U1S8V07 Recitation video.txt

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Welcome back to recitation.
Today, we're going to talk about something you've
been seeing in the lectures.
Specifically, we're going to talk about continuity
and differentiability.
And we're going to use an example
to see a little bit what the difference is, that how
you can be continuous and not necessarily differentiable,
and how it's a little stronger to have differentiability.
So we're going to deal with a piecewise function.
So I'm going to ask a question, I'll
give you a little bit of time to work on it,
and then we'll come back.
So the question is the following.
For what values of a and b is the following function either
first continuous, or second differentiable?
So the function is defined in the following way.
f of x is going to be equal to the function x squared
plus 1 when x is bigger than 1, and it's
going to be equal to a linear function
where you get to pick the a, which is the slope,
and you get to pick the b, which is the y-intercept, if x
is less than or equal to 1.
So what I've drawn so far, so that you can see,
is that I've drawn the part of x squared
plus 1 if x is bigger than 1.
So I've taken the x squared function that starts at
(0, 0)-- or the vertex is at (0, 0).
I've shifted it up one unit and then
I've chopped off everything left of and including the value
x equals 1.
So my question, again, is what choices do I have for a and b
to either--
first, figure out what choices for a and b
allow this function to be continuous
when I put in the left part, so the values when x is less than
or equal to one.
And what values for a and b allow this function
to be differentiable?
So I'm going to give you a moment to work on it yourself,
and then we'll come back and I'll work through them for you.

OK, so what we were doing, again,
is we were trying to answer this question,
choose values for a and b that make-- first, let's
look at the continuity equation.
OK, so what we really need is, we
need to have a linear function that
ultimately goes through this point, whatever this point is.
So we have to figure out what that point is, and then we
can figure out what values of a and b will allow that to work.
So if I want a continuous function, these-- a and b,
again, they have to be such that the line goes straight
through that point.
So what is that point?
Well, the x-value is 1.
So what would the y-value have to be in order
to fill in that circle?
We can actually look at f of x and we can say, well,
if I wanted it to be continuous, then
the y-value I need here, I need at x equal 1
is going to be whatever the y-value is here at x equal 1.
So let me write that--
make a little space over here.
So to answer question 1, I need a and b so that at x equal 1,
y is equal to--
well if I put in 1 here, what do I get?
I get 1 squared plus 1.
So I get 2.
So I need y to equal 2, when x is equal to 1.
OK?
So in other words, down here--
So again, let me just say before I go on,
this point is 1 comma 2.
So what I need is a line that goes
through the point 1 comma 2.
OK?
So what is that?
That would be, in this case, that would be, the output is 2
and the input is a times--
well, x is 1--
1 plus b.
So that means a plus b has to equal 2.
And that actually represents all the solutions.
So there are an infinite number of solutions I could have.
And if you think about it, I could draw some of these--
I could draw some of these lines.
So let's take, for instance, b equals 2 and a equals 0.
When b is two and a is 0, it's the constant function f
of x equals 2.
It's a straight line.
That would graph-- we're going to graph that here.
That would graph straight across,
and that would fill in right here.
And this would be, if we had on the side,
this is f of x equals 2.
OK?
I could also have chosen a plus b equals 2,
I could also have chosen a equals 1 and b equals 1.
So I could have slope 1 and intercept 1.
Notice, by the way, this one is not differentiable, right?
It's obviously-- there's a significant break there.
We'll see also with a equal 1 and b equals 1,
it's also not differentiable.
So a equals 1 is slope 1 and b equals 1 is intercept is 1.
That's this line.

So this is-- kind of running out of room--
but f of x equals x plus 1.
Hopefully you can see that.
f of x equals x plus 1, f of x equals 2.
This one looks a little bit closer to being almost--
like the tangent lines are going to match up there,
but it's not quite.
And we'll see in a second what we will need.
So this was-- we answered the continuity question.
OK?
So now let's answer the differentiability question.
OK.
Notice again, continuous.
We have a lot of lines that will work.
OK?
But to answer the differentiability question,
I need the derivative as I come in from the left
to be equal to the derivative as I come in from the right.
So actually, I think I did that backwards.
This is, as I come in from the right, I need this derivative,
gives me a certain value.
As I come in from the left, I need a derivative
coming this direction.
I need them to be the same.
That's what it would mean for the function
to be differentiable.
I can write that in words.
The limit as x goes to 1 from the left of f prime of x
has to equal the limit as x goes to 1
from the right of f prime of x.
I'm not sure, I think you saw this notation already.
Limits coming in from the left are designated by a minus sign,
limits coming in from the right are designated by a plus sign.
So we need these two limits to agree in order for the function
to have a well-defined derivative at that point.
So let's figure out what this is, because we
have the function here.
The function to the right of 1 is x squared plus 1.
We know its derivative.
Its derivative is 2x.
So on this side, we have the limit
as x goes to 1 from the right of 2x.
And now we can fill that in.
What is that?
Well I just evaluate at x equal 1, and I get 2.
OK?
And now that means I need the limit
as x goes to 1 from the left of the derivative to also be 2,
so let's fill in the side.
The function on the left--
at the left of 1 is ax plus b, so what is its derivative?
Its derivative is just a, right?
Because the derivative of b-- b is a constant,
that derivative is 0.
The derivative of this is a.
OK?
So we get the limit as x goes to 1 from the left of a.
That's what this left-hand expression is.
Well, if I evaluate a at x equal 1,
I can actually just plug it in.
There's no x there.
This is a constant function, so as x goes to 1
from the left I just get a.
So that tells me that a has to equal 2 here.
OK?
So in order to make the function differentiable,
I only have one value of a I'm allowed to use,
and that's when a equals 2.
OK?
So I didn't draw this one yet.
I didn't draw it on purpose, because I
wanted to save that for last.
So when a is 2, what's b have to be?
Well in order to be differentiable,
it first has to be continuous.
So it has to satisfy this, a equals 2,
and it has to satisfy this, a plus b equals 2.
So that tells me-- a equals 2, that tells me b has to equal 0.
That comes from our work in number 1.
What does that represent?
That's a line with intercept 0 and slope 2.
So I'll draw that one last.
Goes through 0, has slope 2, so it goes through this point.
And if I draw those, if I connect those up,
we see-- if we continued, we see that the tangent lines there
agree exactly.
But the function itself is just this part.
The right-hand side as is x squared plus 1,
the left-hand side is y equals 2x.
So we've now figured out how to make
this piecewise function both continuous and differentiable.
And we'll stop there.