### Video Transcript

Find the absolute maximum of the function π of π₯ is equal to the natural logarithm of π₯ to the fifth power divided by two π₯ between π₯ is equal to one over π and π₯ is equal to π squared.

The question is asking us to find the absolute maximum of this function between the values of π₯ is equal to one over π and π₯ is equal to π squared. And we know the absolute maximum means the largest possible output of our function π of π₯. And in this case, since weβre only looking at values of π₯ between one over π and π squared, weβre only interested in values of π₯ in the closed interval from one over π to π squared.

So, to find this absolute maximum, we first notice that π of π₯ is equal to the natural logarithm of π₯ to the fifth power divided by two π₯. We notice the numerator of π of π₯, the natural logarithm of π₯ to the fifth power, is the composition of continuous functions. This means itβs continuous on its domain.

So, the numerator π of π₯ is continuous. The denominator π of π₯ is a linear function, which is also continuous. This means our function π of π₯ is the quotient of continuous functions. And we know the quotient of two continuous functions will be continuous across its entire domain.

We want to show that π of π₯ is continuous on the closed interval from one over π to π squared. So, to check this, we need to find the domain of our function π of π₯. So, letβs check the domain of our function π of π₯. First, we see π₯ cannot be equal to zero because then we would be dividing by zero. Next, we see since weβre taking the natural logarithm of π₯ of the fifth power, we canβt have π₯ to the fifth power is less than or equal to zero. And we can then see that π of π₯ is defined for all other values of π₯. So, the domain of our function π of π₯ is all values of π₯ greater than zero.

In particular, this means our function π of π₯ is continuous on the closed interval from one over π to π squared. So, the question is asking us to find the absolute maximum of a continuous function on a closed interval. And we know how to do this. We can find the absolute maximum of a continuous function on a closed interval by using the following three steps.

First, we want to find all the critical points of our function π of π₯. And we recall we say thereβs a critical point of π of π₯ when π₯ is equal to π if π prime of π is equal to zero or the derivative does not exist when π₯ is equal to π. Second, we want to evaluate π of π₯ at all of our critical points. Third, we need to evaluate π of π₯ at the endpoints of our closed interval. Then, we just compare all of these evaluations. The largest one will be the absolute maximum of our function π of π₯ on this closed interval.

So, to find the absolute maximum of π of π₯, weβre going to need to find an expression for π prime of π₯. Since π of π₯ is the quotient of two functions, weβll do this by using the quotient rule. We recall the derivative of π’ of π₯ divided by π£ of π₯ with respect to π₯ is equal to π’ prime of π₯ times π£ of π₯ minus π’ of π₯ times π£ prime of π₯ all divided by π£ of π₯ squared.

In our case, the function in our numerator, π’ of π₯, will be the natural logarithm of π₯ to the fifth power. And weβll simplify this expression by using the power rule for logarithms. The natural logarithm of π₯ to the fifth power is equal to five times the natural logarithm of π₯. Now, we can easily find an expression for π’ prime of π₯ since we know the derivative of the natural logarithm of π₯ with respect to π₯ is just equal to one over π₯. So, we get π’ prime of π₯ is five over π₯.

We then have that π£ of π₯ will be the function in our denominator. In this case, thatβs two π₯. And since this is a linear function, its derivative with respect to π₯ will just be the coefficient of π₯. So, we get π£ prime of π₯ is equal to two. Weβre now ready to find an expression for π prime of π₯ by using the chain rule. Substituting in our expressions for π’ of π₯, π’ prime of π₯, π£ of π₯, and π£ prime of π₯. We get that π prime of π₯ is equal to five over π₯ times two π₯ minus five times the natural logarithm of π₯ multiplied by two all divided by two π₯ all squared.

And we can simplify this expression. First, five over π₯ multiplied by two π₯ is equal to 10. Next, five times the natural logarithm of π₯ multiplied by two is equal to 10 times the natural logarithm of π₯. We can also simplify our denominator. Two π₯ all squared is equal to four π₯ squared. Next, weβll cancel out the shared factor of two in our numerator and our denominator. This gives us five minus five times the natural logarithm of π₯ all divided by two π₯ squared.

The last thing weβll do is take out the factor of five in our numerator. And this gives us π prime of π₯ is five times one minus the natural logarithm of π₯ all divided by two π₯ squared. Now that we found an expression for π prime of π₯, weβre ready to apply the first of our three steps to find the absolute maximum of our function over this closed interval.

The first step tells us we need to find the critical points of our function. Thatβs where the derivative is equal to zero or where the derivative does not exist. Letβs start by finding all the places where our derivative does not exist. We see from our definition of π prime of π₯ if π₯ is equal to zero, then weβre dividing by zero. So, our derivative will not exist. And we also see since weβre taking the natural logarithm of π₯, we canβt have π₯ is less than or equal to zero.

Equally we can see if π₯ is greater than zero, then we can take the natural logarithm of π₯. We can then subtract this from one, multiply it by five, and then divide it by two π₯ squared. So, π prime of π₯ exists for all values of π₯ greater than zero. In particular, this tells us that our derivative exists for all values of π₯ in our closed interval from one over π to π squared. So, the only critical points we could possibly get are when the derivative is equal to zero.

So, letβs now look for critical points where our derivative is equal to zero. A quotient of two numbers can only be equal to zero if the numerator is equal to zero. So, letβs solve our numerator is equal to zero. We get five times one minus the natural logarithm of π₯ is equal to zero. And we can then just solve this equation for π₯.

First, we divide through by five. We get one minus the natural logarithm of π₯ is equal to zero. Then, we add the natural logarithm of π₯ to both sides to get one is equal to the natural logarithm of π₯. Finally, we just raise π to the power of both sides of this equation. And since π to the power of the natural logarithm of π₯ is just equal to π₯, this gives us that π₯ is equal to π.

So, the numerator of our quotient when π₯ is equal to π is zero. Itβs worth noting if we were to substitute π into our expression for π prime of π₯, the denominator would not be equal to zero. So, weβve shown that π of π₯ has a critical point when π₯ is equal to π.

Remember, we still need to check that π is in our closed interval from one over π to π squared. And of course, we know π is in this closed interval; one over π is smaller than π, and π squared is bigger than π. So, weβve completed the first step. Weβve found all of the critical points of our function π of π₯ on our closed interval. There was only one when π₯ was equal to π.

Next, we need to evaluate our function π of π₯ at all of our critical points. And we need to evaluate π of π₯ at the endpoints of our closed interval. So, we need to find π evaluated at π, π evaluated at one over π, and π evaluated at π squared.

Letβs start with π evaluated at π. We substitute π₯ is equal to π into our expression for π of π₯. We get the natural logarithm of π to the fifth power divided by two π. Using the power rule for logarithms, the natural logarithm of π to the fifth power is equal to five times the natural logarithm of π. But then, the natural logarithm of π is just equal to one. So, our numerator simplifies to give us five. So, π evaluated at π is equal to five divided by two π.

Letβs now find π evaluated at one over π. Using our laws of exponents, weβll write one over π as π to the power of negative one. Substituting this into our expression for π of π₯, we get the natural logarithm of π to the power of negative one raised to the fifth power all divided by two times π to the power of negative one. We now want to simplify this expression. Weβll start by multiplying both the numerator and the denominator by π.

In our denominator, we get π to the power of negative one times π is equal to one. So, our denominator simplifies to give us two. Next, by using our laws of exponents, π to the power of negative one all raised to the fifth power is the same as saying π to the power of negative five. So, our numerator simplifies to give us the natural logarithm of π to the power of negative five times π all divided by two. Finally, by using our laws of logarithms, the natural logarithm of π to the power of negative five is just equal to negative five. So, weβve shown π evaluated at one over π is equal to negative five π divided by two.

Lastly, we need to evaluate π at π squared. We get the natural logarithm of π squared raised to the fifth power divided by two π squared. And we simplify this as we did before. π squared raised to the fifth power is π to the 10th power. And then, the natural logarithm of π to the 10th power is 10. Finally, 10 divided by two π squared is equal to five divided by π squared.

The largest of these three terms will be the absolute maximum of our function on our closed interval. We could do this by using our calculator. However, we can also notice it canβt be π evaluated at one over π since this is a negative value. Next, we can see the only difference between π evaluated at π squared and π evaluated at π is the denominator. And since we know π is bigger than two, we also know that π squared is greater than two π.

Well, what does this mean? Since the numerators are equal and positive, when we divide by π squared, weβre dividing by a bigger positive number. This makes our number smaller. Therefore, since five divided by two π is the biggest of these three values. It must be the absolute maximum of our function π of π₯ is equal to the natural logarithm of π₯ to the fifth power divided by two π₯ between π₯ is equal to one over π and π₯ is equal to π squared.