asymptotic discontinuity discontinuity and intuitively you have an asymptote here if you it’s a vertical asymptote at x equals two if I were to try to trace the graph from the left I’d be I would just keep on going in fact I would be doing it forever because it would be it’d be infinitely it’d be unbounded as. How To Discontinuity Calculator Symbolab! how to find the points of discontinuity tutorial, step by step. A discontinuity is a point at which a mathematical function is not continuous. Given a one-variable, real-valued function y= f (x) y = f ( x), there are many discontinuities that can occur.

## Removable Discontinuity 1

Click on the graph either to the left or to the right of the removable discontinuity (hole). Drag toward the removable discontinuity to find the limit as you approach the hole.The discontinuity is said to be of infinite type if at least one of the two limits is infinity. We illustrate this concept with the help of following examples The discontinuity is said to be oscillatory when the limits oscillate between two finite quantities.The Canon Discontinuity trope as used in popular culture.discontinuity to find how removable. Who we are Nest is your destination removable discontinuity how to find for design. rock band 3 squier stratocaster discontinued.

To find the value, plug in into the final simplified equation. is the point of discontinuity. Listing Of Websites About how to find points of discontinuity.The point of discontinuity refers to the point at which a mathematical function is no longer continuous. If you are in an Algebra II class, it is likely that at a certain point in your curriculum, you will be required to find the point of discontinuity.Hi, I have a parser and a function: const parser = new math.parser() parser.Find 22 ways to say DISCONTINUITY, along with antonyms, related words, and example sentences at Thesaurus.com, the world’s Roget’s 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group. TRY USING discontinuity. See how your sentence looks with different synonyms.

## How do you find discontinuity of a piecewise function?

In most cases, we should look for a discontinuity at the point where a piecewise defined function changes its formula. You will have to take one-sided limits separately since different formulas will apply depending on from which side you are approaching the point.Decimal to Fraction Fraction to Decimal Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time. Function Discontinuity Calculator. Find whether a function is discontinuous step-by-step.Details: How to find discontinuity of a piecewise function Because the functions are often used to model the phenomena of the real world How. Details: If the discontinuity is removable, find a function g that agrees with f for x ≠ x 0 and is continuous on R. (i) f (x) = (x3 + 64)/ (x + 4), x0 = -4.Discontinuity definition: Discontinuity in a process is a lack of smooth or continuous development . | Meaning, pronunciation, translations and examples. The text is good in parts, but suffers from discontinuity. There may appear to be discontinuities between broadcasts.

to how points discontinuity find of and continuity. Can i wear contact lenses after pupil dilation? These are excellent methods to find out important figures. You are just going to pay alot more for quality meats in the city.removable to non find discontinuity how. Reeds Ingevulde Belastingaangifte 2013 Downloaden.asymptotic discontinuity discontinuity and intuitively you have an asymptote here if you it’s a vertical asymptote at x equals two if I were to try to trace the graph from the left I’d be I would just keep on going in fact I would be doing it forever because it would be it’d be infinitely it’d be unbounded as.T

## Is a horizontal asymptote considered a discontinuity? : learnmath

Horizontal asymptote? No, how could it? Where’s the discontinuity? I know the best way to learn is doing tons of exercises and i will keep doing that. But i really need to find a way to take good and helpful notes.All points of discontinuities are divided to points of discontinuities of first and second kind. Our online calculator, built on the basis of the Wolfram Alpha system, calculates the discontinuities points of the given function with step by step solution.Classification of Discontinuities. Similarly to the one-sided limits, we can define one-sided continuity. Infinite or Essential Discontinuity: if one or both of the one-sided limits don’t exist or are infinite.Typical discontinuities found in ingots are non-metallic inclusions – bubbles, pockets, or cavities on the surface of the ingot.

Click again to stop watching or visit your profile/homepage to manage your watched threads.Classification of discontinuities. Quite the same Wikipedia. Just better. in an essential discontinuity, oscillation measures the failure of a limit to exist.limit is constant.How to find discontinuity? Asked by: Marshall Jones. Score: 4.3/5 (57 votes). Start by factoring the numerator and denominator of the function.Google will attempt to find other copies of the same image on the internet, as well as provide visually similar images.

## How To Find Discontinuities – 09/2021

VRoumy immobilier 35. How to find the discontinuity of a function.. What are the symptoms of streptococcus pyogenes. Adam ondra change video over cellular. Why do i hate getting up in the morning. How to find the discontinuity of a function..Similar to point discontinuities, jump discontinuities exist when entire portions of the curve jump instead of a normal curve having a single-point To find the point of a discontinuity, factor the function’s denominator and numerator. How to Tell If a Function Is Continuous or Discontinuous.How to Get Participants For Your Study. Regression-Discontinuity Design. Other Quasi Designs. The regression-discontinuity design. What a terrible name! In everyday language both parts of the term have connotations that are primarily negative.

to how continuity a of find discontinuity and function. Shop new and classic BJJ gear from 93brand. How to get a discount on Cobone.How To Discontinuity Calculator Symbolab! how to find the points of discontinuity tutorial, step by step. A discontinuity is a point at which a mathematical function is not continuous. Given a one-variable, real-valued function y= f (x) y = f ( x), there are many discontinuities that can occur.Our professor wants us to find the discontinuity in an array of data points that he gives us. We have to find the points where the discontinuity occurs and show the result in the command window without using the diff() function.The point of discontinuity refers to the point at which a mathematical function is no longer continuous. If you are in an Algebra II class, it is likely that at a certain point in your curriculum, you will be required to find the point of discontinuity.

## Vertical Asymptotes… How? (NancyPi)

Hi guys! I'm nancy.. And i'm going to show you how to find the vertical asymptotes of a rational function. If you don't know what i mean by rational function just ignore that, you're probably in the right place. Actually all this stuff is a lot easier than it sounds. So don't worry. It's gonna be fine.

let me show you. Ok, so say you need to find the vertical asymptote or vertical asymptotes, if there's more than one. First of all.. What is a vertical asymptote? It's basically an invisible vertical line that your graph approaches and gets really really close to but never actually touches. Never quite reaches.

so how do you find the vertical asymptote? Well, if you have a rational function, the fraction form type, which is the most common kind for this, you can use the same 3 steps always to find the vertical asymptote. Ok. So here are the 3 steps you can always use to find the vertical asymptotes of a rational function. The first is to factor the top and bottom. If you can and all that you can.

i'll show you what i mean. The second is to cancel any common factors from the top and bottom to simplify. And the last step is to take the bottom, the denominator and set it equal to zero to find the vertical asymptotes. So let's try it for this function.. The first step is to factor the numerator and denominator..

if you can. And all that you can. So you can't factor the top.. In any way, but the bottom you can factor. It's a quadratic.

and if you want more help on how to factor quadratics i have a video for that. So you can check that out. But basically the idea is you want to break this into.. 2 separate factors that start with x. So x plus a number times x plus another number.

ok, and what you want are 2 numbers here. 2 numbers here that.. Multiply to 6 and add to 5. And those 2 numbers would be 2 and 3. Since 2 times 3 is 6 and 2 plus 3 is 5.

so in here we have 2 and 3. And that's all that you can factor for this rational function. So that's the first step. You factored. Second step..

see if there are any similar terms on the top and bottom that you can just cancel out. In this example, there are not any. The top has different factors from the bottom completely. So you can't do that here, in this particular case. So we move on to the last step.

thankfully.. Which is to take the bottom, the whole denominator and set it equal to 0. And solve for the real zeroes of the denominator. So take the whole bottom and set it equal to 0. And to solve something like this..

you may already know this but you have to take each factor separately and set each of these equal to 0 individually. So take each of these and set them equal to 0. And solve for x. Get x alone. So these are your vertical asymptotes.

if you solve for x in each case. Get x alone. Subtract a 2 here.. Subtract a 3 there. You have x = -2 and x = -3..

these are your answers. The vertical asymptotes, if you have any will always be at x = some number or numbers. And they correspond to invisible vertical lines on the graph. So this is your answer. You don't need to do anything more.

but i want to show you what this means. So let me show you that on a graph. Ok, so again you don't have to graph this but just so you know what the vertical asymptotes mean these answers.. X = -2 and x = -3 correspond to dotted invisible lines vertical lines on your graph at x = -3 and x = -2. And your curve or your graph will never cross those invisible vertical lines.

or touch them. Reach them. So like up here in this branch.. Each side is shooting toward positive infinity if you go up but when it does that it will never go beyond x = -3 or -2 on either side. And for these branches down here, same idea they're each shooting down to negative infinity.

here and here. But they will not cross to the left of -2 or to the right of -3. So those vertical lines are.. The asymptotes are where the function is not defined for x. Where it does not exist for x.

and i know that there's another dotted line on here. This horizontal one at y = 0. I'm not trying to confuse you! That is a horizontal asymptote, so it's different. And you find it in a different way from the vertical ones. Actually, i have a video on how to find the horizontal asymptote so you can check that out if you want.

but basically in a nutshell if you want to find the vertical asymptotes if your function is in its most simplified form the vertical asymptotes will be at x = the real zeros of the denominator. Of the bottom. And now i want to show you a kind where factors do cancel. Ok, so let's look at another one. Say you need to find all the vertical asymptotes of this another rational function.

it's the same three steps to keep in mind. So first, try to factor as much as you can. If you try to factor the top, this is a quadratic that you can break into two factors. When you do that, when you factor you'll want two numbers that multiply to -10 and add positive 3. Those two numbers would be positive 5 and -2.

since 5 times -2 is -10. And 5 plus -2 is positive 3. And now see if you can also factor the denominator. X^2 – 4. I know that this looks different from the top.

it's not a trinomial. There's no middle x term. But it is a difference of squares. 4 is a perfect square so it fits into the difference of squares formula. So you can break this into two factors that start with x.

and using the difference of squares formula since 4 is a perfect square that's 2 squared this breaks into (x + 2) times (x – 2) and if that is an issue for you, feel free to look that up the difference of squares. But you can also check this.. If you multiply this out and foil this out, you would end up getting this back. So that works. Anyway, you've factored the top and bottom as much as possible, so that was the first step.

second step is, see if there's anything the same on top and bottom that you can just get rid of. Just cancel. Here there is. The (x – 2) is on top and bottom so we can get rid of it. Simplify by canceling it out, and write what's left.

and only after simplifying like we did do you take the denominator and set it equal to zero. That's the last step. Take the bottom, set it equal to zero. Solve for x. Which i'm sure you know how to do, to solve for x and get it alone.

when you do, you get x = -2. And that is the vertical asymptote. You just have one this time. That's fine. It came from what was left in the denominator after simplifying like we did.

those cancelled terms, those other ones did not end up giving us a vertical asymptote. So, by the way, what does this look like on a graph? And what happened to those cancelled terms? What was that about? Let me show you.. Ok, so here's what this looks like. Your answer was just this.. X = -2 for the vertical asymptote.

on the graph, that corresponds to this dotted vertical line here. And for the curve, the graph.. It will get very close to that dotted vertical line, but never cross it or touch it. On this side and on this side. This other dotted vertical line is a horizontal asymptote and so that's another topic.

but i want to show you what happened to those cancelled terms. Just so you know. When that happens, it creates a hole in the graph. A "removable discontinuity. " because at that point at x = 2 it made this equal 0 and this equal 0 and whenever you get a 0 / 0 form, any time, it is indeterminate as to the value. 0/0, by the way, does not equal 1! Does not! It's a special indeterminate form and it will be this gap in your graph if you zoomed in close enough.

so that's just good to know if you have to graph the whole thing. It doesn't matter for just answering what the vertical asymptote is. Now i want to show you one last kind of vertical asymptote problem. K, here's one last kind. This kind looks much simpler than the ones before it just has a number on top, just a constant, but that could throw you off that it's so simple.

so i want to show this to you. Still try the same three steps, so.. Try to factor the top and bottom. Can't factor the top. It's just a number.

try to factor the bottom.. X^2 + 1. Turns out you cannot factor that. If it had been (x^2 – 1) you could use the difference of squares to factor. Using the difference of squares formula, x + 1, x – 1, but since this is a sum of squares, x^2 + 1, you can't factor.

go ahead and try. It's not factorable. So move on.. There's nothing you can cancel either. So really, if you can't factor anything, can't cancel just take the bottom and set it equal to 0.

so take the denominator, x^2 + 1, set it equal to 0 and solve for x. Ok, when you try to solve for x.. You get x^2 = -1 and then at this point to get just x alone typically you'll just.. You would square root both sides. Here you get plus and minus the square root of -1.

that is imaginary, complex number, and.. This means that you get no vertical asymptote. If you get no "real" solution, in the real number system so like a square root of a negative number it is not defined in the real number systemso like a square root of a negative number it is not defined in the real number system so you have no vertical asymptote. And that's fine. You don't have to have a vertical asymptote.

that happens sometimes. Let me show you what it looks like. Ok, so this is what the graph looks like here there is no vertical asymptote so there's no dotted invisible line where the function is undefined for those x values. There is a horizontal asymptote, but that's another story. So that's the graph.

and then in general, to find the vertical asymptotes just stick to these three steps for finding them for rational functions. Ok. So if you're up for more.. I also have videos on how to find the horizontal asymptote and how to find the domain of a rational function. But i hope this video helped you understand vertical asymptotes.

i'm sure learning about asymptotes is exactly what you wanted to be doing right now.. It's ok, you don't have to like math.. But you can like my video! So if you did.. Please click 'like' or subscribe! . Hi guys! I'm nancy.. And i'm going to show you how to find the vertical asymptotes of a rational function. If you don't know what i mean by rational function just ignore that, you're probably in the right place. Actually all this stuff is a lot easier than it sounds. So don't worry. It's gonna be fine.

let me show you. Ok, so say you need to find the vertical asymptote or vertical asymptotes, if there's more than one. First of all.. What is a vertical asymptote? It's basically an invisible vertical line that your graph approaches and gets really really close to but never actually touches. Never quite reaches.

so how do you find the vertical asymptote? Well, if you have a rational function, the fraction form type, which is the most common kind for this, you can use the same 3 steps always to find the vertical asymptote. Ok. So here are the 3 steps you can always use to find the vertical asymptotes of a rational function. The first is to factor the top and bottom. If you can and all that you can.

i'll show you what i mean. The second is to cancel any common factors from the top and bottom to simplify. And the last step is to take the bottom, the denominator and set it equal to zero to find the vertical asymptotes. So let's try it for this function.. The first step is to factor the numerator and denominator..

if you can. And all that you can. So you can't factor the top.. In any way, but the bottom you can factor. It's a quadratic.

and if you want more help on how to factor quadratics i have a video for that. So you can check that out. But basically the idea is you want to break this into.. 2 separate factors that start with x. So x plus a number times x plus another number.

ok, and what you want are 2 numbers here. 2 numbers here that.. Multiply to 6 and add to 5. And those 2 numbers would be 2 and 3. Since 2 times 3 is 6 and 2 plus 3 is 5.

so in here we have 2 and 3. And that's all that you can factor for this rational function. So that's the first step. You factored. Second step..

see if there are any similar terms on the top and bottom that you can just cancel out. In this example, there are not any. The top has different factors from the bottom completely. So you can't do that here, in this particular case. So we move on to the last step.

thankfully.. Which is to take the bottom, the whole denominator and set it equal to 0. And solve for the real zeroes of the denominator. So take the whole bottom and set it equal to 0. And to solve something like this..

you may already know this but you have to take each factor separately and set each of these equal to 0 individually. So take each of these and set them equal to 0. And solve for x. Get x alone. So these are your vertical asymptotes.

if you solve for x in each case. Get x alone. Subtract a 2 here.. Subtract a 3 there. You have x = -2 and x = -3..

these are your answers. The vertical asymptotes, if you have any will always be at x = some number or numbers. And they correspond to invisible vertical lines on the graph. So this is your answer. You don't need to do anything more.

but i want to show you what this means. So let me show you that on a graph. Ok, so again you don't have to graph this but just so you know what the vertical asymptotes mean these answers.. X = -2 and x = -3 correspond to dotted invisible lines vertical lines on your graph at x = -3 and x = -2. And your curve or your graph will never cross those invisible vertical lines.

or touch them. Reach them. So like up here in this branch.. Each side is shooting toward positive infinity if you go up but when it does that it will never go beyond x = -3 or -2 on either side. And for these branches down here, same idea they're each shooting down to negative infinity.

here and here. But they will not cross to the left of -2 or to the right of -3. So those vertical lines are.. The asymptotes are where the function is not defined for x. Where it does not exist for x.

and i know that there's another dotted line on here. This horizontal one at y = 0. I'm not trying to confuse you! That is a horizontal asymptote, so it's different. And you find it in a different way from the vertical ones. Actually, i have a video on how to find the horizontal asymptote so you can check that out if you want.

but basically in a nutshell if you want to find the vertical asymptotes if your function is in its most simplified form the vertical asymptotes will be at x = the real zeros of the denominator. Of the bottom. And now i want to show you a kind where factors do cancel. Ok, so let's look at another one. Say you need to find all the vertical asymptotes of this another rational function.

it's the same three steps to keep in mind. So first, try to factor as much as you can. If you try to factor the top, this is a quadratic that you can break into two factors. When you do that, when you factor you'll want two numbers that multiply to -10 and add positive 3. Those two numbers would be positive 5 and -2.

since 5 times -2 is -10. And 5 plus -2 is positive 3. And now see if you can also factor the denominator. X^2 – 4. I know that this looks different from the top.

it's not a trinomial. There's no middle x term. But it is a difference of squares. 4 is a perfect square so it fits into the difference of squares formula. So you can break this into two factors that start with x.

and using the difference of squares formula since 4 is a perfect square that's 2 squared this breaks into (x + 2) times (x – 2) and if that is an issue for you, feel free to look that up the difference of squares. But you can also check this.. If you multiply this out and foil this out, you would end up getting this back. So that works. Anyway, you've factored the top and bottom as much as possible, so that was the first step.

second step is, see if there's anything the same on top and bottom that you can just get rid of. Just cancel. Here there is. The (x – 2) is on top and bottom so we can get rid of it. Simplify by canceling it out, and write what's left.

and only after simplifying like we did do you take the denominator and set it equal to zero. That's the last step. Take the bottom, set it equal to zero. Solve for x. Which i'm sure you know how to do, to solve for x and get it alone.

when you do, you get x = -2. And that is the vertical asymptote. You just have one this time. That's fine. It came from what was left in the denominator after simplifying like we did.

those cancelled terms, those other ones did not end up giving us a vertical asymptote. So, by the way, what does this look like on a graph? And what happened to those cancelled terms? What was that about? Let me show you.. Ok, so here's what this looks like. Your answer was just this.. X = -2 for the vertical asymptote.

on the graph, that corresponds to this dotted vertical line here. And for the curve, the graph.. It will get very close to that dotted vertical line, but never cross it or touch it. On this side and on this side. This other dotted vertical line is a horizontal asymptote and so that's another topic.

but i want to show you what happened to those cancelled terms. Just so you know. When that happens, it creates a hole in the graph. A "removable discontinuity. " because at that point at x = 2 it made this equal 0 and this equal 0 and whenever you get a 0 / 0 form, any time, it is indeterminate as to the value. 0/0, by the way, does not equal 1! Does not! It's a special indeterminate form and it will be this gap in your graph if you zoomed in close enough.

so that's just good to know if you have to graph the whole thing. It doesn't matter for just answering what the vertical asymptote is. Now i want to show you one last kind of vertical asymptote problem. K, here's one last kind. This kind looks much simpler than the ones before it just has a number on top, just a constant, but that could throw you off that it's so simple.

so i want to show this to you. Still try the same three steps, so.. Try to factor the top and bottom. Can't factor the top. It's just a number.

try to factor the bottom.. X^2 + 1. Turns out you cannot factor that. If it had been (x^2 – 1) you could use the difference of squares to factor. Using the difference of squares formula, x + 1, x – 1, but since this is a sum of squares, x^2 + 1, you can't factor.

go ahead and try. It's not factorable. So move on.. There's nothing you can cancel either. So really, if you can't factor anything, can't cancel just take the bottom and set it equal to 0.

so take the denominator, x^2 + 1, set it equal to 0 and solve for x. Ok, when you try to solve for x.. You get x^2 = -1 and then at this point to get just x alone typically you'll just.. You would square root both sides. Here you get plus and minus the square root of -1.

that is imaginary, complex number, and.. This means that you get no vertical asymptote. If you get no "real" solution, in the real number system so like a square root of a negative number it is not defined in the real number systemso like a square root of a negative number it is not defined in the real number system so you have no vertical asymptote. And that's fine. You don't have to have a vertical asymptote.

that happens sometimes. Let me show you what it looks like. Ok, so this is what the graph looks like here there is no vertical asymptote so there's no dotted invisible line where the function is undefined for those x values. There is a horizontal asymptote, but that's another story. So that's the graph.

and then in general, to find the vertical asymptotes just stick to these three steps for finding them for rational functions. Ok. So if you're up for more.. I also have videos on how to find the horizontal asymptote and how to find the domain of a rational function. But i hope this video helped you understand vertical asymptotes.

i'm sure learning about asymptotes is exactly what you wanted to be doing right now.. It's ok, you don't have to like math.. But you can like my video! So if you did.. Please click 'like' or subscribe! .

## HOW TO CHECK THE CONTINUITY OF FUNCTION || WHAT IS CONTINUOUS FUNCTION || LIMIT AND CONTINUITY

If you are visiting this channel 1st time then “subscribe" it like share and comments on this video subscribe this channel for more such videos on maths share this video with your friends through whatsapp or facebook to benefit them if you have not subscribed channel then subscribe it. If you are visiting this channel 1st time then “subscribe" it like share and comments on this video subscribe this channel for more such videos on maths share this video with your friends through whatsapp or facebook to benefit them if you have not subscribed channel then subscribe it.

## How To Find The Domain Of Any Function (NancyPi)

Hi guys, i'm nancy. And i'm going to showyou how to find the domain of any function. The domain is kind of an annoying questionthat you'll get about a function, but it can be very simple. And all it means is all possiblex values that you can have in your function. So all allowable x values that you can havein your function. To do this you should already know how to solve a quadratic by factoring. And how to solve an inequality.

okay so these are the five types of functions you mighthave to find the domain for. I'm going to go through all of them. If you want to skipahead you can. Just look below in the description for the time to skip to. The first case isreally easy.

if you have a polynomial – just an x expression like x squared plus 3x plus1, no square roots, no fractions, your domain answer is just all real numbers. And you canwrite that phrase "all real numbers" or you can write interval notation: parenthesis negativeinfinity comma positive infinity, end parenthesis. Now if you don't have just a polynomial andyou have something more complicated like a fraction or a square root or possibly both,it's going to be one of these four cases. It's either just a fraction, it's a squareon its own, it could be fraction with a square root in the bottom, or a fraction with a squareroot in the top (the numerator). And i'm going to explain how to find the domain for allof those cases.

okay, say you have just a fraction. To find the domain all you haveto do is take the bottom, whatever that may be, and set it not equal to zero. So in thisexample we have x squared plus 5x plus 6 "not equal to zero. Notice that it doesn't matterwhat the top is, just take the bottom and set it not equal to zero. And you can solvethis just as if it were a normal equality.

and that involves factoring. If you need helpwith factoring or solving by factoring you can look at one of my other videos. So inthis equation we get two restrictions two solutions: x cannot equal negative 3 and xcannot equal negative 2. So when we want to write the domain, the final answer for domain,you can write it as all real numbers but x cannot equal negative 3 and cannot equal negative2. Or you can use interval notation and write it as everything from negative infinity, allthe way up to negative 3, not including negative 3, combined with everything from negative3 to negative 2, not inclusive of either number, combined with everything from negative 2 allthe way up to infinity.

so either of these notations is fine – they mean the same thing. Let's look at another example. What if you're given something like this? Where the denominatoris x squared plus 3. Same thing: take the bottom, set it not equal to zero because wecannot have the denominator equal zero. It's not defined when it is. Then when you solvethis equation, you end up with x squared cannot equal negative 3.

but x squared can neverequal negative 3, so we just can't have that ever. There's no x value that would ever makethat happen, so you don't need to worry about that restriction. All you need to write inthat case would just be your normal all real numbers domain. So you can say all real numbers. You could say capital r symbol for all real numbers. Or you could say, with interval notation,everything from negative infinity to positive infinity, with parentheses on either side. So that would be your three ways of writing the domain for this fraction.

now let me showyou what to do if you have just a square root in your function. I'm going to focus on asquare root, but this is true for any even root – fourth root, sixth root, etc. It'snot true for odd roots like third root. If you have a square root, or an even root, youtake the expression underneath the radical sign, which is the radicand (everything that'sunderneath the root symbol), so in this case x plus 1, and you need it to be greater thanor equal to zero, because we can't have a negative number underneath a root. You canonly have a positive number or zero.

so what's underneath must be greater than or equal tozero. Now you're going to solve that normally, as you would solve any inequality, and youget the restriction x is greater than or equal to negative 1. So your domain would be allreal numbers, x greater than or equal to negative 1. Or you can write it in interval notationas everything from negative 1, all the way up to infinity. Notice that i used a squarebracket for negative 1 because you're including negative 1, and everything all the way upto infinity, parenthesis.

so either of these ways is fine, as notation for the domain. Okay let me show you quickly also an example with a quadratic expression under the root. So here we have x squared plus 5x plus 6 underneath the square root symbol. You still do justtake the expression underneath the root symbol and require that it be greater than or equalto zero. Okay you need to solve this inequality. It's a quadratic inequality, which is morecomplicated than the last example. If you don't know how to solve that, you can lookit up somewhere else.

once you have solved this quadratic inequality, you will get thatx can be less than or equal to negative 3, or x can be greater than or equal to negative2. So your domain you can write either as all real numbers but x less than or equalto negative 3, x greater than or equal to negative 2. If you were going to write thisin interval notation, it would look like everything from negative infinity all the way up to negative3, including negative 3, and everything from negative 2 all the way to positive infinity. So either of those ways are fine to write your domain answer. Okay, if you have a squareroot in the bottom of your fraction, all you need to do is take the expression under theroot, x plus 1, and set it greater than zero. Now usually with a square root, we would setit greater than or equal to zero, but we can't have that the denominator equal zero, so weleave that part out.

so if you see the square root in the bottom, all you need to do istake the expression under the root and set it greater than zero, and solve. So in thiscase, you subtract one from each side, and you would get that x is greater than negative1, so your domain would beall real numbers, x greater than negative 1, or if you wanted to use intervalnotation, you would write parenthesis negative 1, everything from negative 1, not includingnegative 1, all the way up to infinity. So either of these ways is fine to write yourdomain answer. Okay, if your square root is in the top of your fraction, you'll have tworestrictions. The first one will be that what's under the square root has to be greater thanor equal to zero.

so x plus 1 greater than or equal to zero. And when you solve thatby subtracting 1 from each side, minus 1, minus 1, you get x greater than or equal tonegative 1. The other restriction will come from the bottom. That the bottom can't bezero. So we have x squared minus 4 cannot equal zero.

now to solve that as before, youwould have to factor, so you have x plus 2, times x minus 2, can't be zero. So i've re-writtenthat as a product of two factors, and when you solve each of them separately, set eachof them not equal to zero, you get x cannot be negative 2, and x cannot be 2. Okay sowe have this restriction from the root, and this restriction from the bottom. What youhave to do in this case is take the intersection of the two. So if you think about this, xhas to be greater than or equal to negative 1, which already takes care of negative 2- if x is greater than or equal to negative 1, we can't have negative 2 anyway, so youcan ignore that one, throw it away.

all you need to write for your domain: x greater thanor equal to negative 1, except it cannot be 2, positive 2, and in interval notation thatwould be that you can have everything from negative 1, including negative 1, up to 2but not including 2, and then everything from 2, not including 2, up to infinity. So eitherof these notations is fine, and that's your answer for the domain. So i hope that wassuper fun, finding the domain of your function. It's okay if you didn't have fun. You don'thave to like math.

but you can like my video, so if you did, please click like below. . Hi guys, i'm nancy. And i'm going to showyou how to find the domain of any function. The domain is kind of an annoying questionthat you'll get about a function, but it can be very simple. And all it means is all possiblex values that you can have in your function. So all allowable x values that you can havein your function. To do this you should already know how to solve a quadratic by factoring. And how to solve an inequality.

okay so these are the five types of functions you mighthave to find the domain for. I'm going to go through all of them. If you want to skipahead you can. Just look below in the description for the time to skip to. The first case isreally easy.

if you have a polynomial – just an x expression like x squared plus 3x plus1, no square roots, no fractions, your domain answer is just all real numbers. And you canwrite that phrase "all real numbers" or you can write interval notation: parenthesis negativeinfinity comma positive infinity, end parenthesis. Now if you don't have just a polynomial andyou have something more complicated like a fraction or a square root or possibly both,it's going to be one of these four cases. It's either just a fraction, it's a squareon its own, it could be fraction with a square root in the bottom, or a fraction with a squareroot in the top (the numerator). And i'm going to explain how to find the domain for allof those cases.

okay, say you have just a fraction. To find the domain all you haveto do is take the bottom, whatever that may be, and set it not equal to zero. So in thisexample we have x squared plus 5x plus 6 "not equal to zero. Notice that it doesn't matterwhat the top is, just take the bottom and set it not equal to zero. And you can solvethis just as if it were a normal equality.

and that involves factoring. If you need helpwith factoring or solving by factoring you can look at one of my other videos. So inthis equation we get two restrictions two solutions: x cannot equal negative 3 and xcannot equal negative 2. So when we want to write the domain, the final answer for domain,you can write it as all real numbers but x cannot equal negative 3 and cannot equal negative2. Or you can use interval notation and write it as everything from negative infinity, allthe way up to negative 3, not including negative 3, combined with everything from negative3 to negative 2, not inclusive of either number, combined with everything from negative 2 allthe way up to infinity.

so either of these notations is fine – they mean the same thing. Let's look at another example. What if you're given something like this? Where the denominatoris x squared plus 3. Same thing: take the bottom, set it not equal to zero because wecannot have the denominator equal zero. It's not defined when it is. Then when you solvethis equation, you end up with x squared cannot equal negative 3.

but x squared can neverequal negative 3, so we just can't have that ever. There's no x value that would ever makethat happen, so you don't need to worry about that restriction. All you need to write inthat case would just be your normal all real numbers domain. So you can say all real numbers. You could say capital r symbol for all real numbers. Or you could say, with interval notation,everything from negative infinity to positive infinity, with parentheses on either side. So that would be your three ways of writing the domain for this fraction.

now let me showyou what to do if you have just a square root in your function. I'm going to focus on asquare root, but this is true for any even root – fourth root, sixth root, etc. It'snot true for odd roots like third root. If you have a square root, or an even root, youtake the expression underneath the radical sign, which is the radicand (everything that'sunderneath the root symbol), so in this case x plus 1, and you need it to be greater thanor equal to zero, because we can't have a negative number underneath a root. You canonly have a positive number or zero.

so what's underneath must be greater than or equal tozero. Now you're going to solve that normally, as you would solve any inequality, and youget the restriction x is greater than or equal to negative 1. So your domain would be allreal numbers, x greater than or equal to negative 1. Or you can write it in interval notationas everything from negative 1, all the way up to infinity. Notice that i used a squarebracket for negative 1 because you're including negative 1, and everything all the way upto infinity, parenthesis.

so either of these ways is fine, as notation for the domain. Okay let me show you quickly also an example with a quadratic expression under the root. So here we have x squared plus 5x plus 6 underneath the square root symbol. You still do justtake the expression underneath the root symbol and require that it be greater than or equalto zero. Okay you need to solve this inequality. It's a quadratic inequality, which is morecomplicated than the last example. If you don't know how to solve that, you can lookit up somewhere else.

once you have solved this quadratic inequality, you will get thatx can be less than or equal to negative 3, or x can be greater than or equal to negative2. So your domain you can write either as all real numbers but x less than or equalto negative 3, x greater than or equal to negative 2. If you were going to write thisin interval notation, it would look like everything from negative infinity all the way up to negative3, including negative 3, and everything from negative 2 all the way to positive infinity. So either of those ways are fine to write your domain answer. Okay, if you have a squareroot in the bottom of your fraction, all you need to do is take the expression under theroot, x plus 1, and set it greater than zero. Now usually with a square root, we would setit greater than or equal to zero, but we can't have that the denominator equal zero, so weleave that part out.

so if you see the square root in the bottom, all you need to do istake the expression under the root and set it greater than zero, and solve. So in thiscase, you subtract one from each side, and you would get that x is greater than negative1, so your domain would beall real numbers, x greater than negative 1, or if you wanted to use intervalnotation, you would write parenthesis negative 1, everything from negative 1, not includingnegative 1, all the way up to infinity. So either of these ways is fine to write yourdomain answer. Okay, if your square root is in the top of your fraction, you'll have tworestrictions. The first one will be that what's under the square root has to be greater thanor equal to zero.

so x plus 1 greater than or equal to zero. And when you solve thatby subtracting 1 from each side, minus 1, minus 1, you get x greater than or equal tonegative 1. The other restriction will come from the bottom. That the bottom can't bezero. So we have x squared minus 4 cannot equal zero.

now to solve that as before, youwould have to factor, so you have x plus 2, times x minus 2, can't be zero. So i've re-writtenthat as a product of two factors, and when you solve each of them separately, set eachof them not equal to zero, you get x cannot be negative 2, and x cannot be 2. Okay sowe have this restriction from the root, and this restriction from the bottom. What youhave to do in this case is take the intersection of the two. So if you think about this, xhas to be greater than or equal to negative 1, which already takes care of negative 2- if x is greater than or equal to negative 1, we can't have negative 2 anyway, so youcan ignore that one, throw it away.

all you need to write for your domain: x greater thanor equal to negative 1, except it cannot be 2, positive 2, and in interval notation thatwould be that you can have everything from negative 1, including negative 1, up to 2but not including 2, and then everything from 2, not including 2, up to infinity. So eitherof these notations is fine, and that's your answer for the domain. So i hope that wassuper fun, finding the domain of your function. It's okay if you didn't have fun. You don'thave to like math.

but you can like my video, so if you did, please click like below. .

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