Video: Finding the Absolute Maximum and Minimum Values of a Function in a Given Interval

Find the absolute maximum and minimum values of the function 𝑦 = π‘₯/4 + 1/(π‘₯ βˆ’ 4) on the interval [1, 3].

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Video Transcript

Find the absolute maximum and minimum values of the function 𝑦 equals π‘₯ over four plus one over π‘₯ minus four on the interval from one to three inclusive.

We have a closed interval that is an interval which includes both of its endpoints, one and three. And we want to find the absolute maximum and minimum values of the function given on this interval. So for values of π‘₯ between one and three inclusive, what’s the greatest and smallest that 𝑦 can be.

To solve this problem, we’re going to use the closed interval method which tells us how to find the absolute maximum and minimum values of a continuous function 𝑓 on a closed interval π‘Ž, 𝑏. We evaluate 𝑓 at its critical numbers in the open interval from π‘Ž to 𝑏 and also at the endpoints π‘Ž and 𝑏. The largest value obtained in this way is the absolute maximum and the smallest is the absolute minimum.

Before applying this method, we should check that it is applicable. This method finds the absolute maximum and minimum values of a continuous function. Is our functions continuous? As a rational function, it is continuous wherever it is defined. However, it’s not defined when this denominator polynomial is zero. So it’s not defined at π‘₯ equals four which is the only value of π‘₯ for which π‘₯ minus four is zero. And so it’s not continuous at π‘₯ equals four. So what do we do?

Well, for this problem, it’s not an issue because we’re restricting our attention to the closed interval from one to three. And π‘₯ equals four does not lie in this interval. And so our function is continuous on the smaller domain, the closed interval from one to three. This allows us to apply the closed interval method. So let’s apply the method.

We first have to evaluate 𝑓 at its critical numbers in the open version of the interval on which it is defined. Our function is π‘₯ plus four plus one over π‘₯ minus four. And we’re looking for the critical numbers in the open interval from one to three. But what are the critical numbers of this function in this interval? In fact, more generally, what are critical numbers of a function? Period.

A number 𝑐 in the domain of a function f is a critical number of 𝑓 if the derivative 𝑓 prime evaluated at 𝑐 is zero or if 𝑓 prime evaluated at 𝑐 does not exist. In other words, 𝑐 is not in the domain of 𝑓 prime. We’re looking for a critical number of our function and this depends on the derivative of the function. So let’s differentiate. We can do this term by term.

The derivative of π‘₯ over four or a quarter π‘₯ is straightforward. It’s a quarter. The derivative of one over π‘₯ minus four is slightly more complicated. So let’s give it a bit of thought. If we let 𝑧 equal π‘₯ minus four, then this is the derivative with respect to π‘₯ of one over 𝑧. And we can apply the chain rule 𝑑 by 𝑑π‘₯ is 𝑑𝑧 by 𝑑π‘₯ times 𝑑 by 𝑑𝑧. So now we can consider 𝑑𝑧 by 𝑑π‘₯ and 𝑑 by 𝑑𝑧 of one over 𝑧 separately. What is 𝑑𝑧 by 𝑑π‘₯? 𝑧 is π‘₯ minus four by definition. And so differentiating this with respect to π‘₯, we see that 𝑑𝑧 by 𝑑π‘₯ is one.

How about 𝑑 by 𝑑𝑧 of one over 𝑧. We can write one over 𝑧 as 𝑧 to the negative one. And then we can apply the fact that 𝑑 by 𝑑 whatever of whatever to the power of 𝑛 is 𝑛 times whatever to the power of 𝑛 minus one. With 𝑛 equal to negative one, we get that 𝑑 by 𝑑𝑧 of 𝑧 to the negative one is negative one times 𝑧 to the negative two. Let’s clear some room. We can write 𝑧 to the negative two as one over 𝑧 squared. Here, we’ve used the fact that adding one times negative one times something is just the same as subtracting that something.

Now, we use the fact that 𝑧, the auxiliary variable, is π‘₯ minus four to get our derivative 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯. Let’s clear away some working to make room for some more. Let’s remind ourselves why we differentiated our function. It was to find the critical numbers of our function, where the derivative is zero or does not exist. Let’s start with where the derivative does not exist.

Because of the denominator π‘₯ minus four squared, we can see that the derivative doesn’t exist when π‘₯ is four. But this isn’t an issue because four wasn’t in the domain of our original function. And it certainly isn’t in the open interval from one to three. The derivative does in fact exist everywhere in this open interval. Therefore, the only critical numbers come from when the derivative is zero. Let’s find out when this happens by solving 𝑑𝑦 by 𝑑π‘₯ equals zero.

We solve this by writing the derivative as one fraction. It’s simple to write the two fractions we have with a common denominator, and so combine them. Now, this fraction is going to be equal to zero exactly when the numerator is equal to zero, assuming that the denominator is not also equal to zero at the same time. We expand and simplify the numerator getting π‘₯ squared minus eight π‘₯ plus 12, which we can factor by inspection.

Now that the numerator is factored, we can see that the derivative is zero when π‘₯ is six or π‘₯ is two. These are therefore our critical numbers. But remember, we are only interested in the critical numbers in the open interval from one to three. Six is not in the interval. And so the only critical number in the open interval from one to three is two. Our first line of working becomes evaluate π‘₯ over four plus one over π‘₯ minus two at two. Let’s clear some room so we can do this.

To find the value of 𝑦 when π‘₯ is two, we just substitute two for π‘₯. So 𝑦 is two over four plus one over two minus four. And this simplifies to zero. We have completed the first step of the closed interval method. We have evaluated our function at its critical numbers in the open interval from π‘Ž to 𝑏. We also need to evaluate our function 𝑓 at the end points π‘Ž and 𝑏 of the closed interval.

π‘Ž and 𝑏 are one and three, respectively. Let’s evaluate then, starting with π‘₯ equals one. 𝑦 is then one over four plus one over one minus four which is negative a twelfth. How about for the other endpoint when π‘₯ is three? There, 𝑦 is three over four plus one over three minus four which simplifies to negative a quarter. What do we do now? Well, we look at the closed interval method. And it says that the largest value obtained is the absolute maximum and the smallest value obtained is the absolute minimum. The largest of our three values β€” zero, negative a twelfth, and negative a quarter β€” is zero. And the smallest value we obtained is negative a quarter.

What do we conclude then? The absolute maximum value of our function 𝑦 equals π‘₯ over four plus one over π‘₯ minus four on the closed interval from one to three is zero. And the absolute minimum of our function on this interval is negative a quarter. This was a relatively straightforward application of the closed interval method. While our function wasn’t continuous on the set of real numbers, it was continuous on the interval that we cared about, the closed interval from one to three. And so we could apply the method.

We found the critical members of our function by differentiating and finding which numbers made the derivative zero. There were no numbers in the interval for which the derivative does not exist. We found that only one of the critical numbers, two, was in the open interval from one to three. And we evaluated our function at this critical number as well as the two endpoints one and three of the closed interval on which the function was defined.

The closed interval method told us that the largest value we obtained, zero, is the absolute maximum of our function over our interval and the smallest, negative a quarter, is the absolute minimum.

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