PROFESSOR: Hi.
Welcome back to recitation.
I have here a little bit of a strange problem for you.
So let me just tell it to you, and then I'll give you
some time to work on it.
So I want to define a function, g of x,
and I want to define it piecewise.
So when x is positive, I just want g of x to be 1 over x.
But when x is negative, I want g of x to be 1 of x plus 2.
So I've got a little graph here of the function.
So you've got, you know, when x is positive,
it's just your usual y equals 1 over x.
But when x is negative, I've taken, I've shifted it up by 2.
So this is a perfectly good function.
It's not defined at 0.
OK?
So what I would like you to do is
to compute the derivative of this function, wherever
it's defined.
And you'll notice, when you get there, that you'll have some..
you'll get some answer, and maybe you'll notice something
a little weird about that answer.
So if you notice something weird about it, what I want you to do
is try and explain why this is true.
And if you don't notice something weird,
then, you know, come back and we'll talk about it together.
So why don't you pause the video, go do that computation,
and think about what, if there's something strange going on
here.
And then come back and we can talk about it together.
Welcome back.
Hopefully you had some fun working on this problem
and thinking about it.
So let's do the first part, which
is just the computational part.
Let's have a go at it.
So, because this function is defined piecewise,
when we compute a derivative, we can just compute the derivative
on the different pieces.
So the function isn't defined at 0,
so of course, it doesn't have a derivative at 0.
But then we can compute a derivative when x is positive,
and we can compute a derivative when x is negative.
So when x is bigger than 0, g prime of x, well,
that's just d over dx of 1 over x.
So that's something we're familiar with.
Its minus 1 over x squared.
So that's for x positive.
When x is less than 0, g prime of x is d over dx of 1
over x plus 2, because that's what g of x is.
And, OK, and so this is, well, the plus 2 gets killed,
and so then we have the derivative of 1 over x.
That's minus 1 over x squared.
So one thing you've noticed is that this is minus 1
over x squared here, and it's minus 1 over x squared here.
So although we defined this piecewise,
we could, we can summarize this by saying,
so the derivative is minus 1 over x squared always,
so for all x not equal to-- you know,
it doesn't have a derivative at x equals 0.
It's not defined at 0, it can't have a derivative there.
So, but we don't need the piecewise definition, anymore.
So that was kind of interesting, that we
can summarize the derivative of this piecewise function
in a non-piecewise way.
Now, the thing is, we've learned what the anti-derivative
of this function is.
So we know that the anti-derivative of minus 1
over x squared dx is 1 over x plus a constant.
So we know that the functions whose derivative is minus 1
over x squared are of the form, 1 over x plus a constant.
The thing is, this function g that we just talked about,
this function g isn't of that form.
Right?
You don't get this function by taking the function
1 over x and just shifting it up or down.
You-- something weird happens.
You've shifted it up on one piece
and not on the other piece.
And yet, it's still true that the derivative of g
is equal to minus 1 over x squared.
So this is a little bit of a head-scratcher.
And I wanted to talk about why this happens.
And the thing is that there's a sort of theoretical reason
for this, which is that you remember
that the reason that we know that anti-derivatives have
this form, a function plus a constant,
is because we know that constants
are the functions with derivative 0.
And so we were able to apply the mean value theorem in order
to show that if two functions have the same derivative, then
they differ by each other, differ from each other
by a constant.
If two functions have the same derivative,
they differ by a constant.
And we used, as a really crucial step in that proof,
the mean value theorem.
Now the thing is, the mean value theorem
has, as one of its assumptions, as one of its hypotheses,
that the functions that you're working with
are continuous and differentiable
in some interval.
OK?
So what's happened here is that the functions that we're
talking about, the function 1 over x and the function minus 1
over x squared, those functions are
continuous and differentiable on certain intervals.
So if we look-- if we go back to this picture
here we see that this function g of x, just like the function 1
over x, it's continuous and differentiable for positive x,
it's continuous and differentiable for negative x,
but at 0, there's a discontinuity.
So there's no interval that crosses
0 on which this function is continuous or differentiable.
As a result, the mean value theorem
can't tell us anything about intervals that cross 0.
So if the mean value theorem doesn't tell us
anything in that case, it means the conclusion isn't true
and we get a situation-- sorry.
I should rephrase that.
It means the conclusion doesn't have to be true.
Our proof doesn't work in a case where we have a discontinuity.
And what happens, in fact, is right
what we have here, which is that when you have a function that
has a discontinuity and you look at its anti-derivatives, what
you can do is that, in addition to shifting the whole thing up
and down, you can shift the pieces
on either side of the discontinuity separately.
Just like in this case we can shift the piece
to the left of 0 separately from the piece to the right of 0
and get a function whose derivative is still
what we started with.
So this function g of x, we get by shifting part of 1
over x up, and it gives us a function whose derivative is
still minus 1 over x squared.
So this is true anytime you have a function
with a discontinuity.
So one consequence of this-- I'm going to go back over here
and just write down one special case of this-- is that to say,
we say that the anti-derivative of 1 over x dx
is equal to ln of the absolute value of x plus c.
What this really means is that when x is positive,
we have a single kind of anti-derivative,
and they're of the form, ln x plus a constant.
And when x is negative, we have a single anti-derivative,
that's-- or single family of anti-derivatives,
of the form ln of minus x-- remember,
absolutely value of x is minus x when x is negative-- plus c.
But if we consider x to be positive and negative
at the same time, the two constants
don't necessarily have to agree.
You can have the same situation that you
had before where one side can shift up and down independently
of the other, because there's that discontinuity at 0 there.
So this is just something to keep in mind.
It also means you have to be careful
with certain substitutions.
You don't want to do substitutions
that have discontinuities.
If you do substitutions that have discontinuities,
you might accidentally introduce a discontinuity
and bad things can happen that I won't go into now.
You can make-- end up with statements
that don't make any sense by making a substitution where
the function that you're substituting
has a discontinuity in it.
So you-- or another way of saying it is you
have to restrict to some interval on which it really
is continuous.
And then on each of those intervals it makes sense,
but bad things could happen when you
cross those discontinuities.
So this is a little bit theoretical,
but I think it's a nice thing to be aware of,
a nice thing to keep in mind when you're working
with some of these expressions.
So I'll end there.