WEBVTT

00:00:00.000 --> 00:00:08.820
PROFESSOR: Welcome
back to recitation.

00:00:08.820 --> 00:00:10.850
In this video, what
I'd like us to do

00:00:10.850 --> 00:00:13.400
is answer the
following question.

00:00:13.400 --> 00:00:16.730
Suppose that f is a continuous,
differentiable function.

00:00:16.730 --> 00:00:19.930
And if it's derivative,
if f prime is never 0,

00:00:19.930 --> 00:00:23.010
and a is not equal to
b, then show that f of a

00:00:23.010 --> 00:00:24.640
is not equal to f of b.

00:00:24.640 --> 00:00:26.640
I'm going to let you think
about it for a while,

00:00:26.640 --> 00:00:28.490
see if you can come up with
a good reason for that,

00:00:28.490 --> 00:00:30.323
and then I'll be back
to explain my reasons.

00:00:38.010 --> 00:00:38.510
OK.

00:00:38.510 --> 00:00:42.300
Our object, again,
is to show, if f

00:00:42.300 --> 00:00:46.290
is continuous and differentiable
and its derivative is never 0

00:00:46.290 --> 00:00:49.200
and you're looking at two
x-values that are different,

00:00:49.200 --> 00:00:51.217
show that their y-values
have to be different.

00:00:51.217 --> 00:00:52.800
Show that if the
inputs are different,

00:00:52.800 --> 00:00:54.256
the outputs have
to be different.

00:00:54.256 --> 00:00:55.880
Now, this might remind
you of something

00:00:55.880 --> 00:00:58.882
you saw in lecture about if
the derivative has a sign,

00:00:58.882 --> 00:01:01.396
show the function-- if the
derivative is positive,

00:01:01.396 --> 00:01:03.020
show the function is
always increasing,

00:01:03.020 --> 00:01:04.394
or if the derivative
is negative,

00:01:04.394 --> 00:01:06.160
show the function is
always decreasing.

00:01:06.160 --> 00:01:10.340
So this is a similar
type of problem to that.

00:01:10.340 --> 00:01:13.010
So what we're going to use
is actually the mean value

00:01:13.010 --> 00:01:13.880
theorem.

00:01:13.880 --> 00:01:15.690
If you'll notice, I have f.

00:01:15.690 --> 00:01:18.380
It does satisfy the
mean value theorem

00:01:18.380 --> 00:01:20.220
on an interval from a to b.

00:01:20.220 --> 00:01:22.420
I haven't even specified
which is bigger, a or b.

00:01:22.420 --> 00:01:25.350
But it doesn't
matter in this case.

00:01:25.350 --> 00:01:26.280
So what do we know?

00:01:26.280 --> 00:01:31.290
The mean value theorem tells
us that if we look at-- well,

00:01:31.290 --> 00:01:37.990
let's just write it out-- f of
b minus f of a over b minus a

00:01:37.990 --> 00:01:40.830
is equal to f prime
of c-- and what

00:01:40.830 --> 00:01:43.200
do we know-- for
c between a and b.

00:01:46.100 --> 00:01:49.300
So we want to know whether
or not f of b minus f of a

00:01:49.300 --> 00:01:50.320
can ever be 0.

00:01:50.320 --> 00:01:52.740
We're trying to show
that it cannot be 0.

00:01:52.740 --> 00:01:57.960
So we're going to isolate this
expression and show that this

00:01:57.960 --> 00:01:59.530
subtraction cannot be 0.

00:01:59.530 --> 00:02:00.530
Well, how do we do that?

00:02:00.530 --> 00:02:03.427
Let me come over here to
give us a little more room.

00:02:03.427 --> 00:02:05.260
I'm going to rewrite
the mean value theorem.

00:02:05.260 --> 00:02:07.331
I'm going to multiply
through by b minus a.

00:02:12.250 --> 00:02:18.460
So we get f prime
of c time b minus a.

00:02:18.460 --> 00:02:20.090
Now, we just want
to show, again,

00:02:20.090 --> 00:02:22.530
that f of b minus
f of a cannot be 0.

00:02:22.530 --> 00:02:24.570
What's the only thing--
well, not only thing,

00:02:24.570 --> 00:02:26.570
we know two things-- what
two things do we know?

00:02:26.570 --> 00:02:29.350
We know f prime of c is not 0.

00:02:29.350 --> 00:02:30.380
That was given to you.

00:02:30.380 --> 00:02:33.360
f prime is never 0, so
certainly at any fixed value,

00:02:33.360 --> 00:02:35.360
f prime of c is not 0.

00:02:35.360 --> 00:02:37.860
So we know this term is not 0.

00:02:37.860 --> 00:02:39.720
We also know that b
is not equal to a,

00:02:39.720 --> 00:02:42.160
so we know b minus a is not 0.

00:02:42.160 --> 00:02:45.170
The only way to get a product
of two numbers to be 0

00:02:45.170 --> 00:02:46.380
is if one of them is 0.

00:02:46.380 --> 00:02:51.007
So this in fact, this
product is not equal to 0.

00:02:51.007 --> 00:02:52.840
The fact that this
product is not equal to 0

00:02:52.840 --> 00:02:55.950
tells us f of b minus f
of a is not equal to 0.

00:02:55.950 --> 00:02:59.600
And that alone is enough
to conclude that f of b

00:02:59.600 --> 00:03:03.024
is not equal to f of a.

00:03:03.024 --> 00:03:04.690
So, again, let me
just point out of this

00:03:04.690 --> 00:03:08.140
is probably reminds you very
much of the type of thing

00:03:08.140 --> 00:03:11.860
you've seen where you were
showing if f prime had a sign,

00:03:11.860 --> 00:03:15.400
then you could determine whether
f was increasing or decreasing.

00:03:15.400 --> 00:03:17.200
It's the same type
of problem as that.

00:03:17.200 --> 00:03:20.180
It's exploiting what the
mean value theorem tells you.

00:03:20.180 --> 00:03:21.835
So I think we'll stop there.