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in this lesson we're going to focus on
identifying the location of the absolute
maximums and minimums of a function so
let's start with a graph so you can
visually see it and let's say if we have
a function that looks like this so let's
call this point a point B C and D
identify the location of any absolute
and relative extrema of this particular
graph so the absolute maximum is the
highest point on a graph and the
absolute minimum is the lowest point now
this peak right here it's not the
highest point but it's the highest point
in this local area so it's called a
local Max or a relative maximum and this
point is a relative or a local minimum
now to identify the location of the
absolute extrema you simply have to
compare the Y values so you may need to
make a data table to do so to identify
the relative extrema you need to
identify the locations where the slope
is 0 since we have a horizontal tangent
line f prime of B will be equal to 0 and
F prime of C will also be equal to 0 so
you need to use the first derivative to
find the location of any relative
extrema now let's work on a practice
problem so let's say that f of X is
equal to x squared minus 4x plus 9 and
identify any absolute and relative
extreme values of the function anywhere
in the interval between 1 and 4 so the
first thing I would recommend doing is
identifying the critical points that's
where the first derivative is equal to 0
at those locations you could find any
relative extrema values like a relative
min or max the derivative of x squared
is 2x the derivative of negative 4x is
negative 4 and the derivative of a
constant is 0 so let's set this equal to
0 now let's take out a 2 so this is
going to be X minus 2 if we divide both
sides by 2 and 0 is equal to X minus 2
and if we add 2x is equal to 2 so we
have 2 as the critical number and we
need to identify if this is going to be
a relative maximum or a relative minimum
so we need to create a sign chart so
let's make a number line now we need to
determine what the slope is going to be
to the left and to the right of 2 so
let's pick a number that's greater than
2 let's try 3 if we plug in 3 into the
first derivative 3 minus 2 is positive
so the result will be positive if we
plug in a number that's less than 2 like
1 into the first derivative 1 minus 2
will be negative and then if we multiply
that by 2
that will give you a negative result so
notice that the slope changes from
negative to positive across the critical
point so a to do we have a relative and
Maxima or a relative minimum when a
slope is negative the function is
decreasing and when it's positive the
function is increasing and so this is a
minimum so we have a relative minimum at
x equals 2 so now let's move on to the
absolute extreme values so once you have
all of the critical points make a table
now on the table you want to put the
endpoints of the interval so run in 4
and then in between that any critical
numbers that you have now we need to
determine the Y values
for each of these x-values so we need to
plug in the values into this function to
get one so let's start with one F of one
is going to be 1 squared minus 4 times 1
plus 9 so this is 1 minus 4 plus 9 1
plus 9 is 10 10 minus 4 is 6 so f of 1
is equal to 6 now let's evaluate the
function when X is 2 so this is 2
squared minus 4 times 2 plus 9 2 squared
is 4 4 times 2 is 8
4 minus 8 is negative 4 negative 4 plus
9 is 5 now let's evaluate the function
when X is 4 4 squared is 16 4 times 4 is
the same thing 16 minus 16 is 0 0 plus 9
is 9 so f of 4 is equal to 9 so notice
that 9 is the highest Y value therefore
9 is the absolute maximum value of the
function 5 is the absolute minimum value
because 5 is less than 6 and 9 in
addition it's also the relative minimum
so it's both the absolute minimum and
the relative minimum on the interval of
1 to 4 and so that's how you could find
the absolute extreme values of the
function as well as the relative extreme
values now let's work on another problem
so let's use a different function so f
of X will be equal to let's say 4 X
cubed minus 30 9 x squared plus
90 X +2 so identify all of the absolute
and relative extreme values of the
function on the interval 1 to 6 so feel
free to pause the video if you want to
try this problem so let's begin by
finding the critical numbers of the
function so we need to find the first
derivative in order to do that the
derivative of X cube is 3x squared and
the derivative of x squared is 2x the
derivative of X is 1 and the derivative
of a constant is 0 so f prime of X is
going to equal 4 times 3 which is 12 so
that's 12x squared 39 times 2 is 78 and
then plus 90 so let's set the first
derivative equal to 0 now notice that
all of the coefficients are even so we
can divide everything by 2 if we do so
if we divide both sides by 2 on the left
0 divided by 2 is 0 on the right half of
12 is 6 half of 78 is 39 and half of 90
is 45 now 639 and 45 they are divisible
by 3 so let's divide everything by 3 so
this is going to be 0 and then 6 divided
by 3 is 2 39 divided by 3 is 13 45
divided by 3 is 15
so now let's factor the expression that
we have by the way if you want the first
derivative in its factored form we
divided everything by 6 so you can write
it like this if you want to
so how can we factor 2x squared minus
13x plus 15 how can we do that so first
we need to multiply the leading
coefficient by the constant term you
need to do that anytime this number is
anything but 1/2 times 15 is 30 now we
need to find two numbers that multiply
to 30 but add to the middle coefficient
of negative 13 so this is going to be 10
and 3 both of which are negative because
negative 10 times negative 3 is positive
30 but negative 10 plus negative 3 adds
up to negative 13 so what we're going to
do is replace the middle term negative
13x with negative 10x minus 3x and then
we need to factor by grouping so take
out the GCF in the first two terms the
greatest common factor is 2x 2x squared
divided by 2x is X negative 10x divided
by 2x is negative 5 in the last two
terms take out the greatest common
factor which is going to be negative 3
negative 3x divided by negative 3 is X
positive 15 divided by negative 3 and
it's negative 5 so now we need to factor
out X minus 5
so this term divided by X minus five
these two will cancel leaving behind 2x
and if we take this term divided by X
minus five these will cancel giving us
negative three so this is the first
derivative in its factored form
now we need to set each factor equal to
zero so FLE set X minus 5 equal to zero
and add 5 to both sides this is the
first critical number X is equal to 5
and then let's set the other factor to X
minus 3 equal to 0 so add 3 to both
sides and then divide by 2 so the second
critical point is 3 over 2 so now we
need to make a sign chart to determine
which of those critical points represent
the relative maximum and which one is
the relative minimum 3 over 2 comes
before 5 so let's pick a number that's
greater than 5 and then you should
rewrite this expression so let's try 6
if we plug in 6 we'll the first
derivative be positive or negative 6
minus 5 is positive 2 times 6 minus 3 is
also positive so if we multiply two
positive numbers this will give us a
positive result now notice that for each
factor we have an odd exponent that
means the signs will change across each
critical number so if you pick a number
between one point five and five it's
going to be negative so let's try 2 2
times 2 minus 3 that's positive that's
equal to positive 1 2 minus 5 is
negative a negative number times a
positive number will give us a negative
ezel and if we pick something that's a
less than 2 we have a 2 like zero that
will be positive zero minus 5 is
negative two times zero minus 3 is also
negative if you multiply two negative
numbers that will give you a positive
result
now which of these critical numbers
represents a maximum and which one is a
minimum well let's focus on 3 over 2 the
function is increasing and then it's
decreasing so that represents a maximum
and for five its decrease in first and
then is increasing so that's a minimum
so now let's make a table
so let's put the endpoints so we have
one and six and make sure that the
critical numbers are in this interval if
they're not you can't use them so 3 over
2 is 1 point 5
that's between 1 and 6 and so it's 5 now
let's identify the y-values for each of
the X values so let's start with one
this is going to be 4 minus 39 plus 90
plus 2 and I'm going to use the
calculator for this one so I got 57 when
X is 1 now when X is 3 over 2 or 1.5
this is going to be 4 times 1.5 to the
third power minus 39 times 1.5 squared +
90 times 1.5 plus 2 so hopefully you're
allowed to use calculators on a test
because this will take a while we could
do it without a calculator but I'm not
gonna do it in this video
so I got as the decimal 60 2.75 now for
F of five I'm just going to type it in I
got negative 23
and then for F of 6 I got two
now which of these points represent the
absolute maximum in this interval sixty
two point seven five is the highest
value so this is the absolute maximum
what about the absolute minimum negative
twenty three is the lowest Y value
listed so five comma negative
twenty-three that's the absolute minimum
so the absolute minimum of the function
is located at x equals five but the
absolute minimum value that represents
the y-coordinate the value is negative
twenty three so typically the location
is associated with the x coordinate but
the value is associated with the y
coordinate now we said that three over
two was also the relative maximum it was
a critical number and five was a
relative minimum so sometimes the
endpoints they may be an absolute Max or
an absolute minimum you don't know you
have to check it other times the
absolute max can be the relative Max and
the absolute minimum can be the relative
minimum so you need to make the table
make sure you put the endpoints of the
closed interval and their critical
points in the table and then just
compare the y-values so that's it for
this video
