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Conslder the following piecewise function: 2 +I I <-3 X(c+1)2 ~3<1< -1 1 -1<I<1 f(s) 1 <1<4 I =write the answers in the table; do your scratch ...

Question

Conslder the following piecewise function: 2 +I I <-3 X(c+1)2 ~3<1< -1 1 -1<I<1 f(s) 1 <1<4 I =write the answers in the table; do your scratch work Fill out the table below. Only On another sheet of paper: function g(z) , defined piecewise, with the following properties: (b) Come up with 0 new has jump discontinuity at I has & removable discontinuity at I = 1, has an infinite discontinuity at = = 3, and is continuous everywhere else_ A" ( ' | 1 ' 1 0 "

Conslder the following piecewise function: 2 +I I <-3 X(c+1)2 ~3<1< -1 1 -1<I<1 f(s) 1 <1<4 I = write the answers in the table; do your scratch work Fill out the table below. Only On another sheet of paper: function g(z) , defined piecewise, with the following properties: (b) Come up with 0 new has jump discontinuity at I has & removable discontinuity at I = 1, has an infinite discontinuity at = = 3, and is continuous everywhere else_ A" ( ' | 1 ' 1 0 " 26'114/" ' [ L~ : e eligri' ' "else colaoi hl 40 0 FW6;"1 4: '^* aiala c =-3 c=-1 c =1 c =4 lim f (z) I-+ct lin f(c) 1-C7 lim f(c) TeU Jesaul < ; 17C 6 >1- "l &; * i' Jarkw %eudicsje?. f(c) Is f continuous at c? If not what kind of discontinuity is it



Answers

Sketch a graph of a function $f$ that satistics the following conditions: 1. $\operatorname{dom}(f)=[-3,3]$ 2. $f(-3)=f(-1)=1 ; f(2)=f(3)=2$ 3. $f$ has an infinite discontinuity at -1 and a jump discontinuity at 2 4. $f$ is right continuous at -1 and left continuous at 2

Okay, so let's look at this problem number 13 13. So I have four conditions so I have to accord in this for conditions to scratch uh the graph of the function F. Okay, so Maybe 1st, Let's look at the Condition two. Okay, so it is easier to draw. So I have F f minus two, it goes to f f minus one, goes to F one goes to F of two, it goes to zero. Okay. Okay, this 4.4 points are zero. Now the next step I want to do is I want to uh look at the number three. The condition number three. So F has an infinite discontinued This continuity at -2. A drum discontinuity at -2 At -1. Okay. And I dropped discontinuity at one and the infinite discontinuity at two. Okay, So which means at -2 and one mm Is infinite discontinuity and at -1 and one is jump discontinuity. Okay. Mhm. And uh let's look at number four because if we combine the conditions number three and number four it is easier for us to two sketch the graph. Okay, so f is continuous from the right at -1 and continuous from the left at one. Okay, so now and we combine the condition at number three. Okay, this is a trump discontinuity. So maybe you know drums to another molly? I don't know mm And here also jump okay maybe drunk to hear this is open dot you can see and let's see And let's see the # two and the two. Okay, there's infinite discontinuity. So infinite discontinuity. Uh well okay so I have to change in a little bit. Right, okay, so here is a drum discontinuity but here is infinite, so I have to make it to goes to infinite. Okay and here is a jump discontinuity and also it goes to infinite. Okay, further middle from -1-1. I just connecting this. Okay, so this is a this is left For -1 is the right, the right continuity, and for one is left left continuous. Okay. And the domains from -2-2. Okay, so now this is a graph. It's my graph for question # 13.

Okay, we have to graph of a function Y equals f of X. That has quite a number of properties. First of all, the domain Of the function is 0- five. Uh And that means that this function will only be defined Uh for X0 All the way up to AH Excess five. Those are the only X values from 0 to 5 that can go into the function. Then we have some things happening at the number one. Uh The number one is going to be an important number and uh we have some things happening at two and again at three And we might as well squeezing the four. We have some discontinuities are some great continuous left continuous limits approaching negative infinity. Okay, so the domain of the function uh is from 0 to 5. So the function will only be uh graft for X value 0 to 5. Nothing to write a five, nothing to the left at zero As X Approaches one. Uh The limit of the function from the right side and the left side exists. But when we read down further we're going to have a discontinuity at one. So if we have the function having the same limit at one, Uh but it has a removable dis continuity at one. let's do this. Let's make this the value of the function at one. Uh we don't have any particular information at zero. It's more of what's happening near X. two and X is three. So uh and we do have a removable dis continuity At one. But the function, The limit of the function as X approaches one from the right side will equal the limit of the function as it approaches from the right from the left side. So we'll check all these uh you know, conditions against our graph as we uh finish it off to make sure we were ok with everything. Uh Let's see, we're not continuous, set to its left continuous at two. So as X approaches to uh this function is left continuous, meaning that as we close in on two from the left to function will be continuous and uh all the way up until uh the value of the function at two, so left continuous at two. That means all the way up to and including two. Uh it's not continuous at two. We have a jump dis continuity at two. So let's say a jump. This continuity means either jump up or jump down. Uh So we'll say it jump up, we'll fix it if we have to. And then as we approach uh three from the left side, the function is going to go down towards negative infinity. So that's like a vertical assam Tope line. So let's let's draw that Vertical Assam toe at three. So the red line is just a vertical isotope is not part of the actual graph of the function. So it's extra approaches three from the left side. Uh We had the function dropping down to negative infinity. So we picked back up from this open circle point above excess too. And let's see if it'll let me draw it, we're going to have the function as we approach three from the left side. Uh The function is going down towards negative infinity, gets closer and closer to the vertical. Aspirin. Topa doesn't touch it. All right, what else do we have? We are right continuous, but not continuous at three. Right continuous at at three means coming in from the right to function will be continuous up until it actually hits its value. Uh F F three when X is three. So let's put a point when X is three and this will be uh the function And the limit as X approaches three from the right side is too. So we'll say this value right here, let me move that point up a little bit. Uh when excess treaty function value is going to be too. So let's put a couple of uh lines, couple numbers on the Y axis. All right. I'm going to give a quick look, pausing this video to see if we did everything and then I'm going to run through to make sure that we met all the conditions, we had to be right continuous um At three. So, as the function gets uh as X comes in Uh 2, 3 from the right side to function is continuous up until it hits this point. And uh f of three the limit as X approaches straight of the function had to be too. So we are going to have a continuous function up until X. It's three as we're coming in from the right side, so we are right continuous at three. and FF three is to uh the limit of the limit of our function. The limit of F of X as X approaches three. Uh from the right side deposits side is too all right. I'm going to pause, make sure that we checked everything out and then we're gonna run into our throw one little detail. Since the domain of the function is X values from 0 to 5. Uh Let's have uh this curve of this graph of ffx uh extend all the way to x equals five and this could be the value of the function When X is five. Okay. Um I just checked through all the conditions and uh this graph of uh y equals f fx does meet every single condition. First of all, the domain is disclosed, interval from 0-5. So you can see that FFX uh is defined for all the X values from 0 to 5 at one, there's ffx at two years, ffx uh at three. There's ffx. And then since the domain is all the way up until five, we have the graph existing all the way up to and including five. Now the limit as X approaches one from the right side. As we approach one from the right side, the limit of F of X and the limit of F of X as X approaches one from the left side. Uh They both exist and are equal. So as X approaches one from the left side to function approaches one as X approaches one from the right side to function again approaches one. Uh So the limit as X approaches one from the right and left left exist and are equal. Uh F of X is left continuous but not continuous at two. Left continuous at two means as X goes up to two and hits two from the left, the function is continuous all the way up until uh including when access to it's not continuous at two. You can see that F a bex is not continuous at two. Uh It is right continuous but not continuous at three. So you can see that X equals three. The function is not continuous, but it is right continuous meaning as X approaches three from the right side, the function is continuous all the way up to and including when X is three, it hits F of three. Uh let's see right continuous, but not continuously. three. Alright, FFX has removable dis continuity at once. So when X is one uh F of X does have a removable discontinuity. If ffx instead of being defined to be too, When X is one, if ff one instead of being too. If we redefined F1 to be one, we will have removed this continuity by basically putting this point here, we have a jump this continuity at two. You can see that when next is too we have jumped this continuity. All right. And uh two following limits have to hold as X approaches three from the left side, the negative side to function is approaching negative infinity. You can see it's going down towards negative infinity and last but not least. The limit as X approaches three from the right side of F of X is to as X approaches three from the right side, you can see That the function is approaching the value of two. So uh this graph of F of X satisfies every single condition that it needed to satisfy.

For this problem, we are given the function F of X equals coast of one over X where X does not equal zero. And one where X equals zero. Were then asked to determine the points of discontinuity, the type of discontinuity and whether the function is left or right continuous. So for this coast of one over X part that is going to be a uh sandy soil oscillation that is rapidly increasing until as we approach X equals zero, it is oscillating infinitely quickly, though the magnitude of the oscillation shouldn't be changing. That's just my rough sketch there. So the thing is that the limit as X approaches zero from the left, doesn't exist. It's impossible to define and the limit as X approaches zero from the right doesn't exist. So it's impossible to find to define rather. So obviously there is discontinuity at X equals zero, and the type of discontinuity is none of the above. In terms of infinite jump. Removable walkman, it is a sort of unique kind of discontinuity, and then, in terms of whether the function is left or right continuous, it is neither. Since the function cannot be defined at the point X equals zero, it can't have a limit that approaches that from either side, especially because of the limits on the other side don't exist.

So our function F of X is equal to the absolute value of X plus one. And that is the translated version of the absolute value function that will be shifted one unit to the to the left. And so we're gonna have a point here that will be the vertex of that. And we'll have a slip up on this way and try to kind of free hand to make it look like has a slope negative one. That way, By the way, our domain is the set of all real. Let's read at the interval notation and our range the lowest Y value we can have is zero and then it goes up to positive infinity. Now let's write the piecewise function for this. By the way, we have no discontinuities. So we can see that are linear function here has a slope of negative one And it would have a Y intercept of -1. And we would use that part if X is less than -1. And then this function, That linear pattern has a slope of one And has a lie intercept of one, and that would be what we would use. Effects is greater than or equal to negative one. And once you also notice that this is what our linear pattern is when we use the points to the right of the vertex and the opposite of this by that by negative one is what we have here when the X values are less than that vertex points and again, no discontinuities.


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