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# 2011 Calculus AB frit svar #6a

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Problem number six. Let f be defined by f of x is equal to-- and we have two cases. When x is less than or equal to 0, f is 1 minus 2 sine of x. When x is greater than 0, f is e to the negative 4x. Show that f is continuous at x equals 0. So for something to be continuous at x equals 0, let's think about what has to happen. So if I have a function here-- So that is my x-axis. And let's say this is my y-axis. And we care about what happens at x equals 0. So x equals 0 there. And let's say that this is our function. So maybe our function maybe looks something like this. I don't know. Well, this one in particular probably looks something like this. This one in particular might look-- who knows. And then it might look something like that. And the particulars are that important. We just have to think about what they're asking. In order for it to be continuous here, so the limit as we approach 0 from the left should be equal to the value of the function at 0. So the limit of f of x should be equal to f of 0, which should be equal to the limit as we approach from the right, which should be equal to the limit as we approach 0 from the right. So that should be equal to the limit as x approaches 0 from the right of f of x. And the reason why this matters is, if this wasn't the case, if f of 0 wasn't the same as the limit, then we might have a gap right over there. So you could have limits. So you could have a situation like this, where you have a gap right over there. And then it looks something like that. So the two limits from the left and the right both exist, and the limit at that point would exist. But if the function itself does not equal that value there, if it equals something else, then the function would not be continuous. So that's why the limit has to be equal to the value of the function in order for it to be continuous. So let's think about whether all of these things equal each other. First of all, let's just think about the value of the function there. So remember, we're doing part a. F of 0 is equal to-- we're going to use this first case, because that's the case for x is less than or equal to 0. So f of 0 is going to be equal to 1 minus 2 sine of 0. Well, sine of 0 is 0, 2 times 0 is 0. So this whole thing is 0. 1 minus 0 is 1. Fair enough. Now let's think about the limit as x approaches 0 from the left-hand side. So as we approach 0 from the left-hand side of f of x. So as we approach 0 from the left-hand hand side, we're dealing with values of x that are less than 0. So once again, we're dealing with this first case up here. So this is the limit as x approaches 0 from the left-hand side of 1 minus 2 sine of x. Now, sine of x is continuous. It's a continuous function. So this is going to be the same thing as 1 minus 2 sine of 0, which we've already figured out, which is exactly equal to 1. So this is equal to 1. So the value of the limit as we approach from the left is the same as the value of the function. Now let's do it for as we approach from the right-- as we approach from values of x greater than 0. So let's think about the limit as we approach 0 from the right of f of x. So here, we're dealing with values of x that are larger than 0. So we're dealing with this case right over here. So that's going to be the limit as x approaches 0 from the right of e to the negative 4x. And for the x's that we care about, or actually in general, this is a continuous function. This is a continuous function. So this is going to be the same thing as e to the negative 4 times 0, which is just e to the 0, which is just 1. So this, once again, is equal to 1. So the function is equal to 1 at that point, the limit as we approach from the left is equal to 1, and the limit as we approach from the right is equal to 1. So the function is continuous there.