In this problem, we're asked to find the absolute maximum and minimum values of the function over the given interval. Now the function that we're working with here is f(x)=x⋅ln(x) over the given interval from 1 to 2. With both of those endpoints included, this is a closed interval indicated by those square brackets. So based on the extreme value theorem and what I see with this function and its closed interval, I know that this function does have an absolute max and an absolute min, which I can find by starting with my steps here. Now the first thing I want to do is find my critical points.
That's where my derivative is either equal to 0 or does not exist. So the first thing I want to do is find my derivative f′(x). Now since I have one function, x, multiplying another function, the natural log of x, I want to use my product rule, which if we remember our little memory tool here, that's a left d right plus a right d left. So I have my left-hand function, that's x, times the derivative of that right-hand function. The derivative of the natural log of x is 1x.
Then I'm adding that together with my original function over here, natural log of x times the derivative of x, which is just 1. Now I can do some simplification here. These x's cancel, leaving me with 1+ln(x) because natural log of x times 1 is just the natural log of x. So here I have my derivative. Remember, we want to look for where this derivative is equal to 0 or where it does not exist.
Now the place where this derivative does not exist is where x is equal to 0 because we can't take the natural log of 0. Right? But 0 is not within that interval. So even though one of my critical points is x equals 0, it doesn't matter because it's not in that interval. So we now just need to set this equal to 0 and find our other critical points.
So if I go ahead and subtract 1 from both sides, that gives me ln(x)=−1. And if I go ahead and raise both sides with e, these e and natural log cancel out, leaving me with x=e−1, which we could also rewrite this as 1e. So this is our other critical point. We have found both of our critical points. We know that one of them is already not one we're going to use.
And from here, we want to plug our critical points and our endpoints into our original function. Remember, we need to double-check that our critical points are in that included interval. It can be hard to tell if something like one over e is within a specified interval. But if we go ahead and type one over e into our calculator, that's going to give us a value of about 0.37, which is not within that established interval. So I don't need to plug this critical point in either.
And all I need to do here is actually just plug my endpoints in because none of my critical points are within my interval. So I'm going to go ahead and plug my endpoints in. That's x equals 1 and x equals 2. So plugging those in to my original function, which is x times the natural log of x, this gives me here f(1) as a 1 times the natural log of 1, which is going to give me a value of 0. And then for 2, plugging 2 in, that's 2 times the natural log of 2.
Typing that into my calculator gives me about 1.39. So I only have 2 values here. One of them is the largest. One of them is the smallest. That's going to give me my max and my min over this closed interval.
So I can see here that this 1.39 is my largest value at x equals 2. So this represents the global max of my function over this interval. And then at x equals 1, my function value is 0. That's my smallest value. So this is my global minimum.
So we have our global minimum of 0 at x equals 1, our global max of 1.39 at x equals 2. Feel free to let us know if you have any questions, and I'll see you in the next video.