use-the-guidelines-of-this-section-to-sketch-the-curve-yx-ln-x-2

Question

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

Andrew Madea · Accepted Answer

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&#x27;t take the natural log of zero. So it&#x27;s not so. It&#x27;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&#x27;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&#x27;s see what happened to get minus X times the natural log of minus X. It&#x27;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&#x27;s not because you negate the whole function that we don&#x27;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&#x27;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&#x27;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&#x27;s say in this interval, let&#x27;s say one, say 1/10 and then for this one, let&#x27;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&#x27;re decreasing, they&#x27;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&#x27;s just l a X plus one. So the derivative of one&#x27;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&#x27;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&#x27;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&#x27;m not I&#x27;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&#x27;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&#x27;s why I start the equation at zero, because I use limits, and I used Low Patel&#x27;s rules to figure out where it is starting from.

Jamie M · Answer

So for the domain, we know that it&#x27;s gonna be, um X is greater than zero because it has to be first of all, ex cannot be zero because of first part. And the second part tells us that it has to be positive. So x is greater than zero intercepts. There are none symmetry, there is none. And for Assam totes, if we look for a horizontal, ask himto, for example, we have to set limits. Um, as X approaches infinity or limit as X approaches. Negative infinity. We see that there&#x27;s no horizontal. However, there is a vertical ass until at X equals zero. Because we set our denominator for one of her ex. We said the denominator to zero. So x zero is our, ah, vertical ass in two. And now we look for increasing decreasing intervals. So you find the first derivative y prime, and that is one of her ex minus one over X squared. And if we set our first derivative to zero, we confined critical points and we see that X is one is a critical point. So if we do a first derivative test, so this is for F prime and we substitute, um, and research for inside of our derivative. We&#x27;re gonna plug values that are smaller than one. What we have to remember to stay in our domain. So if it&#x27;s smaller than one, it&#x27;s going to be negative. If it&#x27;s greater than one, it&#x27;s gonna be positive. So that means that it&#x27;s decreasing until it gets to one and increasing after that. So if it&#x27;s going from negative to positive, so decreasing to increasing its gonna be men a local minimum value. And if you want to know the why value of our critical point we go F one. So we&#x27;re putting it back into her original equation to find one. So 11 is our minimum local minimum value for our graph. And now we can find, um, we can look at kong cavity for this graph by finding the second derivative. So why double prime equals negative, and this negative is on the outside of the entire fraction. So negative X minus two over X cubed. So if we set that equal to zero, we see that X is equal to two and now our second derivative test, so double prime, and this is too and we have to make sure to test values in our domain. So, for example, if it says 0.5 for over here so smaller than two. But in still inside of her domain, we&#x27;re going to get positive. So Khan came up, and beyond that, it&#x27;s gonna be Khan came down so negative. Okay? And now we can graph with the information that we got. So first thing I&#x27;m gonna put on the graph is the, um, vertical ascent. Tow X equals zero, and we can see that, um, we have a minimum value at 11 So let&#x27;s say this is one. This is one. So that&#x27;s our local minimum value. And, um, we see that we haven&#x27;t Oh, yeah. This is an inflection point because it&#x27;s switching signs from Khan gave up to conquer it down. So we can say that this is an inflection point at two. All right, so, um, say this is about to So this is a rough sketch of our graph. It&#x27;s gonna be decreasing until it hits our minimum value so it cannot touch the ass in tow. So decreasing until hits the minimum value, then it&#x27;s gonna increase from then on. So it&#x27;s so we have con cave up. That&#x27;s correct. And then it&#x27;s gonna be calling came down after it gets to our inflection permit. So it&#x27;s Kong cave up, but still increasing until it&#x27;s our inflection point that it&#x27;s gonna be con cave down but still increasing, and that&#x27;s a rough sketch.

Jamie M · Answer

So the domain for this function is X. It has to be greater than zero. Um, because we can&#x27;t have a zero because of the denominator. And we can&#x27;t have negative numbers because of the numerator intercepts. 10 symmetry. None. Essen totes. We have one X equals zero again because the bottom cannot equal zero. And we also have ah, horizontal one. Because the limit as X approaches infinity, um, of our function. Ellen of X over X squared is zero. So we have, um as X approaches infinity, we have ah, horizontal one at y equals zero. Okay. And next we confined the first derivative. So why praying? My prime equals one minus two ln x over X cute set that equal to zero. And we find, um the ex equals the square root of E, which is approximately 1.6. Um, so we have a max at 1.6 because the first derivative test tells us that, um, 1.6. So values below 1.6 will be increasing above will be decreasing if you test it in the first derivative. So, um, 1.6 is a max. So And by the way, the point is 1.6 comma, 0.18 if we substitute, um into our original function. So 1.60 point 18 that&#x27;s her first derivative test. And now we confined our second derivative, Furqan Cavaney. So our second derivative is why, um, double prime equals Negative. Negative. Six Ellen X plus five over X to the power of four. Set that equal to zero. Um, in Seoul for ex um, so you can see that X equals e to the five over six, which is approximately 2.3. Okay. And, um, f of 2.3 is zero. Oops. Up of 2.3 is about 0.15 zero point. So our second derivative test tells us that 2.3. If we test values, um, into the second derivative that are smaller, we&#x27;ll see it&#x27;s conk it down. So Colin Cave don&#x27;t. And then Khan gave up. If it&#x27;s above 2.3. So this is an inflection point at 2.3 comma, 1.0 point 15. Okay, so that&#x27;s all the information we have. And we can graft now. Okay. So we can graph our, um, Assam totes. So we have when X equals zero. So that&#x27;s our vertical one. And we also have one at why equals zero. That&#x27;s a horizontal one. So we have ah intercept at 10 So let&#x27;s say this is one 10 and we have, um, a Max at 1.6. So that&#x27;s about here. Um, in 0.18. So this is our max. This is just an approximate It&#x27;s the sketch, Um, and we have It&#x27;s gonna be increasing until it hits 1.6. So it&#x27;s gonna be increasing. Go through the, um, intercept. And once it has 1.6, it&#x27;s going to start decreasing. So now it&#x27;s gonna go down, but it&#x27;s still gonna be con cave up until we hit 2.3. So let&#x27;s say, um, if this is one, let&#x27;s say that so 012 this is about two. So 2.3 would be here, So it&#x27;s gonna, um although it&#x27;s gonna decrease its going to still be con cave down until about here, where it switches to con cave up. And although it&#x27;s conclave down, it&#x27;s gonna continue decreasing and approaching but never touching our ass in tow.

Jamie M · Answer

the domain for this function is, um X cannot be pie end. All right, intercepts. Um, there is every Scylla strait, this every, um, pie over to that. We see. Um, there&#x27;s gonna be one in each to pie, period, you know? All right, symmetry. There is none. Assam totes at X equals pi n. So the reason why the domain cannot have this is because is because there isn&#x27;t asking too. Um, at this value. And the reason for that is because sign pi is zero and Ellen zeros undefined, so we cannot have that. So now let&#x27;s do increasing decreasing intervals. So why prime is co sign Oh, CO sign of X over a sign of X. Is that that equal to zero? And we see, um, ex every pie over too, plus pi n. So if we do the first derivative tests for pi over too high over to, we&#x27;re going to see, it&#x27;s ah positive here. So increasing here and decreasing there. So it&#x27;s gonna be a increasing to decreasing. So this is a local max at every pie over to cause this is a periodic function or has repeats. So that&#x27;s for of prime. It&#x27;s right that in this is of prime. Okay. And now we can look at f double prime to see Con cavity. So why double prime? The second derivative is negative. Co sign squared X over science. Weird X. All right. Minus one equal zero eso We see an X at every pi over too goes to plan and x three pi over too. Plus two pi n and is an integer Okay, so this is for a double prime f double prime. This is the con cavity test. This is pi over too. This is three Pi over too. It&#x27;s always gonna be con cave down, down, down, always down. Okay, so let&#x27;s graft now. So, um, we have asked me, Toto, every pie. So if this is negative pie, this is pie. If this is to pie, this is negative to pie. Negative. Okay. And this is where our asking totes are as well as here. All right, so now we can see, um, a local max that pie over too. So pi over to zero. Is that point? So pi over too. This is negative player too. I&#x27;m sorry. It&#x27;s not here. Here. Okay, so it&#x27;s gonna be Kong cave down, so see how it&#x27;s increasing and then decreasing. And then also here, it&#x27;s gonna be conquered down con caved in and notice the space. There is nothing here, nothing here.

Jamie M · Answer

All right. So the domain for this one is X has to be greater than negative one, so we can keep our argument positive. And for intercepts, we have 00 So the origin symmetry, we have none. No symmetry. Um, assim toast. We have, um, one X equals negative one. All right, because our domain cannot be, uh, negative numbers because of the natural log. So that&#x27;s why we haven&#x27;t asked into it there. All right. And next up, we have to find increasing and decreasing intervals. So why prime my prime equals three x squared over one plus x cubed set that equal to zero. We have X equals zero is one of our critical points. All right, so first derivative test. So this is a crime. It&#x27;s test points around zero are critical point. And we see that it&#x27;s gonna be increasing always. Okay, So our function is always increasing. And if we find our second derivative now for the con cavity tests, why double prime equals three times to x minus x to the fourth on all over, um, one plus one plus x cubed. And that&#x27;s all squared. Set that equal to zero. And we come to to, um, critical points X zero and X equals the cube root of two, which is approximately 1.3. Okay, so second derivative test. So it&#x27;s double prime. So this is zero. This is, um, the cube root of two. We have, um, conk it down. Oh, my God. That&#x27;s come concave down here. Con cave up here and in Kong Cave down again. Okay, so these air both inflection points. And, um, just so we know the why value so f of the cube root of two. The y value is gonna be, um, ln of three. So that&#x27;s approximately 1.1. All right. And then it was just so we can graph it, leader. Let&#x27;s keep all of that over here. And now we can graft. So we know we haven&#x27;t asked him to put that in in green. It&#x27;s gonna be a negative one. So this is negative one. We have a vertical ass. Until here, we haven&#x27;t intercepted 00 We have, um, it&#x27;s gonna be Kong caves down until it reaches zero. So Kong cave down until it reaches zero, and it&#x27;s always gonna be increasing. And then once it reaches zero is gonna be con cave up until it hits our inflection point. Um, so let&#x27;s say this is, um 1.3. So this is if this is 11.3, be about here and then if this is one 1.1 1.3, that&#x27;s it. Right there. So that&#x27;s our other inflection point. So it&#x27;s gonna be con cave, um, up until it hits her inflection point, and then it will be conquered down the rest of the way up.

Andrew Madea · Answer

so we&#x27;re given the equation. F of X is equal to e equals Teoh X times e to the minus X and weaken. Just rewrite this as X over e to the X sketch this We can easily find out the domain of this function, and you might have an instinct to find to set the denominator equal to zero e to the X is equal to zero and solve for X. But if the X will never equal zero for all X, that&#x27;s why the domain is actually negative. Infinity to infinity or all real numbers. So X is old real numbers. The we can figure out the intercepts so we can solve for the X intercept by setting it equal to zero. And we could essentially just look at the numerator when we set it equal to a number. So we have X is equal to zero. So if I plug in zero into this equation roundup just getting zero back, and if we plug and that will be our only intercept isn&#x27;t we tried to find the y intercept, it will have to plug in zero, which we already 10. If we did so, this will be our only intercept. Now we can focus on the symmetry of dysfunction so we can do the even test. So is our function the same when we replace X with minus X? Well, we have minus X over E to the minus FEC&#x27;s and it&#x27;s not the same thing as the original function. What about the odd test? So it is not even our test is. Do we get the same results from the even tests? If we just negate the entire equation? And if we negate the entire equation, it just becomes negative. X over 82. The X and these two are not the same thing. So it is not even. It is not odd, so there is no symmetry. What about Assam toads? Well, there are no vertical asking totes because our denominator doesn&#x27;t really equal zero. So what about horizontal asking toads? So I&#x27;ll just abbreviated as h A. And you can think of it as sort of take the limit as X tends to infinity infinitely for our equation. X over eighth e x even think about it as what if we just constantly increase the value of X like, let&#x27;s say we have five over E to the fifth. Well, the numerator is going to increase, but the denominator will explode vastly increased much faster than the new than the numerator, because each of the fifth is vastly larger than the number five, and we can continue this trend. So, like if we have X is equal to 20 then we have 20 over E to the 20 which is fastly larger so it&#x27;s can equal to zero, since the denominator will always be will be very larger, then be numerator. So what about what if we decrease the numbers significantly so very large negative numbers? So let&#x27;s say if we want. So let&#x27;s take the limit as extends to negative infinity for our equation. What is this equal to? Well, let&#x27;s just say negative five. Let&#x27;s say exited with a negative five. So it will be negative five over you to the minus five. So Ethan, minus five is going to be well, less than one. So that means negative five over eat. The negative five will be a very via large negative value. In fact, this is actually equal to, well, approximately negative. 700 and 740. And if we do this for negative six, then it is. And if we keep decreasing the number from negative five to negative six, then that is around negative. 2000 420. So you can see that is getting much, much, uh, less. It&#x27;s decreasing significantly, so this limit will be negative. Infinity. So we know the end behavior of this function as well as that. There is a horizontal asking tote on for the essentially the right side of the function. Let&#x27;s focus instead of the original equation. But let&#x27;s try to take the derivative of our equation. So we&#x27;re given this equation. So it is a quotient so we can use the quotient to rule and just a reminder. But the question to rule appear so it&#x27;s F g f prime G minus F g prime over g swear. So G square is e to the X squared. And we could solve this as each of the two X but you might want to keep it like this. So f prime is just ex prime, which is just one. So that is just e to the X minus f then g prime in that just the same thing as e to the X. We noticed that there&#x27;s an E to the exit each term in the numerator, so we could, in fact, to that out. So that&#x27;s one minus X over E to the X squared. If we can cancel out the E to the X in the square, and then this is equal to one minus X over E to the X, and then we can set this equals zero to find our critical point. And then we can just focus on the numerator portion. So it&#x27;s one minus X equals zero, and then it just X is equal to one. So this is our critical point, and with this we can figure out a few intervals. So the interval of negative infinity toe one and then from one to infinity so we can get a test value from each interval. So the test value we could use, let&#x27;s say zero and two. And what is the sign when we plug this value back into the first derivative? Well, we have zero, so it&#x27;s one minus zero. That&#x27;s just 1/1. So that is positive. About two. It&#x27;s one minus two. That&#x27;s a negative and then you know that each the X is always gonna be positive. So it&#x27;s negative over positive, which is negative. So that means it is increasing on this interval and it is decreasing. On this interval, we could plug back in one to the original function to figure out the extreme of it to evaluate any extreme. And it looks like it is increasing and then it is decreasing, so it looks like it&#x27;s going to be a maximum. So we have a local Max at one comma. When we plug one back in its one over E to the one, I&#x27;ll just leave that as e to the minus one. Then we have our intervals and I&#x27;ll just clean this up for enough in a second. After we were done with the first derivative, we can figure out the second derivative. Do you figure out points of inflection, end concave ity And again, this is a quotient. So we&#x27;re going to use the quotient rule. So it is Eat the X squared and you might guessed it. We were going to leave it like this, so f prime is just negative. One times e x minus beaches. I&#x27;ll keep this in parentheses, one minus X, and either the ex prime is just a to the X, then weaken. Distribute at this E to the X first, so it&#x27;s minus E to the X minus a to the X minus x e to the X over e to the X squared. Then we can distribute out this negative. No distribute out here, so it&#x27;s negative and plus, so we have minus E to the X twice. So we have X E to the X minus two easily x over E to the X squared, and then we could factor out and e to the X. So it&#x27;s X minus two over each of the X squared. You can cancel those out and then we are left with X minus two over E to the X and then to find the point of inflection, we can just set it equal to zero. You can only really care about the numerator and then set that equal to zero. Solve for X, which were allowed with X, is equal to two. And then to find concave ity, we can have to have two intervals negative infinity to to to to infinity for test values. We can have zero in three. So our interval are a test value and our sign. So when we plug zero back into our equation here zero minus two, that&#x27;s negative. Over a positive that negative and then three minus two That will be positive over a positive, which is positive. So this will be con cave downwards and this will be con cave upwards. So that means this point and we have to figure out the Y coordinate of that point is a point of inflection of in reflection. So oh, I will organize this. So, for the part of inflection we have to do is put it back into the equation, which is so we have to over e squared and we can just leave it like fat so to over East squared, or is to e to the minus two. And then we can finally graph our equation of X over e to the X so I can just draw that in some blue. So the intercepts, the only intercept is at 00 the local maximum one comment e to the, uh, minus one. So there&#x27;s one. I will say it right there is easily minus one and then we know it will have. It will start at Assam tote when it is very increasing and it is coming from this way because you know, for the end, behavior sort of looks like more linear. So it&#x27;s increasing for infinity to this point. Touches it and then it is decreasing and approaching zero but will never reach zero. Oops. And it will never reach zero. And this is our function f of x legal two x e to the minus x.

Problem

Use the guidelines of this section to sketch the …

Problem 37 Hard Difficulty

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

Answer

$(0, \infty)$

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James Stewart

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$(0, \infty)$

Essential Calculus Early Transcendentals

Anna Marie V.

Kayleah T.

Michael J.

Joseph L.

Precalculus Review - Intro

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James StewartEssential Calculus Early Transcendentals

Anna Marie V.

Kayleah T.

Michael J.

Joseph L.

James Stewart

Essential Calculus Early Transcendentals