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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