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5. LetTI +2 f(z) 2r2 + 2 2 +3 Determine where f(r) is continuous_I <0, 0 < = <2 ~> 2...

Question

5. LetTI +2 f(z) 2r2 + 2 2 +3 Determine where f(r) is continuous_I <0, 0 < = <2 ~> 2

5. Let TI +2 f(z) 2r2 + 2 2 +3 Determine where f(r) is continuous_ I <0, 0 < = <2 ~> 2



Answers

Determine where $f$ is continuous. $$ f(x)=\cos \left(\frac{x^{2}-5 x+1}{2 x}\right) $$

For the given problem, we want to determine the interval um where F of G is continuous, knowing that half of X is equal to five X divided by x minus one, and G of X is equal to x minus two squared if. Okay. Yeah. Okay, so with that in mind we're going to Um see the H of X. or composition is equal to five hands this Over this -1. And then we could simplify things further. Um Or we could consider the fact that if this equals one were in trouble, if this equals one then we see that there is going to be undefined. Uh We made a zero on the dominators of the functions undefined And that's going to occur when we have three or one. So three and one are going to be the areas the points of discontinuity.

And this exercise. We want to find a value for a so that the function f of access continuous. And we know that for a functionary continuous at a certain point, but the limit of ex approaching that X value is going to be equal to the value of the function of that point. So here we see that we have two different functions in our X for when X is less than able to and when x is great and too. So we have to make sure that this entire functions continuous, that if the limit of ex approaching too of X equals off of two then were continuous. But if we're not, then we're not continuous. So we need to find a value of A that makes it so that these two are equal to each other. So f of two. Currently, if you plug it into two actual story because we plug it in as X is less than able to to we have to sew. It goes here two times two is four plus 37 and then right now, our limit of ex approaching too is going to be We have to do it from the right side in the left side to see what they are. So the limit of ex approaching two from the right side is going to be from ex is coming from the positive numbers down. So we're gonna look at X plus one. So if we use X plus one, we're going to see that if we plug that to win, we're going to get to a plus one. Now, if we do our limit from the left side, we're going to use two X plus one. And again, if we plug in that too, we're just going to get seven, as we did for two. So we need to April's 12 equal seven so that this limit is equal to seven equals f of two. We want to a pulse one toe, also be seven. So if we set of equal, you get to a equal six and that gives us that a equals three. So this is the value of a that the functions continuous because the limit from the right and the left for two, where we think that we would have a discontinuity if it was another value, is equal to the volume of of of two, which is seven. So we get a three on the soul for that

So we have the function g of X, and we want to determine, um whether G of X is continuous at X equals negative one. So to determine continuity, the function has to fulfill three criteria. So one G of negative one has to exist or be defined to the limit of G of X. As X approaches, negative one has to exist, and three, they have to be equal to one another. Sergey of negative one equals limit of G of X X approaches, negative one. So let's just go through each step. Someone does G of negative one exists well based on our piece wise equation when X is negative one, we have to use the bottom function because this is when X is greater than or equal to negative one. So we had a negative two times negative one plus one, which is equal to negative two times negative. One is two plus one is three, so you have negative. One exists great to We have to find that the limit exists. So first, let's find the limit from the left so acts approaches negative one from the left hand side of G of X. So since it's approaching from the left hand side. We're gonna be looking at values less than negative ones. So values less than negative one. That's gonna be the top equation two x plus five. We can just plug in negative one to find the limit. So two times negative one plus five equals negative. Two plus five is three now. The limit on the right hand side do the same thing. But this time, because we're looking for X values greater than negative one, we have to use the bottom function. So negative two X plus one x approaches Negative one gam We just plugged in. So, um, negative two times negative Morin plus one equals negative two times night of one that's two plus one is three so great they're equal to one another. So we know that in general, the limit regardless of left or right, that limit its expertise. Native one of g of x exists, and the third criteria on G of negative one equals the limit. As X approaches negative one of G of X equals three. They're equal to each other because they both equal three. And so we've determined that G of X is continuous at X equals negative one

Yeah. In this problem, we have to decide whether or not the function is continuous. If it is not continuous, identify the point at which it is discontinuous. So the human function is f of X is equal to X. You access less than five and we have a piecewise function. Extra care Ive X is greater than are equal to zero is greater than or equal to five. We have to check the continuity of the function and we have to identify the points and with this function is not continuous. So this is a piecewise function as F of X is equal to X. If X is less than five. So f of X is equal to X is a polynomial function. So we know that every polynomial function is continuous everywhere. So in this particular case, access less than five. So for all the values of X, less than five dysfunction is defined so we can say that it's got This function is continuous because fo fact is a polynomial function. Similarly f of X is equal to f of X is equal to X care for X is greater than five. It's again a polynomial function So this function will also be continuous or X is greater than five Now we will take now we will take whether the given function is continuous and accessible to fight or not. So, using the properties of continuity, let us check one by one. The first can first property of the continuities. The function should be defined at that point, so f of X is equal to extra care. For this is the function for X is equal to five. So when we have when we plug X is equal to fight in the function we have 25. So this function is defined at exit equal to five. This is the first condition for a function to re continuous at a point that is, the function should be defined at that point. Now the second condition is we will check the limits. We will check the limit is the limit exists at X is equal to five or not. So for that we have to find the left hand limit and right hand limit. First of all, we will find the left hand limit so limit X approaches to five. Yes, Airport X is equal to S X approaches to fight from left hand side. So we will take this part of the function for access less than why we have X. So when we apply the limit, we have right and for right hand side limit limit X approaches to five from right hand side. Airport X is equal to S X is greater than five. So we take ex occurred from the given function because the value of the function is executed when X is greater than or equal to fight. So we have extra care. So on your on while reading the limit, we have the value 25. As you can see, that left hand side limit is not equal to right and side limit, so we can conclude that limit ex abroad just to five f of X does not exist means that at X is equal to five. Left hand side limit is not able to right hand side limit, so Bellamy does not exist at access equal to five, so the second condition is failed. So the given function does not satisfy the second condition for a function to be continuous. So we can say that the given function is discontinuous at X is equal to five except accessible to five. The human function is continuous everywhere


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