### Video Transcript

Find the intervals on which the function π of π₯ is equal to four π₯ squared times the natural logarithm of π₯ is increasing and decreasing.

Weβre given a function π of π₯, and weβre asked to determine the intervals on which this function is increasing and the intervals on which this function is decreasing. And whenever weβre asked a question like this, the first thing we should always ask is, what is the domain of our function π of π₯? And the reason for this is our function cannot be increasing or decreasing on an interval if itβs not even defined across the entire interval. So weβll start by finding the domain of π of π₯. We can see that π of π₯ is the product of two functions. Itβs four π₯ squared multiplied by the natural logarithm of π₯. Of course, four π₯ squared is a polynomial; we know itβs defined for all real values of π₯.

Next, we know the natural logarithm of π₯ is only defined if π₯ is positive. So the domain of π of π₯ will be π₯ greater than zero. And because weβre working with intervals, we might prefer to write this as the open interval from zero to β. To determine the intervals on which a function which we know how to differentiate is increasing or decreasing, we can use the following rules. We know if π prime of π₯ is positive on an interval πΌ, then π must be increasing on that interval πΌ. However, if π prime of π₯ is negative on an interval πΌ, then π must be decreasing on that interval πΌ.

So to determine the intervals on which our function π of π₯ is increasing or decreasing, we want to determine the intervals on which π prime of π₯ is negative and the intervals on which π prime of π₯ is positive. So weβre going to need to differentiate our function π of π₯. We know π of π₯ is the product of two differentiable functions. So weβll do this by using the product rule. So we need to start by recalling the product rule. The product rule tells us if π’ of π₯ and π£ of π₯ are differentiable functions, then the derivative of π’ of π₯ times π£ of π₯ with respect to π₯ is equal to π’ prime of π₯ times π£ of π₯ plus π£ prime of π₯ times π’ of π₯. We want to use the product rule to differentiate π of π₯. So we need to set π’ of π₯ to be four π₯ squared and π£ of π₯ to be the natural logarithm of π₯.

To apply the product rule, we need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Thatβs the derivative of four π₯ squared with respect to π₯. We can do this by using the power rule for differentiation. We multiply by our exponent of π₯ and reduce this exponent by one. This gives us π’ prime of π₯ is eight π₯. Next, we want to find an expression for π£ prime of π₯. Thatβs the derivative of the natural logarithm of π₯ with respect to π₯, which we know is the reciprocal function one over π₯.

Weβre now ready to find an expression for π prime of π₯ by using the product rule. We just need to substitute our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯ into our product rule. This gives us π prime of π₯ is equal to eight π₯ times the natural logarithm of π₯ plus one over π₯ multiplied by four π₯ squared, and we can simplify this. Remember, we showed the domain of our function π of π₯ is π₯ greater than zero. In particular, this tells us that π of π₯ is not defined when π₯ is equal to zero. So π wonβt have a derivative when π₯ is equal to zero, so we can just assume π₯ is not equal to zero and cancel the shared factor of π₯ in our second term. This gives us that π prime of π₯ is equal to eight π₯ times the natural logarithm of π₯ plus four π₯.

And we can simplify this slightly. Weβll take out the shared factor of four π₯ from our two terms. This gives us that π prime of π₯ is equal to four π₯ multiplied by two times the natural logarithm of π₯ plus one. Remember, to answer this question, we want to determine the intervals on which π prime of π₯ is greater than zero and the intervals on which π prime of π₯ is less than zero. And in particular, remember that π prime of π₯ will only be defined when π₯ is greater than zero. So we know that π₯ is positive. And we can use this information. Letβs take a look at π prime of π₯; we can see it has a factor of four π₯. Since we know that π₯ must be positive, four times π₯ must be the product of two positive numbers, so four π₯ is positive. So, in fact, the sign of π prime of π₯ is entirely defined by the sign of its second factor, two times the natural logarithm of π₯ plus one.

Thereβs a few different ways of determining where this factor is positive and where itβs negative. Since this is a continuous function, weβll just check where itβs equal to zero. So we need to solve the equation two times the natural logarithm of π₯ plus one is equal to zero. Weβll start by subtracting one from both sides of the equation. We get two times the natural logarithm of π₯ is equal to negative one. Next, weβll divide through by two. We get the natural logarithm of π₯ is equal to negative one-half. Then, to solve this equation, we need to recall the natural logarithm function is the inverse function of the exponential function. So if we raise π to the power of both sides of our equation, we just see that π₯ must be equal to π to the power of negative one-half.

And weβll use our laws of exponents to simplify this slightly. Instead of writing π to the power of negative one-half, we can do one divided by π to the power of one-half, and π to the power of one-half is the square root of π. So we get when π₯ is equal to one divided by the square root of π, two times the natural logarithm of π₯ plus one is equal to zero. And one way of visualizing this information would be to sketch a diagram. We want to plot π¦ is equal to two times the natural logarithm of π₯ plus one. We know the general shape of this function. It will have the general shape of the natural logarithm function. However, we stretch it vertically by a factor of two and then translate it vertically one unit.

This means we can sketch the following curve. It will be very similar to the natural logarithm function. And we want to know when this is positive and when this is negative, so when itβs above the π₯-axis or when itβs below the π₯-axis. So we need to know its π₯-intercept, and we already found this value. Itβs when π₯ is equal to one divided by the square root of π. So we can see from our diagram when π₯ is bigger than one divided by the square root of π, our function is positive. However, when π₯ is less than one divided by the square root of π, our function is negative. And remember, there is one small thing. Our function is only defined when π₯ is greater than zero. So we do need to be careful and make sure our intervals only have values where π₯ is positive.

We are now ready to use this information to determine the intervals on which π prime of π₯ is positive and the intervals on which π prime of π₯ is negative. First, we can see if π₯ is greater than one divided by the square root of π, π prime of π₯ is the product of two positive numbers, so π prime of π₯ is positive. However, weβve also seen if π₯ is between zero and one divided by the square root of π, then two times the natural logarithm of π₯ plus one is negative. So π prime of π₯ is the product of a positive number and a negative number, and therefore π prime of π₯ must be negative. Therefore, this just gives us the values of π₯ on which our function is increasing and decreasing.

However, we do have to be careful. The question wants us to write this as intervals. To do this, we just need to recall saying that π₯ is greater than one over root π is the same as saying that π₯ is on the open interval from one over root π to β and saying that π₯ is between zero and one over root π is the same as saying that π₯ is on the open interval from zero to one over root π. And then by combining all of this information, we get our final answer. Therefore, we were able to show the function π of π₯ is equal to four π₯ squared times the natural logarithm of π₯ is increasing on the open interval from one over the square root of π to β and decreasing on the open interval from zero to one over the square root of π.