WEBVTT
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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.