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University of North Texas

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Problem 27 Medium Difficulty

Use the guidelines of this section to sketch the curve.

$ y = \frac{\sqrt{1 - x^2}}{x} $

Answer

see solution

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Video Transcript

we want to sketch the curb. Why is equal to the square root of one minus X squared all over X. So in this structure they give us this laundry list of steps that we should follow any time we want to sketch a graft. So let's just go in and follow those. So the first thing is, they tell us to determine our domain, so this might take a little bit of space. Let's do this on a new page. So one minus X were over X. So their students we need to make sure of. We need to make sure that our demand there is not equal to zero. And we need to make sure that the inside of our radical is greater than or equal to zero. So at least we know X. So we at least know X just can't be zero and define the rest of our domain. We could go ahead and find where one mice expert strictly larger than zero. So doing this would give one sugar, then X squared, and we can do negative one square root of negative one on that side. But a positive one, It's that X plus. Then negative once we square root inside and then just put the negative. All right. So from these two things, since this here tells us all the values that sex can be and we know ex cannot be equal to zero. So we just remove that and we get negative 1 to 0 union, 0 to 1. What? So our domain ends up being negative 1 to 0. Union 02 The next step is to find our intercepts and her lips. So X intercept is when we said function equal to zero. So we'll have zero is equal to one minus x squared square root all over X. And remember, the denominator could never make the function equals zero. So we would do. Zero is equal to the square root of one bias X squared. So the square root each side Add the X word over and takes where roots again, and we'd get extra legal to plus or minus one. And since zero is not included in our domain, we know we will not have a why interested. So we don't need to salt for that. So now symmetry. Well, any kind of rational function like this week don't leave normally think is gonna be periodic. So let's check what f of negative X is to see if this function is even or odd. So we're gonna have a square root of one minus so negative X squared, Will Squaring a negative is just going to be positive. So that would just be X squared all over negative X. So I noticed that this here is equal to negative FX. So we know our function is going to be odd or has symmetry about our origins. So we might be able to use this to help simplify some parts and skip a couple of steps as well. The next thing they tell us to look for his passage tubes. Well, we know we're not. We're gonna have any end behavior as exclusive in your negative infinity since we're bounded between negative 11 or domain. But we do have a horizontal acid toe since when I plugged her a vertical Asako. Since when I plug in zero, the denominator become zero. So let's go ahead and look at what happens as the limit as X approaches zero from the right at first, so ever X. But I know I change that, but someone that in green after banks. So we're going to have one minus zero from the right squared over zero from the right. Now, if I square something slightly to the right of zero, well, that's still going to be a very small number that's close to zero. And if I do one minus that, it's so going to be positive. Um and square bring a positive. It will give us something positive and zero from the right is going to be positive. So positive positive is going to deposits of this function will go to positive and now we can use the fact that we know this function is odd or has symmetry about the origin to see that the limit as X approaches zero fromthe left is going to be negative Infinity, since we would be plugging in numbers closer and closer to zero, but that are negative. So using the fact that it's odd, we can just go ahead and do that in one step and not have to go through those same steps as we just did the next things we want to do, which are steps five and six, or to determine where the function is increasing and decreasing, as well as find any maximums or minimums we may have on, he should be local. All right, so for this we're gonna need to figure out what why Prime is equal to. So let's do this on another page. So again, why is equal to the square root of one minus right that would look better one minus x squared all over x. So to take the derivative of this, we're going to need to use questionable. So remember, caution rule is low d high minus a little all of the square below, where those D's are supposed to be derivatives. And now I'm going to rewrite the square root is 1/2 power. So one minus X squared to the 1/2 and then minus the's in the opposite order, one minus x squared, rebooted, then deep dx of ex all over what we have in the denominator squared now the derivative of one minus X squared to the 1/2 power we need to use power and changeable. So it's gonna be 1/2 one minus X squared now to the negative 1/2 power and the derivative of the inside of one by a sex. Where is going to be negative to X and the derivative of X is just going to be one. Now we can use a little bit of algebra to simplify this down, too. Negative one over X squared times the square root of one minus X squared. Now notice that this will never equal to zero, so we can't find any critical values that way. But we could possibly have something when we set the denominator equal to zero. And doing that would give X squared is equal to zero or one minus. X squared is equal to zero, but that tells us x 00 or x zero to plus or minus one. And so we can get rid of X is equal deserves. Since our original function is undefined at this point and we could get rid of X is equal to plus or minus one due to the fact he's being end points on our domain. So we really don't have any, um, critical points for this one. So let's go ahead and just right. Why prime down so again? That should the negative one over X squared times thes square root of one minus X squared. Now, actually, looking at this, you might notice this is going to be strictly less than zero, since X squared is always going to be bigger than or equal to zero. And the square root of one mind sets were also has to be larger than or equal to zero. And since we have that negative in the numerator will always be negative. Or you can go about how you would normally solved for these to see that as well. But we can just kind of make that little observation and say that our function will always beat it be seen or why prime, it's less than zero on our entire domain. So negative 1 to 0 union, 0 to 1. And then that tells us why Prime is strictly larger than zero or increasing. Never. All right now, the last thing they tell us we should do before we actually start graphing is to find where are functions Conkey about the Concord down and any points of inflection we may get. So we need to know what? Why double time is going to be so we have why prime is equal to so Let's go ahead. Use question Rule again to take this derivative so we'll have low, which is X squared times one minus X squared, minus times the derivative of negative one minus in the office order of negative one ties the derivative the perspective X of X squared times one minus X squared all over what we have in the denominator squared. So they become X to the fourth times, one minus X squared and only when we get rid of that square route we will need to put absolute values here. But we at least know on our domain that will never be less than so. First, the derivative of negative one is going to be zero. So this entire term here is zero. Also, hear these negatives council out. And now we just need to use product will to take the group of expert times the square root of one plus expert. So after need to just write that down here they'll be by DX of X squared times one minus expert to the 1/2 power. So this is going to be so remember product will says, take the derivative of the first multiplied by the second and then add the in the opposite or so two x one minus X squared to the 1/2 power plus on the amount a little bit. Now, if the X squared times the derivative of one mind sex, where to the 1/2 which is going to be 1/2 times one minus X squared now to the negative 1/2 times he drove it on the inside there's gonna be negative two x So we're gonna replace this with that. And once you do that and do a little bit of algebra, we can rewrite this as three x So this is why I double prime and that she'd been Don't find out there. Does he go to three X squared minus two all over X cubed times one minus X square to the three house power. And so we're gonna need to go ahead and set this year equal to zero. And doing that will tell us unless we actually do that in green instead. So that will tell us we need our numerator to be good to zero and solving for that we'd get X is equal to two thirds. This should be, uh, squared three X squared minus two. So actually end up with this being X square root of six over all. Right now we can go ahead and look to see where the functions undefined also. So we want to make sure denominator music with zero at certain points for possible critical values so again that's gonna be excused is equal to 01 minus X squared is equal to 3/2 since he got to zero. But for similar reasons Over here we will end up where these don't actually give critical value since it's either undefined or in points. So the only point of interest we actually get out this is X is equal to the square root of six over three. All right, so we have by double crime is equal to three x flared, minus two all over X cubed times one minus X squared Jude three House power. Now let's look at where the function or where our second river district largest zero or con cave up and this is going to be on and actually let me go back for a pass because I got one thing, so it shouldn't just be X is equal to that square issue B plus or minus. That's since we should have two solutions on our interval. All right, so we can go ahead. And now, actually, put that this interval here where it should be, should be negative. Square root of six over 3 to 0 and the Union Square room six over 321 And our function is going to be con cake down. Or why they will prime It's strictly less than zero on the rest of our film. So negative one to the negatives were routed six over three Union zero to the square root of six over. So now the only possible point inflection that we got was ecstasy. Zero for actually we are possible points of infection with those two. So let me do that. So we had X is equal to negative square root of six over three and X is equal to the square room, six over, so to be left of negative square root of six over three. The functions conch aid down and from zero to the square root of six over three. Function is also calm. Kate down and to the right of negative square root of three 21 20 The function is Kong Kate up and after the square root of six over three Function is also Khan caged up. So both of these have our points of an election since we have that change in contact a little bit. And now, once we did this, they said we're good to go ahead and actually scheduled ground. So look for her intercepts which end up being negative. 11 negative one and one. We have an ass in tow at X is equal to zero and to the right of zero, the function should be going to infinity and to the left. The function should be going too negative. Infinity, We have no Max is airmen's. But we at least know the function should always be decreasing on our domain. And we have points of inflection at negative square root of six over three and square root of six square root of six over three like that. So both of these are points of an election. All right, so now let's go ahead and start graphing. Let's start to the right of this function here, Archer, Let's start at exit one. So we know from Sward of six over 3 to 1. The function it needs to be con cave. Oh, so that means it's going to look something kind of like this until we get here. And then after that, it's going to be con cave down as we approach. And it'd be something kind of like that there. Actually, I think I have the ups and downs for this little backwards. So, yeah, I went ahead and wrote them down to help sports. Actually, this should be Khan Cave down from one to your point of inflection and we go up like this, you just go ahead and change these really fast. So this should be down and they should be up. So if you would have did algebra, you should have got the opposite of what I had said in this step here. Uh, but at least when we start to drop it, it really wouldn't have made much sense for what our function was going to look like. So that was the easy way to catch that. What? And then next, when we start from negative activity, we're going to go until our point of affliction, and on this part we know so from negative swear six over three 20 they should be concrete down. So it should look something like this. And then after this point is going to be Khan caged up so we can go ahead and go back and possibly put in what these two points are for the points of inflection. But since we're just sketching it, I think this here is sufficient unknown.