Use the graph of $

Question

Use the graph of

f

in the figure to determine the values of

x

in the interval

open paren negative 3 comma 5 close paren

at which f fails to be continuous. Justify your answers using the continuity checklist.

Accepted Answer

Welcome back, everyone. In this problem, we want to identify the values of X in the interval open bracket -6, 10, close the bracket where F fails to be continuous, and we want to explain our answers. Here we have a graph of F of X, and for our answer choices A says F fails to be continuous at X equals -4 because the limit of F of X as X approaches -4 is not equal to F of -4 and at X equals 0 because the limit of F of X as X approaches 0 is not equal to F of 0. B says it fails to be continuous at X equals -4 because the limit of F of X as X approaches -4 does not exist. At X equals 0 because the limit of FF X as X approaches 0 is not equal to F of 0, and at X equals 4 because F of 4 is undefined. C says it's at X equals -4 because F of -4 is undefined at X E equals 0 because the limit of F of X as X approaches 0 does not exist, and at X E equals 4 because the limit of F of X as X approaches 4 does not exist. And the D says it as it's at X equals 0 because F of 0 is undefined and that X equals 4 because the limit of F of X as X approaches 4 does not exist. Now what is going to make our function fail to be continuous? Well, let's think about what it means to be continuous. There are 3 criteria to be met at a point for a function to be continuous. OK. Now recall, OK, that F is continuous. OK. At X equals C. If first of all. F of C is defined, OK, that is C is in the domain of F. Second, the limit of FF X, OK, the limit of FFX as X approaches C exists, and we know that limit exists if the left hand, the left hand limit is equal to the right hand limit. So I just put that in bracket here just as a reminder, OK. And third, that the limit, OK, as X approaches C of F of X, OK. is going to be equal to FFC. These are the three criteria to be met for our function to be continuous. I know we also know that we can look at our graph and tell where our function is discontinuous. Now, by looking at our graph here of F, we can tell that our function is discontinuous at X equals -4. Let me highlight that in red here. At X equals 0 and at X equals 4. So visually we can tell that it's discontinuous there, but the question is why is it discontinuous at that point? Let's start with X equals -4, OK? Now when we look. Let me put that in red as well, OK. When we look at X equals -4. What makes our function. Discontinuous or what makes our function fail to be continuous. Well, let's look at the limits of our function as it approaches from the left and the right. Now notice here that the limit as X approaches -4 from the left of F of X is equal to 2. And the limit as X approaches -4 from the right of F of X is equal to 6. So now, since the left hand limit, OK. is not equal to the right hand limit. That means the limit of F of X. As X approaches -4 does not exist. So that's why F fails to be continuous at that point. Next, let's look at it at X equals 0, OK? No, at X equals 0. I should probably make a note here just to separate them so it's another star or a point at X equals 0 if we go back to our graph, yeah. Notice here that if we think about it, well, we know that it's defined. OK, we also know the limit exists because it's approaching the same value from the left and the right, but here notice that our limit in this case, which would be equal to 0. Is not the same as F of 0, OK? In other words, at X equals 0, we can tell that the limit of F of X as X approaches 0. Equals 0 while OK f of 0 equals -2 from our graph. So because the limit as x approaches 0 of F of X is not equal to f of 0, that is why the function is discontinuous at that point. In other words, it failed to meet the third criterion. Now let's look at the final point at which it fails to be continuous at X equals 4. What's going on at X equals 4? We'll notice here at X E equals 4, OK, that there's a vertical asymptote at X E equals 4. The function approach is infinity from both left and right sides, suggesting that the function cannot have a have a finite value at X equals 4. Thus, because there's a vertical asymptote there, that means F of 4 is undefined. Recall from our criteria, the first criterion states that FFC would have to be defined. So in this case, FF4 is not defined in our interval, therefore, it fails to be continuous. Thus our function is discontinuous at X equals -4. At X equals 0 and at X equals 4, and if we think about our reasons for it failing to be continuous, then we can tell that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Use the graph of $f$ in the figure to determine the values of $x$ in the interval $open paren negative 3 comma 5 close paren$ at which f fails to be continuous. Justify your answers using the continuity checklist.
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Key Concepts

Continuity of Functions

Types of Discontinuities

The Continuity Checklist

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