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# Evaluating a limit expression for the derivative of a linear function

Video transcript

Let g of x equal
negative 4x plus 7. What is the value of the limit
as x approaches negative 1 of all of this? So before we think about this,
let's just visualize the line. And then we can think about
what they're asking here. So let me draw some axes here. So this is my vertical axis
and this is my horizontal axis. And let's say this is my x-axis. We'll label that the x-axis. I'll graph g of
x. g of x is going to have a positive-- I guess
you would say y-intercept. or vertical axis intercept. It's going to have a
slope of negative 4, so it's going to look
something like this. Let me draw my best. So it's going to look
something like that. And we already know the slope
here is going to be negative 4. We get that right from
this slope intercept form of the equation, slope
is equal to negative 4. And they ask us,
what is the limit as x approaches negative 1
of all of this kind of stuff? So let's plot the
point negative 1. So when x is
negative 1, so that's this point right over here. And this point right over here
would be the point negative 1, g of negative 1. Let me label everything else. So I could call this my y-axis. I could call this graph. This is the graph of
y is equal to g of x. So what they're
doing right over here is they're finding the slope
between an arbitrary point x, g of x, and this point
right over here. So let's do that. So let's take another x. So let's say this is x. This would be the x, g of x. And this expression
right over here, notice it is your change
in the vertical axis. That would be your g of x. Let me make it this way. So this would be your
change in the vertical axis. That would be g of x
minus g of negative 1. And then that's over-- actually,
let me write it this way so you can keep track of the
colors-- minus g of negative 1, all of that over your change
in the horizontal axis. Well, your change in
the horizontal axis is this distance, which is the
same thing as this distance. Notice your change in vertical
over change in horizontal, change in vertical over
change in horizontal, reviewing the green
point as the endpoint. So it's going to be
x minus negative 1. And this is the exact
same expression. These are the exact
same expression. You can simplify them,
minus negative 1, and this becomes plus. This could become a plus 1. But these are the
exact same expression. So this is the
expression, really, for the slope between
negative 1 and g of negative 1 and an arbitrary x. Well, we already know that
no matter what x you pick, the slope between x, g of , and
this point right over here is going to be constant. It's going to be the
slope of the line. It's going to be
equal to negative 4. This thing is going to
be equal to negative 4. It's going to be
equal to negative 4. Doesn't matter how close
x gets, and weather x comes from the right or
whether x comes from the left. So this thing, taking
the limit of this, this just gets
you to negative 4. It's really just the
slope of the line. So even if you were to take
the limit as x approaches negative 1, as x gets
closer and closer and closer to negative 1, well
then, these points are just going to get closer
and closer and closer. But every time you
calculate the slope, it's just going to be the
slope of the line, which is negative 4. Now, you could also
do this algebraically. And let's try to do
it algebraically. So let's actually
just take the limit as x approaches
negative 1 of g of x. Well, they already
told us what g of x is. It is negative 4x plus
7, minus g of negative 1. So that's minus, what
is g of negative 1? Negative 1 times
4 is positive 4. Positive 4 plus 7 is 11. All of that over x plus 1,
all of that over x plus 1. And that's really
x minus negative 1, is you want to think
of it that way. But I'll just write x
plus 1 this way here. So this is going to
be equal to the limit as x approaches negative 1 of,
in our numerator-- let's see. 7 minus 11 is negative 4. We can factor out a negative 4. It's a negative 4 times x plus
1, all of that over x plus 1. And then since we're just
trying to find the limit as x approaches negative 1, so
we can cancel those out. And this is going to be
non-zero for any x value other than negative 1. And so this is going to
be equal to negative 4. So either way, we
get negative 4. But if you just realize,
hey, this is a line. It's going to have
a constant slope. This is just the slope
of between some arbitrary point on the line and the
point negative 1, 11, really. Negative 1, 11, you say, well
that's just going to the same as the slope of a line. It's negative 4.