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## AP Calculus AB 2011 free response

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# 2011 Calculus AB free response #6a

## Video transcript

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.