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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
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