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CHRISTINE BREINER: Welcome
back to recitation.

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Today we're going to talk about
something you've been

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seeing in the lectures.

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Specifically, we're going to
talk about continuity and

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differentiability.

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And we're going to use an
example to see a little bit

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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 give you a little bit of

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time to work on it, 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 plus 1

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when x is bigger than 1.

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

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

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the b, which is a y-intercept,
if x is less

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than or equal to 1.

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

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drawn the part of x squared plus
1 if x is bigger than 1.

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So I've taken the x squared
function that starts at 0, 0--

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or the vertex is at 0, 0--

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I've shifted it up one unit,
and then I've chopped off

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everything left of
and including the

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value at x equal 1.

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

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

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allow this function to be
continuous when I put in the

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left part--

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so the values when x is less
than or equals to 1-- and what

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values for a 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

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through them for you.

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OK.

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So what we were doing, again,
is we were trying to answer

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this question, 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 to have a linear function

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that ultimately goes
through this point,

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whatever this point is.

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We have to figure out what that
point is, and then we can

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figure out what values of a and
b 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 to

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fill in that circle?

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

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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

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here 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 that at x

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equal 1, y is equal to--

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well, if I put in 1 here,
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 y equal 2
when x is equal to 1.

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OK?

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So ,in other words, down here--
again, let me just say

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before I go on, this point
is one comma two.

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So what I need is a line
that goes through the

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point one comma two.

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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

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2, and the input is a times--
well, x is 1--

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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, I

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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, it's the

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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, and that would fill in

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right here, and this would be
if we had on this side, this

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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, it's also not

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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, but f of x

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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 like

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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, I

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need the derivative as I come
in from the left to be equal

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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, I need this

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derivative--

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this gives me a certain value.

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As I come in from the
left, I need a

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derivative 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 to be

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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 has

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to equal the limit as x goes
to 1 from the right

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of f prime of x.

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I'm not sure--

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I think you saw this
notation already.

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Limits coming in from
the left are

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designated by a minus sign.

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Limits coming in from
the right are

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designated by a plus sign.

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

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function to have a well-defined
derivative at

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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 is x goes to 1 from the

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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 as x goes to 1 from the

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left of the derivative to
also be 2, so let's

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fill in this side.

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The function on the left, at
the left of 1 is ax 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, I can actually just

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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

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left, 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, I

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

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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 wanted to save that

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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 it first has to

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be continuous.

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So it has to satisfy this--

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a equals 2-- and it has
to satisfy this--

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a plus b equals 2.

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So that tells me a equals
2, and it tells me b

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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. Goes
through 0, has slope 2,

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so it goes through this point.

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

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we continued we see
that the tangent

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lines there 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, the left

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hand side is y equals 2x.

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So we've now figured out how
to make this piecewise

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function both continuous
and differentiable.

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And we'll stop there.

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