okay here we go the second half of our
worksheet and so now we're going to talk
about something called critical numbers
and then we're gonna kind of tie it all
together with the first half of what
we're looking at with the extreme values
so critical values a critical number of
a function is simply a number at which
the slope is zero so the slope is zero
when F prime if the derivative is equal
to zero so it's a value of x which gives
you a derivative equal to zero which
means the slope is zero or it is a value
which gives you an undefined slope or
where F prime of C the derivative
doesn't exist that those are usually
either one of those cases produce a
critical number so that is what it is
there is a theorem in your book called
Fermat's theorem and it says if and i
might be saying his name wrong cuz I
think he's French
but anyway if F has a local min or a
local max at a value C then the
derivative at that value has got to
equal to zero if F prime of C exists so
that's a little theorem right now we're
just going to practice finding critical
value so number three and number four
say find the critical numbers for this
function so they give us a nice
polynomial f of X is number three it
says f of X is equal to X to the third
plus six x squared minus 15 X F find the
critical values well a critical value is
where the derivative equals to zero so
first take the derivative alright so
what is f prime of X that's equal to
three x squared plus 12x minus 15 and to
find the critical values you simply set
this equal to zero and solve and we're
gonna find the critical numbers alright
so when you have an equation I'm going
you can simplify it by dividing both
sides by 3 by the way when you're
solving for x do not ever divide both
sides by X itself because what that does
is it eliminates
possible solution I've seen students do
that and it's not okay to do you can
divide by a number other than zero so
I'm allowed to divide by three but don't
divide both sides by X that is not
you'll lose a solution when you do that
if I divide both sides by three what I
get here is x squared plus four X minus
five is equal to zero that just
simplifies my quadratic equation nicely
and this one will factor into X plus
five X minus one let me just check x
squared for X minus five and then I get
X is negative five and X is one these
are the critical values that's all we
had to do right there was find the
critical values now we get one with the
trig function because we always have to
remember how to do trick so let's do
this one what if they say find the
critical values for this function and
they give us f of theta is equal to two
cosine of theta plus sine squared theta
all right well what is the critical
number you got to know that a critical
number is where the derivative equals to
zero or where their derivative is
undefined so let me just address was
there any place where the derivative is
undefined here is the derivative of the
function this is a polynomial in fact
it's a it's a it's a parabola if you
imagine cuz it's quadratic it is defined
for all values for X so there's no place
where this is undefined so we didn't
have to consider that let's find the
derivative of f of theta so f prime of
theta is okay two is a constant the
derivative of cosine is negative sine um
this is power this is sine squared if
you want to think of this like this this
is sine of theta squared so its power
rule its and chain rule so it's mm front
times the sine of theta to the first
power times the derivative of the inside
trivet of cosine is I mean if sine is
cosine all right so all of this is the
derivative all right so is there any
place where this is undefined
no because it's made up of sines and
cosines and those are defined for all
values of theta all real numbers all
angles so that's fine so we don't to
check for that but we do have to check
win with this equal to zero so we take
this negative two times the sine of
theta plus two sine of theta cosine of
theta we set it equal to 0 and we solve
for the values of theta that would make
this equal zero I'm going to factor
that's a very nice thing to do when
especially when you're solving trig
equations so I'm gonna I'm gonna factor
out a two sine of theta leaving me with
a negative one plus the cosine of theta
if you ever wondered if you factored it
correctly simply multiply back together
this sons this is this this exist is
this so I feel fine now I'm gonna use
the your product property which says
when you're multiplying stuff and it
equals zero that could only happen when
the things you're multiplying this times
this one of those a zero so this will be
true when the two sine of theta is equal
to zero or when negative one plus the
cosine of theta is equal to zero so I'm
gonna start with this one divide both
sides by two you this will be true when
the sine of theta is zero all right when
will this be true I'm gonna add one to
both sides this will be true when the
cosine of theta is equal to one so
recall your unit circle where is the
cosine of theta equal to one right here
and these are the values of zero and
this is the angle zero or 2pi or four PI
or six PI and so on how about this where
is the sine equal to zero oh the sine is
equal to zero also right here this is
the point 1 0 so this the X values
cosine the y-guy your sine so also here
that's where the sine is your but also
here we're at PI ok so the song is equal
to 0 at 0 pi 2 pi 3 PI 4 PI 5 PI back
and forth so we could say this real
simply whenever
is equal to PI times n where n is an
integer that would capture all of the
critical values for that one all right
what else could we say here why are we
looking for critical values critical
values are possible possible places
where we could find a local max or a
local min that's possibility if you
think about the pictures on the other
page right look at these where are the
local maxes here here here local mins if
you look at the slope of these functions
those are those are where it has a
horizontal slope that's where the
derivative is equal to zero so these are
possibilities of local maxes or local
mins not guaranteed but possibilities so
let's have a look at these that's why we
come up with against the closed interval
method and it helps us identify the
extreme it helps us find absolute Max
and absolute mins on a continuous
function on a closed interval so this is
this is what we can do if you have a
function that is continuous and it has
to be on a closed interval so this is
the conditions that have to be met it
has to be a closed interval and it has
to be a continuous function we already
saw it will achieve its absolute maxes
and mins that happens it will either be
another endpoint or it will be at a
local Max or min so this is a method to
find where it occurs so it says this is
if you want to find the absolute Max or
absolute mins the extreme values this is
what you do you find the critical values
and then you plug in the critical values
and you plug in the end points because
the extreme values will either occur at
an endpoint or at a local Max or min and
then simply the highest value that you
find is the absolute Max and the lowest
value that you find is the absolute min
this is what the closed interval method
it's a way of finding the
man and the absolute max on a closed
continuous function on a closed interval
so here we go we're just gonna try it
out this is how it works
number five goes like this and the
instructions are finding absolute min
and absolute Max of the function and
they give us a function the other thing
they could say would be find the extreme
values of this function that means find
the absolute bags and the absolute min
that Mehcad me it's the exact same thing
so they could word it slightly
differently be prepared for them
throwing the language around just a
little bit all right so let's look at
these two we are tasked with this so the
first thing we're gonna do is we want to
apply if possible to closed interval
method cuz it's super easy we want to
apply that so we have to make sure that
the function that they gave us a meets
the conditions that they gave us a close
interval and that the function is
continuous that's what we have to check
so number five it says find the extreme
values they actually max in the asset
meant for this f of X is equal to X to
the third minus six x squared plus five
okay let's look at this function there
is a theorem and we also are learning to
understand that this is a polynomial and
polynomials are continuous we learned
that a few chapters ago so we are
looking at a continuous function did
they give us an interval well they had
better give us an interval oh yeah yeah
thanks gave a sense
interval on negative three is less than
or equal to X is less than or equal to
five do you see these or equal to that
is a closed interval they could write it
like this or they could say on this from
negative 3 to 5 so I want you guys to
get real familiar with the different
ways they could tell you the same thing
so this meets the conditions for us to
be able to apply the extreme value
theorem so here's what we know an
extreme value will either occur at an
endpoint when X is negative 3 or when X
is 5 or at a critical value where there
might be a local Max or a min so this is
what the steps to do find the values of
F at the critical numbers so find the
critical values let's do that how do we
do that that's when the
zero first find the derivative all right
that's three x squared minus 12x plus
zero okay now set this equal to zero to
find the critical values I'm gonna
factor out a 3x leaving me with the X
minus four let's see three x squared
minus 12x okay set each one equal to
zero and solve this will happen when x
is zero this will happen when X is four
these are the critical values okay these
are the critical numbers values critical
values of that we found the critical
values we have the endpoints the last
thing to do when you're applying the
extreme value theorem is this or using
the closed interval method you plug in
these four values when X is zero when X
is 4 these are my critical values you
plug in these makes these endpoints when
X is negative 3 f of negative 3 and F of
5 you plug these in to not the
derivative but to the original function
and see what the Y values are alright so
if we do this f of 0 is gonna be 5 f of
4 is gonna be negative 27 F of negative
3 it's gonna be negative 76 and f of 5
is negative 20 now I did work these out
ahead of time just to save time on the
video but I plug these values into the
original function to get this alright so
after you get the critical values you
say what is the highest value that
happened we have negative 20 negative 76
negative 27 and 5 at 5:00
obviously is the highest output we just
label him the absolute max as this
function has an absolute max of 5 when x
is 0 that's what we know using this
theorem what is the lowest value that we
see okay of all these numbers neg
seventy-six is the lowest this is the
absolute men this is the absolute man
upset negative 76 one X is negative
three just like that it's a nice handy
theorem okay let's do number six find
the absolute max and absolute min of the
function f of X is equal to X to the
negative two times the natural log of X
on this interval on the interval from
1/2 to 4 ok so now first let's look at
the interval is this in the domain this
guy has a few domain issues is a
polynomial that's defined for everybody
he has some domain issues so let's just
see what does this mean f of X is the
natural log of x over x squared so what
do we have to worry about here division
by zero when will we be dividing by zero
when x is 0 but 0 is not in this
interval so we're okay the next thing is
the natural log you cannot take the log
of a negative number or of 0 it has to
be a positive number look at our
interval those are all positive numbers
so we're still okay so we have a
continuous function on this interval
there's a theorem back I forgot what
chapter but it was back when they said
polynomials are continuous they said
logarithmic functions and and power
functions are and rational functions are
continuous on their domain and we're in
the domain so they're continuous so
we're okay we have on endpoints a closed
interval so we've met the conditions to
apply the extreme value theorem or and
use the closed interval method so here
we go
all we got to do is find the places
where the derivative of this is equal to
0 or undefined so that's the critical
value so let's find the critical values
f prime of X and I'm gonna look at it in
this form because this is you can either
use quotient rule or you can use power
rule we use power first X to the
negative 2 times the derivative
the second one over X plus second the
natural log of x times the derivative of
the first negative 2x to the negative 3
is that right let's see here first
derivative second plus second derivative
of first yeah okay all right so this
becomes 1 over X cubed plus the natural
log of X Oh nonono minus 2 times the
natural log of x over X cubed okay this
is what at prime of x equals okay so
this is basically 1 minus 2 natural log
of X over X cubed okay so this is it
know how we saw this as we set it equal
to 0 and find the critical values so
let's set this equal to 0 and solve now
if you have a fraction equal to a
fraction you can always cross multiply
let's cross multiply and what we get
here is X cubed times 0 is equal to 0
and when we get here is this is equal to
1 minus 2 natural log of X okay so we're
gonna solve here for X now this will
tell us the values of x that will make
the derivative 0 when the derivatives
you're okay the other is when will the
derivative be undefined this will be
undefined when we're trying to divide by
0 or when we have a negative right
because that's what's out of the domain
so when X is equal to 0 this will be
undefined or if we had something
negative we don't have to worry about
those values because that's not in the
interval so we don't don't worry about
that one
let's just see what what values we get
for critical values in this interval
when we solve this equation so we'll
subtract 1 minus 1 is negative 2 natural
log of X divided by negative 2 1/2 is
equal to the natural log of X I'm gonna
raise both sides by E each of this each
of that what I just did
apply an inverse function this undoes
this so e to the one-half is X so this
is a critical value all right so what's
left to do plug in F of the endpoints f
of 1/2 as a 4 and f of our critical
value the square root of a basically
alright that's what we have to to plug
in and so I did this we we throw this
one out because it's not on there
interval right there so f of 1/2 I
already did this this is about negative
two point seven seven F of 4 is equal to
0.87 and ask of e to the 1/2 or the
square root of e is equal to point one
eight for extreme value theorem says the
highest the biggest value which occurs
right here I'm sorry the smallest value
is negative two point one seven is the
absolute men and it occurred at this
endpoint okay the biggest value we get
on this interval is e to the square root
8 f of the square root of e which gave
us 0.18 which is the highest value that
we see this is the absolute max and this
did not occur to end point it occurred
somewhere in between there okay so that
is an application of the closed interval
method got to have a continuous function
it has to be on a closed interval so
that's how you do that for point one
