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Use the guidelines of this section to sketch the curve.$$y=x \ln x$$

$(0, \infty)$

Calculus 1 / AB

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 4

Curve Sketching

Derivatives

Differentiation

Applications of the Derivative

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we are given the equation F of X is equal to x Ln x, and we need to sketch this function. So the first part we can think about is the domain. And since we have the natural log of X in our function, our function sort of inherits the domain of this function. And the main of this is from zero to infinity. The next thing we could talk about is the intercepts. So you can set this equation equal to zero and then sort of think about what values can this equal to zero, you might think What about X is equal to zero? Because we have X times Ellen of X and zero times. Anything is zero right? Well, zero is not included in our domain because if we plug in zero, we have zero times the natural log of zero. And we can't take the natural log of zero. So it's not so. It's not X. What about l N of X? Is equal to zero. Well, if we you can solve for X and we get X is equal to one because if we plug in X is equal to one didn't get one times the natural log of wine, which is zero. So it looks like we have one X intercept here, so you have X equal to one and then y is equal to zero. So what about the sea symmetry? We can test to see if it is even which the function is, even if the original function is the same as if we were placed x with minus X. So let's see what happened to get minus X times the natural log of minus X. It's not the same thing. So it is. The function is not even. What about odd is the function, so we can see if the result from our even function for even test if that is the same thing as negating the whole function. And it's not because you negate the whole function that we don't have this negative in the natural log. So it is not odd either. So from our results, it is not even and not odd. So we can. We have found a few bits of information from the original function. Now it is time to find the first derivative. Since this is a product, we're going to use the product rule F prime G plus f g prime So F prime f prime is just X primacy just one so that it is l a X plus f That's X times g prime. That is Ellen X, which is just I just one over x and became cancel. The X is out, so our derivative is L A X plus one. Now we can set this equal to zero to find any critical values. So that is Ln of X is equal to negative one. And to solve for X, we could just have X is equal to e to the minus one. So this is our point and that is just approximately. This is just approximately 0.368 So we can set up two intervals. We have negative infinity too e to the minus one and then from well, technically it would be from zero to ive to the minus one. Then from E to the minus one to infinity. Then we could have a test value. So let's say in this interval, let's say one, say 1/10 and then for this one, let's say to So this is our interval. This is our test and then This is the sign of though, when we played this back into the first derivative. So this the result from this this is negative. And for a two, this is positive. That means we are decreasing on intervals zero to e to the minus first. And then we are increasing on the interval E to the minus one to infinity. And then since we're decreasing, they're decreasing and increasing. That means we have a local minimum. So we have a have a local minimum at X value. Waas What was the X value e to the minus one? And then if we plug this back into our original function Well, technically B minus e to the minus one. Let me clean this up a bit. So let us focus on the second derivative. Now that is Oh, it's just l a X plus one. So the derivative of one's just zero Ln of X is just going to be one over X. And, of course, this has the same domain as our original function. And if we try to equal this 20 we a tent because one over X does not equal to zero for all for the values of X in our domain or really any value. Really. But we can also try Teoh Phil drive his konkey of upwards or downwards and it is actually con cave upwards. If you chose any number in our domain which is essentially positive values and you plug it into here, you'll always get conclave upwards. So Khan cave upwards from zero to infinity. So we can now finally graft dysfunction. We are told that there is one intercept 10 So one 10 There is a local minimum at this point. So it is before one so can say it is right here. So it is going to technically start at the origin, decrease until the minimum and then increase that what happens at this origin? Well, so why did I start the function Act zero. But we know that it is not the find at zero. So I am going to leave a hole at zero. But I know that if you take the limit of this function as X tends to zero from the right side. So from the positive of these function, X Ln x Well, this is where it's going to get a big complicated, so I can think of it as the same as limit extends to zero from the from the positive side of Ln X over one over X. Now, I'm not I'm not doing anything because I am doing These two statements are equal. Exelon X is equal to Ellen X over one over X. Now you might be wondering. OK, why? Why on earth am I doing this? But we're going to use a technique called low petals rule, and we can essentially differentiate both the top, the numerator and the denominator of this fraction. So the top of it is l A. Next that is just one over X and one over the derivative of one over X is negative one over X squared. Now we have a fraction over a fraction, so you can just all right over here. So we have the limit as X tends to zero positive. We have one over x times a negative X squared over one, and you might see where this is going. So limit as X tends to zero positive. So we have X, then x square, so we just have negative X and this is this is fine we just have a We just can plug in zero for here and then we are left with. Our answer is zero. That's why I start the equation at zero, because I use limits, and I used Low Patel's rules to figure out where it is starting from.

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