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Question 33 Find the intervals on which the function is continuous.sn (40)discontinuous only when 0 discontinuous only when 0 = 0 discontinuous , only when 0 = I co...

Question

Question 33 Find the intervals on which the function is continuous.sn (40)discontinuous only when 0 discontinuous only when 0 = 0 discontinuous , only when 0 = I continuous 5 everywhere

Question 33 Find the intervals on which the function is continuous. sn (40) discontinuous only when 0 discontinuous only when 0 = 0 discontinuous , only when 0 = I continuous 5 everywhere



Answers

Determine the values of $x$, if any, at which each function is discontinuous. At each number where $f$ is discontinuous, state the condition(s) for continuity that are violated. $$f(x)=\left\{\begin{array}{ll} x+5 & \text { if } x \leq 0 \\ -x^{2}+5 & \text { if } x>0 \end{array}\right.$$

And we recommend the value of X. Or this will be the function is just convenient. The function is is given that prevents that it's this fight. If the exorcisms you know. Yeah. Then access questions And it is my hope that this couple is five. Have access to do first condition F. Zero because it is different. First isn't satisfied the second commission areas specific extent street Luminous that the X-plus five the first five. And the living extents to sleepless minus taxes. Focus five baby crystal fine because of their that experts zoo of F. Because 25 limited exist. LTD in this school. LTD excess and this entire this and then you and that's true Stephen. No offense so because you have off this hill so this technician is fears because is not this is not your fault. But then it's fine man. The value of the hotel juries it is not possible exports to zero into that that that's supposed to do. But something that affects. Yeah this convenience and then there's a third. His body. Thank you what it is. Mhm.

Hello. We have a question from calculus and from the topic. Limits and discontinuity are continuity. Oh, fields were fined in this question, we have to find if affects equal to sine X by X is continuous at X equal to X equal to zero. So given function, it affects sin X by X. We have to check the continued to at C equal to zero. So for any function to be continuous, we have three conditions left and limit should be equal to write and limit should be equal to the value of the function at that point. So let us start with left hand limit left hand limit, which is limit X approaches to see or X approaches to zero from left side effects. Mhm. Okay, Hello. This will become limit accidents to zero from the left side. Sign Xbox. So let us now convert this a limit in terms of debt. So in terms of this, will building has limited abstinence to zero saying minus set. My miners had to well, as his garden zero and it is very small quantity, Very small quantity. Okay, so limit at approaches to zero sign mindset is minus sign edge development mindset. So this will be limit as he approaches to zero sign act at which will be equal to one according to our formula. So left and limit is one right and limit limit X approaches to zero from positive side sign Xbox In terms of X, it will be limit at approaches to zero. Sign ads by it, which will definitely be equal to one. And value function at zero will be equal to signs E O bye zero. Okay, so find zero by 0700 Okay, so this is zero by zero. Mhm. That is indeed when informed so But if you use a squeeze the room it will lead that sin X by X called to sign zero by zero equal to one. We see that left and limit equal to write and limit equal to F zero equal to one. Since everything is every, all the three things are equal. So it has continued. T means it is not discontinuous, not discontinues at X equal to zero. And if we see it graphically, it looks like this. This is one. This is Oh, come on. When X equals zero, I could one So there is no point of this. Continue to over there. Thank you

For this given problem, we want to determine the values of X at which each function is discontinuous and then determine at each number where F. Is discontinuous. And these conditions for discontinuity, So we'll have in this case two x minus four When X is less than or equal to zero And then we'll have um 1. 1 actually. Great. Mm. Yeah. So right here we see that there's a jump discontinuity. And the reason why we would consider this a discontinuity is because the limit doesn't even exist. So you see the limit as we approach zero from the left is negative for and the limit as we approach zero from the right is one. So because those limits equal each other, we know that there is no way that this function can be continuous.

In Brooklyn. 38. We want to get the set at which dysfunction is continuous. The function. If is this fraction off the two polynomial, except for the point, X and Y equals zero and you, let's investigate this function. First we take the polynomial, the two. We have the divide off to polynomial the polynomial, our continuous everywhere. But because we have one parliament in the denominator, then we have the condition that X squared, plus X y plus y squared is not equal to zero. We have the condition that the dominator is not equal to zero. This is the condition from the polynomial. But for the value zero and zero, we have another way to evaluate the function, and it gives us zero. Because of that, we need to check the limit for the blown Amir. When we approach this point to make sure that this function this continues at the point off X and Y equals zero. First, let's evaluate the limit. Name it. Oh, ex way divided by X squared plus X y plus y squared. When X and y approaches zero and zero, we evaluate this limit. Let's take any bus to get the limit. For example, we take the bus. Both were X equals y. Then we substitute everybody by X. We have the limit off X squared, divided by X squared plus X squared plus X squared. When X approaches, zero equals the limit. Oh, one third When x approaches zero equals one third without continuing and getting another boss, we can see that the limit if it exists if it equals toe one third is not equal toe the value off the function. At this point, we have l the limit when x off the function when x and y approach 00 isn't it equal to f off zero and zero this If it exists, then the value is discontinuous at this point and this expression is true. Then the the domain off the function and it's discontinuous or the function is continues for everybody off x on Why applying the condition that X squared plus X y plus y squared is not equal toe zero. This includes the point X and Y equals zero and zero and this is the final answer off our problem


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