00:00:00
chapter 4 deals with curve sketching and
00:00:04
4.1 talks about the increasing and
00:00:06
decreasing functions so at this point
00:00:09
you've done um increasing and decreasing
00:00:12
intervals you did that in grade 12
00:00:14
Advanced functions but in in this case
00:00:17
we're going to be using the derivative
00:00:19
to help sketch functions and we're going
00:00:22
to be finding the Max and minimum values
00:00:24
by setting the first derivative equal to
00:00:27
zero so in this section although the uh
00:00:30
um the theory part is quite simple there
00:00:33
are a number of different types of
00:00:34
questions that I will be covering that
00:00:37
is that are part of your homework okay
00:00:39
so the first thing we want to talk about
00:00:41
is where how do you determine an
00:00:43
increasing
00:00:44
function so the description is that if
00:00:49
X1 is less than x2 in other words say
00:00:51
this was one and this is four that's all
00:00:53
that means and F of X2 is greater than F
00:00:57
at
00:00:58
X1 then the tangent slopes are positive
00:01:01
so you can see that if I drew tangent
00:01:04
points to this curve they would all have
00:01:07
positive slopes and the function is
00:01:12
increasing okay so that's pretty simple
00:01:14
you can read from left to right so in
00:01:16
other words if I go from left to right
00:01:18
and the function gets higher then it's
00:01:21
increasing so sometimes the descriptions
00:01:24
are a little worrier than they need to
00:01:26
be and again this is a case here where
00:01:28
it says X1 is less than X2 so let's say
00:01:31
this is one this is two and F at
00:01:34
X2 is less than F at X1 and the tangent
00:01:39
slopes are negative so here if I drew
00:01:42
tangent slope to this
00:01:43
line um maybe it was increasing a little
00:01:47
bit there that would not be true so from
00:01:49
here to here the function is definitely
00:01:52
decreasing because the tangent slopes
00:01:54
are negative you should also note that
00:01:56
the tangents in a decreasing interval
00:01:59
the tangent lines are above the function
00:02:02
whereas when it was increasing the
00:02:04
tangent lines are underneath the
00:02:07
function and that's going to be
00:02:09
important on um another section that we
00:02:12
do when we talk about points of
00:02:14
inflection so it's decreasing on the
00:02:17
interval X1 to X2 or wherever um you're
00:02:20
asked to describe it
00:02:22
for so I want to do some examples the
00:02:26
first is going to be using the
00:02:28
derivative graph the following function
00:02:31
okay so using your knowledge from grade
00:02:35
12 Advanced functions you know this is a
00:02:38
cubic function you know that it has a
00:02:40
positive leading coefficient and you
00:02:43
know that it's going to start in
00:02:44
quadrant 3 and end in quadrant 1 so
00:02:47
don't just rely on your Calculus
00:02:49
derivative techniques use your knowledge
00:02:52
from the previous course and you can
00:02:54
always flip back and forth to to check
00:02:57
your answers using either calculus or
00:03:00
using your Advanced functions knowledge
00:03:03
so the first thing I would do is take
00:03:05
the derivative so y Prime here is 3x^2 -
00:03:10
3 and don't forget that you must say for
00:03:13
critical values CVS set y Prime equal to
00:03:18
zero so when I set this equal to zero
00:03:21
and I factor out of
00:03:23
three then I can see that I still have a
00:03:26
difference of squares here that I can
00:03:28
Factor again and what I'm trying to do
00:03:30
is find the places where the slope is
00:03:33
zero so if a function has zero slope it
00:03:36
means I'm either at a maximum or a
00:03:39
minimum value right so therefore x = -1
00:03:46
or one and these are the critical values
00:03:49
so critical values at minus one and one
00:03:53
now if I want to know what the height of
00:03:54
the function is and this is where you
00:03:56
have to be careful that you're plugging
00:03:58
these values back back into the original
00:04:01
function to find the critical points so
00:04:05
y when x = -1 so if this had been f of x
00:04:09
I could just say F minus one but it's y
00:04:12
so I'm going to write it like this and
00:04:14
this is quite a legitimate um way to
00:04:17
describe it so -3
00:04:20
-1 oh -1 + 1 so -1 cubed is -1 that's +
00:04:26
3 + 1 so that's 4 - 1 is
00:04:30
three so that means there is a critical
00:04:34
point at the critical point is at uh Min
00:04:40
-1 and 3 and you do the same thing for
00:04:45
when x equals 1 so you would say okay
00:04:48
that means y when xal 1 now y make sure
00:04:52
you're not plugging it back into the
00:04:54
derivative because obviously if I plug
00:04:56
back one into the derivative function
00:04:59
I'm going get zero and that would just
00:05:01
be not right okay so we put in 1 Cub - 3
00:05:06
* 1 + 1 that's 1 + 1 is 2 - 3 is -1 and
00:05:13
the other critical point
00:05:16
is 1 and minus1 so I'm going to put
00:05:20
those on my graph 1 and minus one and
00:05:23
Min - one and 1 2 3 and that's
00:05:27
here and I know I'm going from this
00:05:29
quadrant to this quadrant so my function
00:05:33
would have to go oh might probably be a
00:05:35
good idea to find the Y intercept the Y
00:05:39
intercept for y intercept you're going
00:05:42
to
00:05:43
set and you always write this out set x
00:05:47
equal to zero
00:05:49
so Y
00:05:51
intercept is I put in a 0 here 0 0 Y
00:05:55
intercept is 1 so it's going to go
00:05:58
through here now now I know this
00:06:01
function is going to go like this right
00:06:04
because I know a little bit about cubic
00:06:06
functions I know it had a positive
00:06:08
leading coefficient so what we want to
00:06:12
know though is is this the point was it
00:06:15
a maximum or a minimum and obviously you
00:06:17
know that because of what you know from
00:06:19
Advanced functions but you can also see
00:06:22
that as I go from left to right and I'm
00:06:25
going to write a little derivative graph
00:06:28
here I'm going to to call it y Prime and
00:06:31
this is also a very legitimate way of
00:06:33
doing what we call a first derivative
00:06:35
test now first derivative tests aren't
00:06:38
really talked about until the next
00:06:39
chapter but I think it's critical for
00:06:42
your understanding of what we're doing
00:06:43
here right now that you know what we're
00:06:46
doing so if I put in my critical points
00:06:49
here the critical values right these
00:06:52
ones here I put them on my graph and I
00:06:54
check points to the left and right of
00:06:56
these values so if I put in 2 into the
00:07:01
derivative okay that's why I wrote y
00:07:03
Prime here this is a y Prime because I'm
00:07:06
checking the slopes so if I put in a
00:07:09
number less than -1 into the derivative
00:07:13
here let's say I put in -2 that would be
00:07:16
12 - 3 is 9 all I care about is that it
00:07:19
is positive so the function is
00:07:23
increasing and then if I go from minus
00:07:25
one to one say I put in zero here into y
00:07:28
Prime I would get a negative value which
00:07:31
means the function is
00:07:33
decreasing and then if I put in two
00:07:36
obviously I'm going to get the same
00:07:37
answer as I did when I put in -2 because
00:07:40
I squared it so 12 - 3 9 positive so
00:07:45
this shows that there is a minimum this
00:07:48
is a minimum here and this is a maximum
00:07:52
so these are local mins and Maxs on the
00:07:56
interval so um
00:08:00
this critical point Min
00:08:01
-13
00:08:03
-13 minus one I guess I could have put
00:08:06
those on there so Min
00:08:09
-13 is a
00:08:12
maximum and 1 and minus1 is a minimum
00:08:20
value so that's what you're going to do
00:08:22
with this what we call a first
00:08:24
derivative test first derivative test is
00:08:27
basically checking the derivative
00:08:29
function to the left and right of the
00:08:31
critical values to see if the slope is
00:08:33
increasing or decreasing and again
00:08:36
remember that we're talking about the
00:08:38
slope so sometimes students forget which
00:08:41
equation do I plug these things into so
00:08:44
if you're checking the slopes so I want
00:08:45
to know that was positive that's
00:08:47
negative I'm using Y
00:08:50
Prime okay let's move on to second
00:08:53
question this one is asking you to graph
00:08:56
this function X x^2 + 1 so so we have
00:09:00
all kinds of things that we need to find
00:09:03
when you're doing um a graphing analysis
00:09:07
and later on in the chapter there's a
00:09:09
complete algorithm of the the things
00:09:12
that you need to check for so basically
00:09:15
we're looking to see is there a vertical
00:09:18
ASM toote and you know for a vertical
00:09:20
ASM toote that means the denominator um
00:09:24
it's a value that makes the denominator
00:09:26
zero and in this case there is no value
00:09:29
because if I set this equal to zero
00:09:32
let's see if I did this and this is a
00:09:34
very common mistake that students will
00:09:36
say plus or minus one but look what
00:09:38
happens when I move this to the other
00:09:39
side x^2 = -1 so therefore there is no
00:09:45
vertical ASM toote there's no solution
00:09:47
you can't square a number and get a
00:09:49
negative value so there's no vertical
00:09:51
ASM tootes what about a horizontal ASM
00:09:54
toote this brings back your knowledge
00:09:56
from Advanced functions as well
00:09:59
for horizontal I want to know what
00:10:01
happens as X approaches Infinity in
00:10:05
other words you don't have to say
00:10:06
infinity well you should hear but when
00:10:08
you're thinking about it you're saying
00:10:10
what happens is it X gets really really
00:10:11
really really big so if I squared
00:10:15
something really big and added one to it
00:10:17
this number in the denominator is going
00:10:19
to be significantly larger than the
00:10:22
number in the
00:10:23
numerator even though that will still be
00:10:25
a large number I'm dividing by something
00:10:28
much larger and so that means that Y is
00:10:31
going to approach zero so that means
00:10:34
there's a horizontal ASM toote and it is
00:10:38
y =
00:10:42
0 Okay so we've got our ASM tootes
00:10:44
covered what else do we need to know we
00:10:46
need to know what the X and Y intercepts
00:10:49
are so the x
00:10:53
intercept you set y equal to zero so all
00:10:58
that means is what how do I make the
00:11:00
numerator zero so if I set this equal to
00:11:04
zero that means that x equals
00:11:09
z is going to be the intercept is going
00:11:12
to be zero right and so you normally say
00:11:17
the x intercept is zero you don't say x
00:11:21
equals z because that generally means
00:11:23
you're talking about a line x equals 0
00:11:26
which it isn't the x intercept is zero
00:11:29
um I should have put a four here so for
00:11:31
the Y
00:11:34
intercept I set xal to 0 can't spell
00:11:38
today if I put zero in for X here 0 / 1
00:11:44
is
00:11:45
0 so I would say the Y
00:11:49
intercept is zero so I've got my x
00:11:53
intercept my Y intercept I know there's
00:11:56
no vertical ASM tootes and I found the
00:11:58
horizontal ASM toote okay so now I want
00:12:02
to go ahead and see if I can find any
00:12:05
maximum and minimums are critical values
00:12:07
for the function now critical value can
00:12:10
also be what makes the denominator zero
00:12:13
so vertical ASM tootes are also
00:12:15
considered critical values but in this
00:12:17
case I don't have any okay so I'm going
00:12:21
to take the derivative y Prime now we
00:12:24
have a quotient so I must apply the
00:12:26
quotient rule which is your hod High
00:12:28
rule
00:12:29
so I do
00:12:31
ho D high dtive of the top is 1 minus
00:12:35
high D ho the Der of x^2 + 1 is
00:12:40
2x all over ho^ 2 so I have x 2 + 1 2ar
00:12:48
and I'm going to simplify the
00:12:50
numerator so I have X2 + 1 - 2
00:12:57
x^2 and it's still all all over x^2 + 1^
00:13:02
2 and this gives me in the numerator
00:13:08
ax^2 x^2 +
00:13:11
1 over x^2 + 1^
00:13:15
2 okay so I've got my derivative and I'm
00:13:18
going to say what I always say when I'm
00:13:20
trying to find critical values so for
00:13:23
critical values I want to set y Prime
00:13:27
equal to Z and the only thing thing that
00:13:29
will possibly make y Prime 0 is if the
00:13:32
numerator is zero you can't have zero in
00:13:34
the denominator so I'm going to say X2
00:13:39
is equal to 1 so I'm just bringing this
00:13:41
up here and set it to zero so I brought
00:13:44
this to the other side so X is equal to
00:13:47
plus or minus one don't forget the plus
00:13:50
or minus okay it's a very common
00:13:54
mistake okay so I know these are
00:13:57
critical values and I want to find the y
00:14:00
coordinate very very important that you
00:14:03
think about that as you get to this
00:14:05
point if I want a point I need the
00:14:07
original function a point use the
00:14:10
original function okay so when x equal 1
00:14:16
Y is = to 1/ 1^ 2 + 1 is a
00:14:23
half
00:14:25
so the critical point is going to be
00:14:29
1 and 12 for the y coordinate now I'm
00:14:34
going to do the same thing when x =
00:14:37
-1 Y is = to 1 over -1 2ar I have the
00:14:42
same thing right I get 1
00:14:46
half so I have 1 and a
00:14:50
half um when X is - 1 just a minute here
00:14:54
I had Nega 1 up here I'm sorry -1 over 2
00:14:58
so a half so the other critical
00:15:02
point is 1 and
00:15:07
-2 okay so I want to check to see if
00:15:10
these are
00:15:11
actually critical points if there's a
00:15:14
maximum or minimum is there a change in
00:15:17
the slope so the slope has to change
00:15:20
from negative to positive or positive to
00:15:22
negative in order for it to be a Max or
00:15:25
a Min right so I'm going to write out my
00:15:27
number line like this
00:15:29
I'm going to call it y Prime label it
00:15:32
very
00:15:33
important and my critical points are oh
00:15:36
I didn't put the negative here so I have
00:15:40
one and I have negative 1 and I don't
00:15:43
have a vertical ASM toote so I don't
00:15:44
have to put this on on here so I'm going
00:15:47
to check y Prime so that's this this
00:15:50
equation here you're working with this
00:15:52
one right don't go back up to this one
00:15:55
that gives you the point this is your
00:15:57
test your first first derivative test
00:16:00
let's call it that right here first
00:16:04
derivative
00:16:07
test okay I'm going to plug in -2 -22 is
00:16:11
4 that's -4 + 1 is negative in the
00:16:14
numerator the denominator is squared so
00:16:18
it's going to be positive so this is
00:16:20
negative and I have negative slope
00:16:23
between minus1 and 1 I choose zero
00:16:26
always choose zero if you can it's easy
00:16:28
to work with so I have 1 over 1 is 1
00:16:31
that's positive so I have positive slope
00:16:35
here and after 1 if I put in two and
00:16:38
squared that would give me 4 -4 don't
00:16:41
forget I'm squaring before I apply that
00:16:44
negative so that gives me negative over
00:16:47
a positive is a negative and that means
00:16:50
there is a
00:16:52
maximum at xal 1 so this is a maximum
00:16:57
come back here and state that
00:17:00
and - 1 and2 this is a minimum so you
00:17:03
see how it's a minimum because I went
00:17:06
down I had negative slope and then I
00:17:08
went to positive slope you can actually
00:17:10
even sketch in the graph right it's
00:17:12
going to go like this and then like
00:17:16
this so the other thing is um I have my
00:17:21
minimum value at -1 and -2 I'm going to
00:17:26
do it in green on the graph here
00:17:30
so-1 and a half that's here that's a
00:17:34
minimum I have an ASM toote right here
00:17:37
too don't I remember I said a horizontal
00:17:40
ASM toote of y equal
00:17:43
0 and I have 1 and 1/2 that's a maximum
00:17:48
value and I think you might remember
00:17:51
from Advanced functions that you can
00:17:54
cross a horizontal ASM toote for finite
00:17:57
values of X and and we have an X
00:17:59
intercept Y intercept was right here so
00:18:03
as this approaches Infinity the function
00:18:05
is going to approach
00:18:08
zero and when we go this way this is
00:18:11
going to approach
00:18:13
zero so there's my
00:18:16
graph okay so you have to be careful
00:18:19
that you're checking uh using a first
00:18:21
derivative test again this isn't part of
00:18:24
4.1 but it's definitely something you're
00:18:26
going to do many many times so thought
00:18:29
might as well include that here and that
00:18:31
will help now if you were asked for the
00:18:33
increasing and decreasing intervals so
00:18:36
the function is
00:18:38
decreasing so let's get a a red pen here
00:18:41
I'll show you where it's decreasing so
00:18:43
it's decreasing here this is all
00:18:47
decreasing this is
00:18:49
decreasing so the other point here let's
00:18:54
find another color it's
00:18:56
increasing on this interval
00:19:00
it's not really much different color is
00:19:02
it okay so this is
00:19:05
increasing and these two other areas
00:19:08
were
00:19:09
decreasing
00:19:11
decreasing decreasing so in terms of
00:19:14
interval notation you would
00:19:17
say
00:19:21
increasing um so we have intervals so
00:19:24
this is all part of your um your
00:19:27
algorithm for analyzing a graph so it's
00:19:29
going to be
00:19:30
increasing um
00:19:32
forx is an element of so it's going from
00:19:36
minus one to one and I use round
00:19:40
brackets because I'm not including those
00:19:42
points it's neither increasing nor
00:19:45
decreasing when the slope is zero so you
00:19:48
don't include those and
00:19:51
decreasing on the interval for X is an
00:19:55
element of negative Infinity to minus
00:19:59
one so from here to here and from one to
00:20:03
Infinity so U for Union One to
00:20:08
Infinity okay so that's a that's a
00:20:10
pretty lengthy um
00:20:13
analysis um some of them will be much
00:20:15
more difficult than that but this is a
00:20:17
good introduction one for you to get a
00:20:20
grasp of of what you're going to need to
00:20:23
to look
00:20:24
for okay so the next question I want to
00:20:26
do is a question from the homework
00:20:31
and it says determine where fime X is
00:20:34
equal to zero now you might think that's
00:20:37
pretty easy thing to do but I find that
00:20:39
a lot of students have trouble with um
00:20:42
taking the derivative of this product
00:20:45
that also has an exponent so you're
00:20:48
using the chain Rule and the product
00:20:50
rule together so let's try the
00:20:53
derivative you might want to freeze
00:20:54
frame right now and try this and then
00:20:57
come back so I'll go ahead and do it for
00:21:00
you here so frime X so remember we do
00:21:03
the first so I'm going to leave that
00:21:05
alone times the derivative of the second
00:21:08
so the derivative of this is
00:21:10
2x first * der of the second plus the
00:21:13
second x^2 - 9 times the derivative of
00:21:17
the first so I have to do the exponent
00:21:20
first so I'm going to write two I leave
00:21:23
what's in the bracket alone I decrease
00:21:26
it by one so two goes to one and the
00:21:29
derivative of the inside which is
00:21:32
two and so now I'm going to expand this
00:21:36
um I'm just going to simplify the other
00:21:40
the second part
00:21:42
here and then we're going to factor it
00:21:45
because you're trying to find the zeros
00:21:47
you can't find the zeros when it's in
00:21:49
this format so 2 * 2 that gives me 4 *
00:21:55
x^2 - 9 * 2x -
00:21:59
one so now I want to look around this
00:22:03
plus sign for common factors so I'm
00:22:07
going to pull out a common factor what
00:22:09
is common to both sides of this equation
00:22:12
well the first thing would be that I
00:22:15
have um I have two here right and I have
00:22:20
four here so I can take two out of each
00:22:23
of these so I'm going to pull out a two
00:22:26
and then both sides of the equation have
00:22:28
a 2X minus one this one has two of them
00:22:32
but as long as they each have one I can
00:22:34
pull that out so I'm going to take out a
00:22:36
2 2x
00:22:38
-1 make a big square bracket and then
00:22:42
watch on both sides of this plus sign is
00:22:44
what you're looking for what did I take
00:22:46
out so I took out a two I took out one
00:22:49
of these so I'm left with x * 2x - 1 and
00:22:54
I'm going to write it out long ways for
00:22:55
now 2x - 1 on the right hand side of the
00:22:59
plus sign I've taken out a two so that
00:23:02
leaves me still with two CU I had four
00:23:05
here I took out this 2x - 1 it's right
00:23:08
here and I still have an x^2 -
00:23:13
9 okay so very easy for me to tell you
00:23:16
what makes this zero obviously it's
00:23:18
going to be a half but I need to
00:23:21
simplify this side so that I can tell
00:23:23
you what makes what's in this big
00:23:25
bracket equal to zero so I expand
00:23:29
2x^2 - x sometimes students think
00:23:32
they've done way too much work but you
00:23:35
haven't there's just lots to do here
00:23:38
okay so now I have 2 2x - 1 and in this
00:23:42
bracket I have 4x^
00:23:46
2 -
00:23:48
x -
00:23:51
18 and I want to be able to factor this
00:23:56
so what makes a product of 4 * 18 that's
00:24:00
72 so product of -72 and a sum of 1 and
00:24:05
I think you can see right away that it's
00:24:07
- 9 and 8 so - 9 and 8 - 9 * 8 is - 72
00:24:15
whoops - 9 and this should be a sum of
00:24:17
-1 - 9 + 8 is -1 now it's a complex
00:24:23
trinomial has a four out front here so I
00:24:26
put both of these over four and I reduce
00:24:29
so this becomes 2 over
00:24:32
1 and this one I can't reduce so now I
00:24:37
just write out my little brackets here
00:24:39
so 2x - 1 and my factors are 4x -
00:24:44
9 and x + 2 so if you expanded this you
00:24:50
should get back to that it's always a
00:24:52
good thing to do um to check your
00:24:55
work okay so now I'm going to say for C
00:24:58
IAL values CU they want to know where is
00:25:00
frime x equal
00:25:02
to0 for critical
00:25:05
values set F Prime x equal to 0 so X is
00:25:10
going to be equal to what makes this
00:25:13
bracket zero 1/2 this bracket
00:25:17
9/4 and this bracket
00:25:20
-2 now I'm going to do one more step
00:25:23
here and that is to determine whether
00:25:25
they are minimums or maximums and and
00:25:28
what the critical values are now the
00:25:31
critical values I'll let you figure that
00:25:33
out on your own you had to plug back in
00:25:35
these into the original function right
00:25:38
to find the y-coordinate or the height
00:25:40
of the function when X is each of these
00:25:43
values but I can do a first derivative
00:25:46
test by calling this F Prime X and on
00:25:51
this number line I'm going to put in
00:25:53
these critical values so I have minus 2
00:25:57
I have a half
00:25:59
and I have
00:26:00
94 so into the derivative function this
00:26:04
simplified one here is the one I'm going
00:26:07
to
00:26:08
use I'm going to test some values so I'm
00:26:11
going to try -3 first so some teachers
00:26:15
make you write out all these intervals
00:26:17
but you can pretty much do this on your
00:26:19
head right so if I put in Z put in -3 so
00:26:24
this would be -6 -1 so that's the two is
00:26:29
positive and then this little bracket
00:26:31
here would be negative and if I put in
00:26:34
-3 it would be negative again because I
00:26:37
would have minus 21 and if I put a
00:26:39
negative here this would also be
00:26:42
negative so I have oops I didn't mean to
00:26:45
put it like that so I have a negative
00:26:47
time a negative time a negative time a
00:26:49
positive and that's all
00:26:52
negative so it's downhill here negative
00:26:55
slopes between min-2 and 1 I plug in
00:26:59
zero very easily here put in x is 0 so I
00:27:02
have 2 * -1 * -9 so it's negative * a
00:27:07
negative time a positive and that of
00:27:10
course is a positive so that means at
00:27:13
minus 2 I should have a
00:27:16
minimum a minimum put not a mix but a
00:27:19
minimum and when I go between a half and
00:27:22
9/4 I'm going to plug in one nice and
00:27:25
easy so it's positive * a positive time
00:27:29
a negative time a positive so I had
00:27:33
three positives one negative so that
00:27:35
means this will be negative
00:27:38
downhill and on the other side of 9/4
00:27:41
9/4 is two and a bit so let's try three
00:27:45
so I have positive 6 - 1 that's still
00:27:48
positive 12 - 9 is positive and 3 + 2 is
00:27:52
positive all
00:27:54
positives so my function is going to go
00:27:56
like this now you can see from the
00:27:59
original function that it is a quartic
00:28:01
function 2x^2 * x^2 this is a 4th degree
00:28:06
right fourth degree
00:28:08
function fourth
00:28:10
degree so it means it has it has two
00:28:14
minimums this is going to be a minimum
00:28:16
at
00:28:18
9/4 it's going to be a Max at 1/2 it's
00:28:21
going to be a minimum at minus
00:28:24
2 and of course you still need to to
00:28:28
find you still need to find those Y
00:28:31
coordinates this isn't a critical point
00:28:33
it's a critical value critical point
00:28:36
means X and Y okay so this is getting
00:28:39
pretty long so I'm going to stop here
00:28:40
but I do have two more very important
00:28:42
questions that I'm going to do in a
00:28:45
second part for you keep studying
