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Prove the following statement: Proposition 2. If f : (a,c) 7 R is such that f is uniformly contin- uous Onl both (a,b] and [b,c) for some b € (a,c), then f un...

Question

Prove the following statement: Proposition 2. If f : (a,c) 7 R is such that f is uniformly contin- uous Onl both (a,b] and [b,c) for some b € (a,c), then f uniformly continuous Onl (a,c).

Prove the following statement: Proposition 2. If f : (a,c) 7 R is such that f is uniformly contin- uous Onl both (a,b] and [b,c) for some b € (a,c), then f uniformly continuous Onl (a,c).



Answers

Determine whether the given function F is continuous on the indicated intervals?

Suppose you want to find functions F and G. Such that F. S container was at zero and the value of F zero is zero and F G is not continuous at zero. Since F G of X is the same as F of X time strong backs then to make this not continue us, we want to make function G of X That is not continuous at zero since this will affect the product. So how do we make it this container was we just recall that continuity of a function at a point. Let's say X equals C. This is only possible if it Within the three conditions. So the first condition will be for the function G of C to exist. And then the limit as X approaches, see of this function G fx also exist. And then lastly, for this limit as X approaches C of G of X to equal the value of G etc. So if any one of the following do not hold, then it makes it this continuous. So so if you let F of X equal to X squared. That's just container was at zero. Also the value of F at zero here is zero. Then we can make a G of X. A rational expression in which the exponents in the denominator is larger than the exponents of F. So we can say one over X. Cube. So since it didn't say anything about G, then we can form it like this because when you multiply say F G of X, which is the same as F of X G of X. We will then get x squared times one over X cube, which is equal to one over X, which is not continue was at zero. Therefore our functions are x squared and the other one is one over X cube. Get also used one over X to the fourth. As long as it's exponent in the denominator, Fergie is greater than the exponent of F of X. So that when we multiply them, we will still have an X in the denominator, which will Make It. This container was at zero.

For this problem. You want to find the values of A and B where F is continuous and frangible. So we assume that F is continuous and differentiable over its domain. So if that's the case, then my definition of continuity at a point we say the limit as X approaches negative too of F of X exists. But this also implies that the one sided limits that is limited X approaching negative two from the left of F of X equals the limit. As X approaches negative two from the right of F of X. So you're seeing the given function we have limits As X approaches -2 from the left of F of X. In this case will be The first one sends x values there are less than -2. So we use six execute minus three X squared plus seven. This is equal to the limit as X approaches -2 from the right of F of X. In this case will be a X plus B. Since X Is greater than -2. So we have a X plus B evaluating the limits we have six times the cube of negative 2 -3 times The square of negative two Plus seven. That is equal to eight times negative two Plus B. Simplifying that we have -53. This is equal to negative two A plus B or that's the same as two. A- B equals 53. So let's call this our equation one furthermore by definition of differential ability at a point we have limit as X approaches negative too of the different school ocean. That will be F of X -F of negative two over X minus negative two. That'll be X plus two exists. That means the limit as X approaches negatively from the left of this difference Kocian F of x minus F of negative to over expose to. That's equal to the limit Sex approaches -2 from the right of F of x minus F of negative two over. Express to So from here we should have limit As X approaches -2 from the left of alphabets in this case will be a six X cubed minus three X squared plus seven. Since X values are less than negative two and we have six X cubed minus three X squared plus seven -1. X is -2. It will be six times negative. Two to the third Power -3 times negative two To the second power plus seven. This divided by X-plus two equals the limit as X approaches negative two from the right of F of X. In this case will be a X plus B. So we have a X plus B minus when X is negative two, that will be eight times negative two plus B. All over X-plus two. Now simplifying this we have Limit as X approaches -2 from the left of six execute minus three, X squared plus seven minus negative 53. All over X plus two. This equals the limit as x approaches -2 from the rate of A X plus B minus negative to A that will be plus to a minus B. All over X plus two. Simplifying even further we have limits As extra purchase -2 from the left of six, X cubed minus three X squared plus 60 all over, exposed to. That's equal to the limit. As x approaches -2 from the right in this case you can cancel at B and negative B since that will be zero. And from here we should get A X plus two A all over X plus two. Now if we factor out the numerator as we then have Limit as X approaches -2 from the left of three times two, X squared -5 x plus 10 times X was too All over X-plus two. This is equal to the limit as X approaches -2 from the right of eight times x plus two. All over X-plus two. Now and here we can cancel at the X-plus two and we should get three times 2 times the square negative too -5 times negative two plus in this is equal to a simplifying the left hand side we have 84 equals eight. Now if a is equal to 84. Then from here we should have a question one which is to a minus B equals 53. We will get two times 84 minus B Equals 53 or our B equals 115.

Hello here we have to determine whether the function presented by a system of two equations is continuous at minus two. In order to answer this question, we have to calculate the limit of the function right. When X approaches minus two from the left and from the right, the first equation is represents approaching from the left, and second equation represents approaching from the right. The calculation is shown in the slide, the first function at X equals minus two equals to one, and the second function at X equals minus two also equals to one, so therefore these two limits are the same and that's why F function F is continuous at minus two.

Therefore number 19 in which we have been provided with a function of X equal to works by XQ Bliss eight. And we have to find whether the function is continuous. On the indicated interval, the first interval is minus four comma minus three. The best way is to destroy it. So for drawing effects we should just get some points that is first to his. Is it passing through zero comma zero? So if you plug in X equal to zero, I will become zero. So yes, it is passing to zero comma zero. Does this function have uh the vertical ascent tote? So yes, X cube plus eight equal to zero. So X plus two. Access squired minus go two weeks plus four equal to zero. So X plus two equal to zero. That is X equal to minus two years. X equal to minus two Is the vertical ascent toad. Now we have to find if it has intercepts so when X equal to zero, why call to zero and when ah actually, okay, so there are no intercepts only passing through zero comma zero and X equal to minus two is uh particular symptoms. There will be no horizontally some thought as degree of the numerator is one and the german 33 so degree of the numerator is less than less than the diameter. So there will be no whole generation tote X equal to minus two. Is it? Okay? Now if we plug in X equal to one, it will be positive. If you plug in X equal to minus one, it will be negative. So the craft must look like yeah, okay like this and it will be passing through, it will be passing through zero comma zero. So okay, there will be graph like this. Okay. Yes, yeah. Okay, now we have to find the continued in the interval minus four comma three, so minus four minus four will be somewhere here and three will be minus four comma minus three will be somewhere here. So yes, in this interval it is continuous. So yes it is continuous and minus four comma minus three. Now second interval is minus infinity, second interval is miles and 50 to plus infinity. Okay. And open interval. So since we see that we have a sympathy in X equal to minus two. So no, it is not continuous and minus infinity to plus infinity. Thank you.


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