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CHRISTINE BREINER: Welcome
back to recitation.

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Today we're going to
talk about something

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you've been seeing
in the lectures.

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Specifically, we're going
to talk about continuity

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and differentiability.

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And we're going
to use an example

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to see a little bit
what the difference is.

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That how you can be
continuous and not necessarily

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differentiable.

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And how it's a little stronger
to have differentiability.

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So we're going to deal
with a piecewise function.

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I'm going to ask
a question, I'll

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give you a little bit
of time to work on it,

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and then we'll come back.

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So the question
is the following:

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for what values of a and b is
the following function either

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first, continuous, or
second, differentiable?

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So the function is defined
in the following way:

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f of x is going to be equal
to the function x squared

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plus 1 when x is bigger than 1.

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And it's going to be
equal to a linear function

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where you get to pick the
a, which is the slope,

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and you get to pick the b,
which is a y-intercept, if x

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is less than or equal to 1.

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So what I've drawn so
far, so that you can see,

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is that I've drawn
the part of x squared

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plus 1 if x is bigger than 1.

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So I've taken the x squared
function that starts at

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(0, 0)-- or the
vertex is at (0, 0)--

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I've shifted it up
one unit, and then

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I've chopped off everything
left of and including

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the value at x equal 1.

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So my question, again, is, what
choices do I have for a and b

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to either first, figure out
what choices for a and b

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allow this function
to be continuous

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when I put in the left part-- so
the values when x is less than

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or equal to 1-- and
what values for a

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and b allow this function
to be differentiable?

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So I'm going to give you a
moment to work on it yourself,

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and then we'll come back and
I'll work through them for you.

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OK.

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So what we were
doing, again, is we

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were trying to
answer this question,

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choose values for a and
b that make-- first,

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let's look at the
continuity question.

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OK.

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So what we really
need is, we need

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to have a linear
function that ultimately

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goes through this point,
whatever this point is.

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We have to figure out
what that point is,

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and then we can figure
out what values of a and b

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will allow that to work.

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So if I want a continuous
function, these a and b,

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again, they have to be such
that the line goes straight

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through that point.

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So what is that point?

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Well, the x-value is 1.

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So what would the y
value have to be in order

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to fill in that circle?

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We can actually look at f
of x and we can say, well,

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if I wanted it to be continuous,
then the y-value I need here,

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I need at x equal 1, is going to
be whatever the y-value is here

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at x equal 1.

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So let me write that.

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Make a little space over here.

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So to answer question
one, I need a and b so

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that at x equal 1, y is equal
to-- well, if I put in 1 here,

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what do I get?

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I get 1 squared plus 1.

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So I get 2.

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So I need y to equal 2
when x is equal to 1.

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OK?

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So ,in other words,
down here again,

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let me just say before I go
on, this point is 1 comma 2.

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So what I need is
a line that goes

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through the point 1 comma 2.

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OK?

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So what is that?

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That would be, in this case,
that would be, the output is 2,

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and the input is a times--
well, x is 1-- 1 plus b.

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So that means a plus
b has to equal 2.

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And that actually represents
all the solutions.

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So there are an infinite
number solutions I could have.

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And if you think about it,
I could draw some of these,

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I could draw some
of these lines.

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So let's take, for instance,
b equals 2 and a equals 0.

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When b is 2 and a is
0, it's the constant,

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it's the constant
function f of x equals 2.

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It's a straight line.

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That would graph-- going
to graph that here.

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That would graph
straight across,

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and that would
fill in right here,

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and this would be if
we had on this side,

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this is f of x equals 2.

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OK?

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I could also have chosen
a plus b equals 2.

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I could also have chosen
a equals 1 and b equals 1.

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So I could have slope
1 and intercept 1.

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Notice, by the way, this
one is not differentiable.

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Right?

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It's obviously-- there's
a significant break there.

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We'll see also, with a
equal 1 and b equals 1,

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it's also not differentiable.

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So a equals 1 is slope 1, and
b equals 1 is intercept is 1.

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That's this line.

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So this is kind of
running out of room,

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but f of x equals x plus 1.

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Hopefully you can see that.

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f of x equals x plus 1.

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f of x equals 2.

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This one looks a little
bit closer to being almost

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like the tangent lines are
going to match up there.

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But it's not quite.

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And we'll see in a
second what we will need.

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So this was, we answered
the continuity question.

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OK.

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So now let's answer the
differentiability question.

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OK.

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Notice, again, continuous.

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We have a lot of
lines that will work.

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OK.

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But to answer the
differentiability question,

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I need the derivative as
I come in from the left

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to be equal to the derivative
as I come in from the right.

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So actually, I think
I did that backwards.

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This is, as I come
in from the right,

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I need this derivative-- this
gives me a certain value.

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As I come in from the
left, I need a derivative

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coming this direction.

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I need them to be the same.

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That's what it would
mean for the function

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to be differentiable.

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I can write that in words.

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The limit as x goes to 1
from the left of f prime of x

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has to equal the
limit as x goes to 1

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from the right of f prime of x.

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I'm not sure-- I think you
saw this notation already.

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Limits coming in from the left
are designated by a minus sign.

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Limits coming in from the right
are designated by a plus sign.

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So we need these two limits to
agree in order for the function

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to have a well-defined
derivative at that point.

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So let's figure
out what this is.

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Because we have
the function here.

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The function to the right
of 1 is x squared plus 1.

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We know its derivative.

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Its derivative is 2x.

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So on this side,
we have the limit

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is x goes to 1 from
the right of 2x.

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And now we can fill that in.

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What is that?

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Well, I just evaluate at
x equal 1 and I get 2.

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OK?

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And now that means
I need the limit

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as x goes to 1 from the left
of the derivative to also be 2,

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so let's fill in this side.

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The function on the left, at
the left of 1 is a*x plus b.

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So what is its derivative?

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Its derivative is just a.

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Right?

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Because the derivative
of b-- b is a constant.

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That derivative is 0.

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The derivative of this is a.

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OK?

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So we get the limit is x
goes to 1 from the left of a.

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That's what this
left-hand expression is.

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Well, if I evaluate
a at x equal 1,

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I can actually just plug it in.

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There's no x there.

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This is a constant function, so
as x goes to 1 from the left,

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I just get a.

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So that tells me that
a has to equal 2 here.

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OK?

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So in order to make the
function differentiable,

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I only have one value
of a I'm allowed to use,

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and that's when a equals 2.

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OK?

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So I didn't draw this one, yet.

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I didn't draw it on
purpose because I

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wanted to save that for last.

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So when a is 2,
what's b have to be?

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Well, in order to
be differentiable

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it first has to be continuous.

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So it has to satisfy
this, a equals 2,

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and it has to satisfy
this, a plus b equals 2.

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So that tells me a equals 2, and
it tells me b has to equal 0.

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That comes from our
work in number one.

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OK?

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What does that represent?

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That's a line with
intercept 0 and slope 2.

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So I'll draw that one last.

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Goes through 0, has slope 2,
so it goes through this point.

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And if I draw those,
if I connect those up,

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we see, if we continued we see
that the tangent lines there

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agree exactly.

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But the function itself
is just this part.

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It's just, the right-hand
side is x squared plus 1,

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the left hand side
is y equals 2x.

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So we've now figured
out how to make

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this piecewise function both
continuous and differentiable.

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And we'll stop there.