WEBVTT

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PROFESSOR: Hi.

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Welcome back to recitation.

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I have here a little bit of
a strange problem for you.

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So let me just tell it to
you, and then I'll give you

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some time to work on it.

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So I want to define
a function, g of x,

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and I want to
define it piecewise.

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So when x is positive, I just
want g of x to be 1 over x.

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But when x is negative, I want
g of x to be 1 of x plus 2.

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So I've got a little graph
here of the function.

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So you've got, you know,
when x is positive,

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it's just your usual
y equals 1 over x.

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But when x is negative, I've
taken, I've shifted it up by 2.

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So this is a perfectly
good function.

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It's not defined at 0.

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OK?

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So what I would
like you to do is

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to compute the derivative
of this function, wherever

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it's defined.

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And you'll notice, when you get
there, that you'll have some..

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you'll get some answer, and
maybe you'll notice something

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a little weird
about that answer.

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So if you notice something weird
about it, what I want you to do

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is try and explain
why this is true.

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And if you don't
notice something weird,

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then, you know, come back and
we'll talk about it together.

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So why don't you pause the
video, go do that computation,

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and think about what, if there's
something strange going on

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here.

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And then come back and we
can talk about it together.

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Welcome back.

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Hopefully you had some fun
working on this problem

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and thinking about it.

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So let's do the
first part, which

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is just the computational part.

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Let's have a go at it.

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So, because this function
is defined piecewise,

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when we compute a derivative, we
can just compute the derivative

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on the different pieces.

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So the function
isn't defined at 0,

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so of course, it doesn't
have a derivative at 0.

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But then we can compute a
derivative when x is positive,

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and we can compute a
derivative when x is negative.

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So when x is bigger than
0, g prime of x, well,

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that's just d over
dx of 1 over x.

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So that's something
we're familiar with.

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Its minus 1 over x squared.

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So that's for x positive.

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When x is less than 0, g
prime of x is d over dx of 1

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over x plus 2, because
that's what g of x is.

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And, OK, and so this is,
well, the plus 2 gets killed,

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and so then we have the
derivative of 1 over x.

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That's minus 1 over x squared.

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So one thing you've noticed
is that this is minus 1

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over x squared here, and it's
minus 1 over x squared here.

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So although we defined
this piecewise,

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we could, we can
summarize this by saying,

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so the derivative is minus
1 over x squared always,

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so for all x not
equal to-- you know,

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it doesn't have a
derivative at x equals 0.

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It's not defined at 0, it
can't have a derivative there.

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So, but we don't need the
piecewise definition, anymore.

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So that was kind of
interesting, that we

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can summarize the derivative
of this piecewise function

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in a non-piecewise way.

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Now, the thing is, we've
learned what the anti-derivative

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of this function is.

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So we know that the
anti-derivative of minus 1

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over x squared dx is 1
over x plus a constant.

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So we know that the functions
whose derivative is minus 1

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over x squared are of the
form, 1 over x plus a constant.

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The thing is, this function
g that we just talked about,

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this function g
isn't of that form.

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Right?

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You don't get this function
by taking the function

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1 over x and just
shifting it up or down.

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You-- something weird happens.

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You've shifted it
up on one piece

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and not on the other piece.

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And yet, it's still true
that the derivative of g

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is equal to minus
1 over x squared.

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So this is a little bit
of a head-scratcher.

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And I wanted to talk
about why this happens.

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And the thing is that there's
a sort of theoretical reason

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for this, which is
that you remember

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that the reason that we know
that anti-derivatives have

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this form, a function
plus a constant,

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is because we know
that constants

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are the functions
with derivative 0.

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And so we were able to apply
the mean value theorem in order

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to show that if two functions
have the same derivative, then

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they differ by each other,
differ from each other

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by a constant.

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If two functions have
the same derivative,

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they differ by a constant.

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And we used, as a really
crucial step in that proof,

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the mean value theorem.

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Now the thing is, the
mean value theorem

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has, as one of its assumptions,
as one of its hypotheses,

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that the functions that
you're working with

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are continuous
and differentiable

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in some interval.

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OK?

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So what's happened here is
that the functions that we're

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talking about, the function 1
over x and the function minus 1

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over x squared,
those functions are

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continuous and differentiable
on certain intervals.

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So if we look-- if we
go back to this picture

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here we see that this function
g of x, just like the function 1

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over x, it's continuous and
differentiable for positive x,

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it's continuous and
differentiable for negative x,

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but at 0, there's
a discontinuity.

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So there's no
interval that crosses

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0 on which this function is
continuous or differentiable.

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As a result, the
mean value theorem

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can't tell us anything about
intervals that cross 0.

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So if the mean value
theorem doesn't tell us

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anything in that case, it
means the conclusion isn't true

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and we get a situation-- sorry.

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I should rephrase that.

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It means the conclusion
doesn't have to be true.

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Our proof doesn't work in a case
where we have a discontinuity.

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And what happens,
in fact, is right

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what we have here, which is that
when you have a function that

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has a discontinuity and you look
at its anti-derivatives, what

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you can do is that, in addition
to shifting the whole thing up

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and down, you can
shift the pieces

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on either side of the
discontinuity separately.

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Just like in this case
we can shift the piece

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to the left of 0 separately
from the piece to the right of 0

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and get a function whose
derivative is still

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what we started with.

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So this function g of x, we
get by shifting part of 1

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over x up, and it gives us a
function whose derivative is

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still minus 1 over x squared.

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So this is true anytime
you have a function

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with a discontinuity.

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So one consequence of this--
I'm going to go back over here

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and just write down one special
case of this-- is that to say,

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we say that the
anti-derivative of 1 over x dx

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is equal to ln of the
absolute value of x plus c.

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What this really means is
that when x is positive,

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we have a single kind
of anti-derivative,

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and they're of the form,
ln x plus a constant.

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And when x is negative, we
have a single anti-derivative,

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that's-- or single family
of anti-derivatives,

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of the form ln of
minus x-- remember,

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absolutely value of x is minus
x when x is negative-- plus c.

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But if we consider x to
be positive and negative

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at the same time,
the two constants

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don't necessarily have to agree.

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You can have the same
situation that you

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had before where one side can
shift up and down independently

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of the other, because there's
that discontinuity at 0 there.

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So this is just something
to keep in mind.

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It also means you
have to be careful

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with certain substitutions.

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You don't want to
do substitutions

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that have discontinuities.

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If you do substitutions
that have discontinuities,

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you might accidentally
introduce a discontinuity

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and bad things can happen
that I won't go into now.

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You can make-- end
up with statements

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that don't make any sense by
making a substitution where

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the function that
you're substituting

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has a discontinuity in it.

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So you-- or another
way of saying it is you

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have to restrict to some
interval on which it really

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is continuous.

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And then on each of those
intervals it makes sense,

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but bad things could
happen when you

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cross those discontinuities.

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So this is a little
bit theoretical,

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but I think it's a nice
thing to be aware of,

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a nice thing to keep in
mind when you're working

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with some of these expressions.

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So I'll end there.