))โ€ฒ=๐‘ขโ€ฒ(๐‘ฅ)๐‘ฃ(๐‘ฅ)+๐‘ข(๐‘ฅ)๐‘ฃโ€ฒ(๐‘ฅ).

Applying this to our function ๐‘“, where ๐‘ข(๐‘ฅ)=5๐‘ฅ and ๐‘ฃ(๐‘ฅ)=๐‘’๏Šฑ๏—, ๐‘“โ€ฒ(๐‘ฅ)=5๐‘’โˆ’5๐‘ฅ๐‘’.๏Šฑ๏—๏Šฑ๏—

This is defined for all of the values of ๐‘ฅ in our interval [0,4], so the only critical points will be when this is equal to 0: 5๐‘’โˆ’5๐‘ฅ๐‘’=05๐‘’(1โˆ’๐‘ฅ)=0.๏Šฑ๏—๏Šฑ๏—๏Šฑ๏—

We know ๐‘’>0๏Šฑ๏— for all values of ๐‘ฅ, so the only critical point of ๐‘“(๐‘ฅ) is when ๐‘ฅ=1. Now, all that is left to do is to compare the value of the function at the critical point and at the endpoints of our interval [0,4]: ๐‘“(1)=5(1)๐‘’โ‰ˆ1.84,๐‘“(0)=5(0)๐‘’=0,๐‘“(4)=5(4)๐‘’โ‰ˆ0.37.๏Šฑ๏Šง๏Šฑ๏Šฆ๏Šฑ๏Šช

Therefore, to two decimal places, the absolute maximum is 1.84, and the absolute minimum is 0.

Example 5: Finding the Absolute Maximum and Minimum Values of a Root Function in a Given Interval

Find, if they exist, the values of the absolute maximum and minimum points for the function ๐‘“(๐‘ฅ)=โˆš3๐‘ฅ+10, where ๐‘ฅโˆˆ[โˆ’2,5].

Answer

We want to find the absolute extrema of a function on a closed interval; the first thing we can notice is that ๐‘“(๐‘ฅ) is the composition of two continuous functions, so ๐‘“(๐‘ฅ) will be continuous across its entire domain. We want to check that ๐‘“(๐‘ฅ) is defined when ๐‘ฅโˆˆ[โˆ’2,5]; to do this, we need to make sure we are not taking the square root of a negative number. We can see that 3๐‘ฅ+10โ‰ฅ0 for these values of ๐‘ฅ, so our function is defined for all ๐‘ฅ in the interval [โˆ’2,5].

Since our function is continuous, the extrema on this interval will either be critical points or the endpoints of our interval. We can check all of these individually.

Let us start by finding the critical points; these will be when the derivative of ๐‘“(๐‘ฅ) is zero or where it does not exist.

Using either the chain rule or the general power rule, we can see that ๐‘“โ€ฒ(๐‘ฅ)=32โˆš3๐‘ฅ+10.

For ๐‘“โ€ฒ(๐‘ฅ) to be equal to zero, the numerator must be equal to zero, but the numerator must remain constant at 3. Therefore, there is no value of ๐‘ฅ for which ๐‘“โ€ฒ(๐‘ฅ)=0. For ๐‘“โ€ฒ(๐‘ฅ) to be undefined, we must either divide by zero or take the square root of a negative number; however, none of these values of ๐‘ฅ will be in the domain of ๐‘“(๐‘ฅ). This means there are no critical points for our function ๐‘“(๐‘ฅ).

Therefore, we only need to compare the values of ๐‘“ at the endpoints of our interval [โˆ’2,5]: ๐‘“(โˆ’2)=โˆš3(โˆ’2)+10=2,๐‘“(5)=โˆš3(5)+10=5.

This tells us the function has an absolute minimum value of 2 and an absolute maximum value of 5.

We can see this information if we plot the graph of this curve over the interval [โˆ’2,5].

The function is increasing on the entire interval, so its absolute minimum is the start point of the interval and its absolute maximum is the endpoint of the interval.

Let us finish by recapping the key points about finding the absolute maxima and minima of functions on a closed interval.

Key Points

  • The extreme value theorem tells us that a continuous function on a closed interval achieves its absolute maximum and minimum in that interval.
  • We can find the absolute maximum and minimum values of a continuous function on a closed interval by comparing the values of all of the critical points and the endpoints.
  • We can even extend this to discontinuous functions by checking the points of discontinuity separately.

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