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in this video we're going to focus on

finding the critical numbers of a

function

so what are critical numbers

well let me give you a visual

illustration

so let's say if we have this function f

and it looks like this

let's call this

point a

b

c

d

and e

so notice that at point c we have a

local maximum

and we also have

a horizontal tangent line

so at f prime of c

it's equal to zero c is a critical

number

and the same is true for d

this time we have not a local maximum

but

a local

minimum

and at d

the derivative is equal to zero that is

the first derivative so d and c are both

critical numbers

now another point that's a critical

number

is at

point e

so notice that we have a cusp

it's not differentiable at this function

but f of e has a value e exists in the

function

but because it's not differentiable f

prime of e

does not exist

nevertheless e is still a critical

number

now it's important to know that

a function will not exist if there's a

zero in the denominator

so if you have some function let's say

in a fraction format

if you set the numerator equal to zero

you're going to have

this situation a horizontal tangent line

and so

you'll get a critical number but if you

set the denominator equal to zero

then

it's not differentiable at that point

but it's still a critical number

so let's say if

this is equal to f prime of x

if we set the numerator equal to x

x will equal negative four

so that's going to be a critical number

if we set the denominator equal to zero

x will equal three

so these two are both critical numbers

keep that in mind

now let's start with this problem

let's say that f of x

is equal to 4x squared plus 8x

find the critical numbers of the

function so what we need to do is find

the first derivative set it equal to 0

and solve for x

so what's the derivative of 4x squared

using the constant multiple rules can be

4 times the derivative of x squared

which is 2x

and the derivative of 8x is going to be

8 times the derivative of x which is 1.

so that's we have 4 times 2x which is 8x

so 8x plus 8.

so we need to set the first derivative

equal to zero

and solve for x

so let's factor out an eight

if we do so we're going to get x plus

one

dividing both sides by eight zero is

equal to x plus one

so x is equal to negative one

so this is the critical number

of the function

f prime of negative one is equal to zero

now let's work on another example

let's say if we have this function

f of x is equal to 2 x cubed

minus 15 x squared

plus

36 x plus 10.

if you want to try this problem feel

free to pause the video

so go ahead and find the first

derivative

set it equal to zero and solve for x

the derivative of 2 x cubed is going to

be 2

times the derivative of x cubed which is

3x squared

the derivative of x squared is 2x and

the derivative of x is one

and the derivative of a constant is zero

so f prime of x is equal to six x

squared fifteen times two is thirty

and then plus thirty six now we're gonna

replace f prime of x with zero

and now let's solve for x

so we need to factor this expression

notice that every coefficient is

divisible by six

so let's take out the gcf which is 6.

6x squared divided by 6 is x squared

negative 30x divided by 6 is negative 5x

36 divided by 6 is 6.

so now we need to factor this trinomial

what two numbers multiply to six but add

to negative five

two times three is six but negative two

plus negative three adds up to negative

five

so it's going to be x minus two times x

minus three

so now we need to set each factor

equal to zero

and so we're going to get two answers

two

and positive three

so we have two critical numbers in this

particular example

here's another one that you can try

let's say that f of x

is equal to four x raised to the one

third

minus eight x

raised to the four thirds power

now

let's use the power rule

to find the first derivative so the

derivative of x to the one-third is

going to be one-third x

and then it's going to be one-third

minus one

now one is the same as three over three

and one minus three is negative two

so this is going to be one third x to

the negative two thirds power

and then minus eight times the

derivative of x to the four thirds so

using the power rule it's four thirds x

raised to the four over three minus one

which is four over three minus three

over three and that becomes

one over three

now for this particular type of problem

what you need to do right now

is take out the gcf you need to factor

it out

so from both terms what is the greatest

common factor

well both terms contain a four

and a three

so we can take out four thirds

how many x variables can we take out

so let's say if we have x squared plus x

to the fifth power

if you have to factor this expression

what would be the gcf

of these two terms

it's going to be the lowest of the two

numbers so it's going to be x squared

and you'll get 1 plus x cubed

now the way you get these numbers is you

need to divide you divide x squared by x

squared that gives you 1.

and for the second one if you take x to

the fifth power divided by x squared

that will give you x cubed

well we'll need to apply the same

principle here

so which of these two exponents is lower

in value negative two-thirds or

one-third

compare it on a number line the numbers

on the left are lower than the numbers

on the right so negative numbers are

lower than positive numbers on a number

line

so we're going to take out x raised to

the negative two thirds

now if you take this term and divide it

by that term

what will it give you

because they're exactly the same it's

going to be one

and if you take this term and divide it

by that term what will you get

the four thirds will cancel

so it's going to leave behind negative

eight and then if you divide x to the

one third by x to the minus two thirds

you need to subtract so it's going to be

one over three minus

negative two over three

which is basically

one over three plus two over 3 which is

3 over 3

and

3 divided by 3 is 1 so that's simply

just x

which works out nicely

so now this x i'm going to bring it to

the bottom so that we can convert the

negative exponent into a positive one

and so we now have four times one minus

eight x

divided by three

x raised to the two thirds

so that's the first derivative

now recall that to find a critical

number

f prime of c has to equal to zero

or

f prime of c

does not exist or is equal to one over

zero

so whenever you have a fraction

to find a horizontal tangent line or the

first critical number set the numerator

equal to zero

and for the critical numbers where it's

not differentiable set the denominator

equal to zero

that's what we're going to do in this

example anytime you have a fraction

so if we set the numerator equal to zero

we really don't need to worry about the

four

because if you divide both sides by four

zero over four is zero

so we just gotta set one minus eight x

equal to zero so if we add eight x to

both sides

one is equal to 8x and then if we divide

by 8 the first critical number is 1 over

8.

now

let's determine the next

critical number

so let's set the denominator equal to

zero

so if we divide both sides by three

zero divided by three is still zero and

so we can clearly see that x has to be

zero

if you raise both sides to the three

over two

zero raised to the three over two is

zero

and here the twos will cancel the threes

will cancel so x is equal to zero so

there's two critical numbers for this

particular problem

zero and one divided by eight

here's another example

let's say f of x is equal to

the absolute value of five x plus eight

identify the critical numbers in that

function

now this is just a generic shape for an

absolute value function

the absolute value of x looks like this

and the critical number

occurs right at the middle

because the function is not

differentiable at that point

notice that for this particular function

the slope changes from negative one to

positive one instantly

and so

you don't know what the slope is at this

point

it's not differentiable so anytime you

have an absolute value function or if

the

curve changes abruptly

you have a critical number f prime of c

does not exist at that point

nevertheless it's still a critical

number at that point

and to find it all you need to do is set

the inside equal to zero

and so for this graph the critical

number is x equals zero

so what about the example that we have

the absolute value of five x plus eight

all we need to do is set five x plus

eight equal to zero

subtracting both sides by eight we have

five x is equal to negative eight and

then we need to divide by five

so the critical number is negative eight

over five

so if you were to graph

that particular function

around negative one point six

is where the y value will be equal to

zero

and so we have a graph that looks like

this

the slope is five so

it should be rising quickly

so notice that this point is the

critical number

at

x equals negative eight over five

here's another problem

let's say that f of x is equal to x

times the square root of 9 minus x

go ahead and find the first derivative

and set it equal to 0

and determine the critical numbers

so in this example to differentiate it

we need to use the product rule

and just as a reminder the derivative of

f times g

is going to be the derivative of the

first part f prime times the second g

plus the first f

times the derivative of the second g

prime

so in this case we can say that f

is x

and g

is

the square root of nine minus x

but before we differentiate it we need

to rewrite it as x

times nine minus x raised to the one

half

so now let's begin

so the derivative of the first part x is

one

times the second

plus

the first part

times the derivative of the second which

is going to be if you use the power rule

and the chain rule it's going to be one

half

keep the inside the same

and then subtract the exponent by one

one half minus one is negative one half

and then differentiate the inside

function

the derivative of nine minus x is

negative one

so now

like before we need to factor out the

gcf

the only thing that we could take out

is nine minus x

and negative one half

is less than one half

so the gcf is going to be

nine minus x raised to the negative one

half

now

one half minus

negative one half if you divide those

two terms

that's going to be one half plus one

half so

the first one is going to be 9 minus x

raised to the 1

or simply just 9 minus x

and then

if we divide this term

by this one

these two will cancel

leaving behind negative one-half x

to make the negative exponent positive

i'm going to move this to the

denominator

so on top we have 9

minus x

minus 1 half x

and on the bottom

nine minus x raised to the positive one

half

so what is negative x minus one half

negative one minus one half

negative one is the same as negative two

over two

and negative two minus one is negative

three so it's negative three over two

so right now we have nine

minus three over two

x

divided by

i'm to rewrite this back into its

radical form the square root

of 9 minus x

now i want to get rid of this fraction

so what i'm going to do is multiply the

top and the bottom by 2.

so f prime of x

is equal to nine times two which is

eighteen

three over two times two is three so

this is gonna be negative three x and on

the bottom

2 times the square root of 9 minus x

now at this point

we need to set the numerator equal to 0

so 18 minus three x is equal to zero

adding three x to both sides eighteen is

equal to three x

and if we divide by three

the first critical number

x is equal to six

now for the second critical number

we need to set the denominator equal to

zero

so if we divide both sides by two

zero minus zero divided by two is zero

and then if we

that should not be a two that should be

zero

if we square both sides

zero squared is zero

and then we need to add x to both sides

so we can see that x

is equal to nine

so the critical numbers are six

and nine

now let's work on one last problem

let's say that f of x is equal to

sine squared x plus cosine

so we have a trigonometric function

go ahead and find the first derivative

set it equal to zero

and find all the critical numbers

now the first thing i'm going to do is

rewrite this function

because i'm going to have to use the

chain rule

on sine squared x i'm going to write it

like this sine x squared

so now let's determine the first

derivative

oh by the way

the function is restricted

from

zero to two pi

because sine and cosine are periodic

functions and they go on forever so

you may want to add that to

the problem

so if you want to try it feel free to

pause the video and work on it

now let's use the chain rule so i'm

going to move the two to the front

according to the power rule

keep the inside function the same

subtract the exponent by one two minus

one is one and then take the derivative

of the inside function

the derivative of sine is cosine

now we need to differentiate cosine the

derivative of cosine

is negative sine

now let's set the first derivative equal

to zero

and we need to factor the gcf which is

sine

so if we take out sine x

this term divided by that one

the signs will cancel leaving behind two

cosine x

and if we take the negative sign divided

by positive sign

that will give us

negative one

now we need to set each factor

equal to zero

so on the right side we need to add one

to both sides

so we can see that two cosine x

is equal to one and then if we divide

both sides by two

cosine x is one half

now when is sine x equal to zero

sine of zero is zero sine of pi is zero

and sine of two pi

is zero however

x does not include zero or two pi so we

need to get rid of those two answers

so we only have one answer in the

interval and that's pi so far

now when is cosine positive one half

cosine is positive in quadrants one and

four

cosine 30 is equal to the square root of

three over two but cosine 60 is one half

so therefore

x

is going to be 60 degrees which is the

same as pi over 3.

pi is 180 180 over 3 is 60.

and in quadrant 4 it's going to be 5 pi

over 3

which is 300 degrees

that has a reference angle of 60 in

quadrant 4 because 360 minus 300 is 60.

so if you need to review your

trigonometry

you can look up my new trigonometry

playlists you can find it in my channel

and you can review this stuff if you

need to

so these are the answers

pi over 3

pi

and 5 pi over 3 all of which are in this

interval

so that's it for this video now you know

how to find the critical numbers of a

function

so find the first derivative set it

equal to zero and solve for x

you