Video: Absolute Extrema

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

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Video Transcript

In this video, we’ll will learn how to find the absolute maximum and minimum values of a function over a given interval using derivatives. At this stage, you should feel confident in finding relative or local minima and maxima in evaluating the nature of these by considering derivatives. We’ll now look to extend these ideas into finding absolute minima and maxima, in other words, the largest and smallest values that a function will take.

To begin, we’ll recall the extreme value theorem. This says that if 𝑓 of π‘₯ is continuous over some closed interval π‘Ž to 𝑏, then there are two numbers 𝑐 and 𝑑 which are greater than or equal to π‘Ž and less than or equal to 𝑏. Such that, 𝑓 of 𝑐 is an absolute maximum for the function, and 𝑓 of 𝑑 is an absolute minimum for the function over that closed interval.

In other words, if we have a continuous function over some closed interval, then we’re guaranteed at some point in this interval to have both an absolute maximum and an absolute minimum. This theorem is important to finding absolute extrema on a closed interval. It means we’re never going to be looking for something that doesn’t exist. We can say that these extreme values are obtained either at the place or places where critical points occur, or at the end points of the interval.

This means we have a few steps that can help us to find absolute extrema for some continuous function 𝑓. We begin by finding all critical points in the closed interval π‘Ž to 𝑏. We then find the values of our function at these critical points. And then, we evaluate our function at the end points of our interval. We’re looking here for values that might be smaller than any relative minima or larger than any relative maxima. Let’s have a look at how we might apply this process to an example.

Determine the absolute maximum and minimum values of the function 𝑓 of π‘₯ equals two π‘₯ to the fourth power minus eight π‘₯ squared minus 13 in the closed interval negative one to two.

Remember, to find absolute extrema for some continuous function 𝑓 of π‘₯, we follow three steps. We start by finding all critical points in the closed interval. We check the values of 𝑓 of π‘₯ at these critical points. And then, we check the end points for any absolute extrema, values that might be smaller than any relative minima or larger than any relative maxima. Critical points occur when the derivative of the function is equal to zero, or does not exist. So, let’s begin by finding the derivative of our function.

The derivative of 𝑓 of π‘₯ is written as 𝑓 prime of π‘₯, and it’s four times two π‘₯ to the third power minus two times eight π‘₯. And, of course, the derivative of negative 13 is zero. So, we see that the derivative of our function is eight π‘₯ cubed minus 16π‘₯. We’ll set this equal to zero and solve for π‘₯. Here, we factor the expression on the right-hand side to get eight π‘₯ times π‘₯ squared minus two.

And we see that for eight π‘₯ times π‘₯ squared minus two to be equal to zero, either eight π‘₯ must be equal to zero, which means π‘₯ is equal to zero. Or we can say that π‘₯ squared minus two must be equal to zero. And solving this, we see that π‘₯ is equal to both the positive and negative square root of two. So, we have critical points on our function at π‘₯ equals zero, π‘₯ equals negative root two, and π‘₯ equals root two.

Our second step is to evaluate the function at these critical points. That’s 𝑓 of zero, 𝑓 of root two, and 𝑓 of negative two. 𝑓 of zero is two times zero to the fourth power minus eight times zero squared minus 13, which is negative 13. 𝑓 of root two is two root two to the fourth power minus eight root two squared minus 13, which is negative 21. And, actually, 𝑓 of negative root two is also negative 21. Now this by itself doesn’t help us much. It certainly looks like negative 21 and negative 13 might be local extrema, but we need to know whether they are absolute extrema.

So, we’re going to evaluate our function at the ends of our interval. Remember, this is because we know that if we have a continuous function over some closed interval π‘Ž to 𝑏, then we’re guaranteed at some point in this interval to have both an absolute maximum and an absolute minimum. And these extreme values are obtained either at the place or places where local extrema occur, or at the end points of the interval.

So, we’re going to evaluate 𝑓 of negative one, and 𝑓 of two. 𝑓 of negative one is two times negative one to the fourth power minus eight times negative one squared minus 13, which is negative 19. And 𝑓 of two is two times two to the fourth power minus eight times two squared minus 13, which is negative 13. We can quite clearly see then that the absolute maximum value of our function in the closed interval negative one to two is negative 13. And the absolute minimum value is negative 21. This example involves some fairly simple differentiation, so we’ll now look at an example which involves a little more work.

Determine the absolute maximum and minimum values of the function 𝑦 equals π‘₯ over two π‘₯ plus eight on the closed interval two to six.

Remember, to find absolute extrema for our function 𝑓 of π‘₯ over some closed interval, we follow three steps. We begin by finding any critical points in our closed interval. We then find the values of 𝑓 of π‘₯ at these critical points. And then, we check the end points for absolute extrema, in other words, values that are smaller than the relative minimum or larger than the relative maximum.

Remember, the critical points are the points on our curve where the derivative is equal to zero or possibly does not exist. So, we need to find the derivative of our function and set it equal to zero. But how do we differentiate π‘₯ over two π‘₯ plus eight? In fact, we have a number of methods we could use. But since it’s the quotient of two differentiable functions, we can use the quotient rule.

This says that the derivative of the quotient of two differentiable functions 𝑒 and 𝑣 is 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ all over 𝑣 squared. The numerator of our fraction is π‘₯, so we’re going to let 𝑒 be equal to π‘₯ and 𝑣 be equal to two π‘₯ plus eight. Then, d𝑒 by dπ‘₯ is equal to one. And d𝑣 by dπ‘₯ is equal to two.

So, the derivative of 𝑦 with respect to π‘₯ is two π‘₯ plus eight times d𝑒 by dπ‘₯, which is one, minus 𝑒 times d𝑣 by dπ‘₯, that’s π‘₯ times two, all over 𝑣 squared, that’s two π‘₯ plus eight all squared. So, we can say that d𝑦 by dπ‘₯ is equal to eight over two π‘₯ plus eight all squared. Next, we’re going to set this equal to zero and solve for π‘₯. But look at what happens when we do.

For an algebraic fraction to be equal to zero, the numerator must itself be equal to zero. In this case, we end up with the statement zero equals eight, which we know to be rubbish. This means d𝑦 by dπ‘₯, in this case, cannot be equal to zero. There are no turning points for us to evaluate. And we, therefore, jump straight to step three and evaluate the function at the end points of the interval. That’s 𝑓 of two and 𝑓 of six.

𝑓 of two is two over two times two plus eight, that’s two twelfths, which simplifies to one-sixth. 𝑓 of six is six over two times six plus eight. That’s six twentieths, which simplifies to three-tenths. Three-tenths is greater than one-sixth, so we can say the absolute maximum value of our function is three-tenths and the absolute minimum value is one-sixth. Now let’s backtrack a little though.

We said that critical points occur at places on the function where the derivative does not exist, and there is one point on our function where the derivative doesn’t exist. That’s the point where two π‘₯ plus eight is equal to zero, or π‘₯ is equal to negative four. Now since this was outside the closed interval two to six, we actually didn’t need to worry about this critical point. And we could focus solely on the end points of the intervals 𝑓 of two and 𝑓 of six.

In our next example, we’ll look at how to apply the process to finding absolute extrema for piecewise functions.

Determine the absolute maximum and minimum values of the function 𝑓 of π‘₯ is equal to six π‘₯ minus three squared, if π‘₯ is less than or equal to two and two minus nine π‘₯, if π‘₯ is greater than two in the closed interval one to six.

Remember, to find absolute extrema for continuous functions, we follow three steps. We begin by finding any critical points in the closed interval we’re looking at. We find the values of our function 𝑓 of π‘₯ at these critical points. And then, we check the end points for absolute extrema, values that are smaller than the relative minimum or larger than the relative maximum.

Now we have a slight problem here. This is a piecewise function. And we don’t yet know if this piecewise function is itself continuous. We do have a note that the function six π‘₯ minus three all squared is continuous, and the function two minus nine π‘₯ is continuous. So, what we’ll do is consider that π‘₯ equals two might be a critical point. And to test this, we’re going to evaluate the right-hand and left-hand derivative of our function at π‘₯ equals two.

We’ll begin by evaluating the right-hand derivative of 𝑓 at π‘₯ equals two. This is given by the limit as β„Ž approaches zero from the right of 𝑓 of two plus β„Ž minus 𝑓 of two all over β„Ž. We’re looking at the right-hand derivative, so we’re interested in the function that applies when π‘₯ is greater than two. That’s 𝑓 of π‘₯ equals two minus nine π‘₯.

So, we’re looking for the limit as β„Ž approaches zero from the right of two minus nine times two plus β„Ž minus two minus nine times two over β„Ž. That’s two minus 18 minus nine β„Ž minus two plus 18 all over β„Ž. This quite quickly simplifies to negative nine β„Ž over β„Ž. And then, we simplify further, and we see that we’re looking for the limit as β„Ž tends to zero from the right of negative nine. But this is independent of β„Ž, so we know that this is just going to be equal to negative nine. And so, our right-hand derivative is negative nine.

We’ll now repeat this process for the left-hand derivative. This time, we’re evaluating 𝑓 of two plus β„Ž minus 𝑓 of two over β„Ž as β„Ž approaches zero from the left. And so, we’re interested in the part of 𝑓 of π‘₯ where π‘₯ is less than or equal to two. So, we have the limit as β„Ž approaches zero from the left of six times two plus β„Ž minus three all squared minus six times two minus three all squared all over β„Ž. This simplifies to nine plus six β„Ž all squared minus nine squared over β„Ž.

And then, if we distribute the parentheses, we see that we’re left with the limit as β„Ž approaches zero from the left of 108β„Ž plus 36β„Ž squared. We can do a little bit of simplification, and this becomes 108 plus 36β„Ž. And then, we see that as β„Ž approaches zero from the left, we’re left with 108.

We can see that our left-hand and right-hand derivatives are not equal. And so, 𝑓 prime of π‘₯, our derivative, doesn’t actually exist at π‘₯ equals two. And so, we know that we have a critical point at π‘₯ equals two. And we know we’re going to need to find 𝑓 of π‘₯ at this point. We should also though check for any critical points on each part of our piecewise function. So, we’ll differentiate each part with respect to π‘₯ and set that equal to zero.

We can use the general power rule to differentiate six π‘₯ minus three all squared with respect to π‘₯. It’s two times six π‘₯ minus three. And then, we reduce the power by one. And then, we multiply that by the derivative of six π‘₯ minus three, which is just six. So, the derivative of this bit is 12 times six π‘₯ minus three. And the derivative of two minus nine π‘₯ is negative nine.

It should be quite clear that there’s no way for negative nine to be equal to zero. The derivative of this part of our function is always negative nine. But we can set 12 times six π‘₯ minus three equal to zero. And when we solve for π‘₯, we get π‘₯ equals 0.5. So, we have one more critical point at π‘₯ equals 0.5. We’re going to evaluate our function at the points π‘₯ equals two and π‘₯ equals 0.5 then. Let’s clear some space.

0.5 is less than two, so we evaluate the function at this point by using six π‘₯ minus three all squared. And when we substitute 0.5 in, we get zero. So, 𝑓 of 0.5 is zero. We use the same part of our piecewise function to evaluate 𝑓 of two. And when we do, we get 81. So, we’ve found 𝑓 of π‘₯ at the critical points on our function. Next, we need to check the end points.

These are 𝑓 of one and 𝑓 of six. We use six π‘₯ minus three all squared once again to evaluate 𝑓 of one. And that gives us nine. But six is greater than two. So, to evaluate 𝑓 of six, we use two minus nine π‘₯ and we get negative 52. We can see that the absolute maximum value of our function is 81 and the absolute minimum value is negative 52. The key point to remember here is that if we’re dealing with a piecewise function, we must check the behaviour at the end of each piece of our function. In our final example, we’ll consider how we can apply ideas about finding absolute extrema to exponential functions.

Find the absolute maximum and minimum values rounded to two decimal places of the function 𝑓 of π‘₯ equals five π‘₯𝑒 to the negative π‘₯, given that π‘₯ is a part of the closed interval zero to four.

Remember, to find absolute extrema for our function 𝑓 of π‘₯, we follow three steps. We find all critical points in our closed interval. We then find the values of 𝑓 of π‘₯ at these critical points. And then, we check the end points for absolute extrema. The critical points are the points where the derivative is equal to zero or does not exist. So, we’re going to find the derivative of our function and set that equal to zero.

And here, we notice that this itself is the product of two differentiable functions. So, we’re going to use the product rule. This says that the derivative of the product of two differentiable functions 𝑒 and 𝑣 is 𝑒 times d𝑣 by dπ‘₯ plus 𝑣 times d𝑒 by dπ‘₯. So, we let 𝑒 be equal to five π‘₯ and 𝑣 be equal to 𝑒 to the negative π‘₯. Then, d𝑒 by dπ‘₯ is equal to five and d𝑣 by dπ‘₯ π‘₯ is equal to negative 𝑒 to the negative π‘₯. This means the derivative of our function is five π‘₯ times negative 𝑒 to the negative π‘₯ plus 𝑒 to the negative π‘₯ times five, which we can simplify to five 𝑒 to the negative π‘₯ times one minus six

Let’s set this equal to zero. Now there’s no way for five 𝑒 to the negative π‘₯ to be equal to zero. So, for the statement five 𝑒 to the negative π‘₯ times one minus π‘₯ equals zero to be true, we know that one minus π‘₯ itself must be equal to zero, which means that π‘₯ equals one is a critical point. We’re, therefore, going to evaluate our function at this critical point and at the end points of the function, so 𝑓 of one, 𝑓 of zero, and 𝑓 of four.

𝑓 of one is five times one times 𝑒 to the negative one, which is 1.8393 and so on, or correct to two decimal places as required 1.84. 𝑓 of zero is zero. And 𝑓 of four is five times four times 𝑒 to the negative four, which is 0.37 correct to two decimal places. And we can, therefore, say that the absolute maximum value of our function is 1.84 and the absolute minimum value is zero.

In this video, we’ve learned that if we have a continuous function over some closed interval, then we’re guaranteed at some point in this interval to have both an absolute maximum and an absolute minimum. We also saw that these extreme values are obtained either at the place or places where local extrema occur or at the end points of the interval. We saw that we can apply these ideas to exponential functions and functions which are both products and quotients of other differentiable functions, but that we need to be careful with piecewise functions to evaluate the ends of each piece of the function.

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