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AP Calc: LIM‑2 (EU), LIM‑2.C (LO), LIM‑2.C.1 (EK), LIM‑2.C.2 (EK)

The function, f of x is equal
to 6x squared plus 18x plus 12 over x squared minus
4, is not defined at x is equal to
positive or negative 2. And we see why that is, if x is
equal to positive or negative 2 then x squared is going
to be equal to positive 4, and 4 minus 4 is
0, and then we're going to have a 0
in the denominator. And that's not defined. We don't know what that happens
when you divide-- well we've never defined what happens
when you divide by 0. So they say, what value should
be assigned to f of negative 2 to make f of x
continuous at that point? So to think about
that, let's try to actually simplify f of x. So f of x-- I'll just rewrite
it-- is equal to-- Actually let me just start simplifying
right from the get go. So in the numerator
I can factor out a 6 out of every one of those terms. So it's 6 times
x squared plus 3x plus 2 over-- and
the denominator, this is a difference of squares. This is x plus 2
times x minus 2. And then we can factor
this expression up here. So this is going to
be equal to 6 times-- let me do it a different color. So we think of two numbers
and if I take their product I get 2. If I take their sum I get 3. The most obvious one is 2 and 1. So this is 6 times x
plus 2 times x plus 1. When you take the
product there you'll get x squared plus 3x
plus 2, and then all of that over x plus
2 times x minus 2. Now, if we know that x
does not equal negative 2. Then we can divide both the
numerator and the denominator by x plus 2. The reason why I'm
making that constraint is that if x were to
be equal to negative 2 then x plus 2 is going
to be equal to 0. And you won't be
able to do that. You can't. We don't know what it means
divide something by 0. So we could say
that this is going to be equal to-- so we
can divide the numerator and denominator by x plus 2 but
we have to assume that x is not equal to negative 2. So this is equal to 6 times--
we're going divided by x plus 2 in the numerator, x plus
2 in the denominator-- so it's going to be 6 times
x plus 1 over x minus 2. And we have to put
the constraint here because now we've changed it. Now this expression
over here is actually defined at x equals negative 2. But in order to be equivalent
to the original function we have to constrain it. So we will say for x
not equal to negative 2. And it's also obvious that
x can't be equal to 2 here. This one also isn't
defined at positive 2 because you're dividing by 0. So you could say, for x does not
equal to positive or negative 2 if you want to make
it very explicit. But they ask us, what could
we assign f of negative 2 to make the function
continuous at the point? Well the function is completely
equivalent to this expression except that the function is not
defined at x equals negative 2. So that's why we have to
put that constraint here if we wanted this to be the same
thing as our original function. But if we wanted to
re-engineer the function so it is continuous at that
point then we just have to set f of x equal to
whatever this expression would have been when x is
equal to negative 2. So let's think about that. Let's think about that. So 6 times negative 2 plus
1 over negative 2 minus 2 is equal to-- this is
6 times negative 1. So it's negative 6 over negative
4, which is equal to 3/2. So if we redefine f of x, if
we say f of x is equal to 6x squared plus 18x plus 12
over x squared minus 4. For x not equal
positive or negative 2, and it's equal to 3/2
for x equals negative 2. Now this function is going
to be the exact same thing as this right over here. This f of x, this new one. This new definition--
this extended definition of our original one-- is now
equivalent to this expression, is equal to 6 times x
plus 1 over x minus 2. But just to answer
their question, what value should be
assigned to f of negative 2 to make f of x
continuous to that point? Well f of x should be-- or f
of negative 2 should be 3/2.

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