Finding maxima and minima Using First Derivative Test (with Example) Video Lecture | Mathematics (Maths) Class 12 - JEE

in this video we're talking about the
first derivative test and in this
particular problem we've been asked to
use the first derivative test to find
any extrema of the function f of X is
equal to X to the fourth minus 2x
squared so what is the first derivative
test well we call it the first
derivative test because we use the first
derivative to test the critical points
of the function to say where the
function is increasing and decreasing
and therefore to draw conclusions about
extrema which are local and global
maxima and minima so when we want to use
the first derivative test we're going to
follow the singing process every time so
you're always going to use the same
method to find critical points of the
function and then use the first
derivative test to test them and then
draw conclusions about extrema so here's
what we're going to do first we're
always going to start with our original
function f of X and we're going to take
its first derivative which is of course
f prime of X so the first derivative of
this particular function if we take the
derivative of the right hand side term
by term the derivative of x to the
fourth is 4x cubed the derivative of
negative 2x squared is negative 4x this
is our first derivative our next step is
always going to be to set the first
derivative equal to zero so we say zero
equals and then whatever we got here on
the right hand side then we want to
solve this equation for x and usually
that's going to involve factoring this
right hand side so in this particular
case we can factor out a 4x so we're
going to say 4x times x squared minus 1
we can factor x squared minus 1 so we
get 4x times quantity X plus 1 times
quantity X minus 1 and now the right
hand side is factored as completely as
possible we can use 0 theorem to set
each factor equal to 0 individually
because if 0 is equal to this right hand
side if 4x is equal to 0 then the entire
right hand side is equal to 0 since we
have all three of these factors
multiplied together if X plus 1 equals 0
then the right hand side will equal 0 if
X minus 1 is equal to 0
the right hand side will equal 0 so we
can set these equal to 0 individually we
can say 0 is equal to 4x we can say zero
equal to x plus 1 and we can say 0
equals X minus 1 all of these equations
make this equation true so the results
then that we get here for this first
equation if we divide both sides by 4 we
get x equals 0 here if we subtract 1
from both sides we get x equals negative
1 and here if we add 1 to both sides we
get X is equal to positive 1 these then
because they're the solutions to setting
the derivative equal to 0 these are the
potential critical points of the
function remember that a critical point
is a point at which the function changes
direction so if the function is
increasing and that hits a critical
point it will start decreasing or if the
function is decreasing and hits a
critical point at that point it will
start increasing so it changes direction
from increasing to decreasing or vice
versa but we can't conclude that these
three values are in fact critical points
until we've used the first derivative
test to test them so here's how we test
them here's what the first derivative
test looks like once we get these values
of critical points we want to set up a
simple number line so here's what that's
going to look like we just draw a simple
number line like this and we're going to
call the ends of the number line always
negative infinity and positive infinity
and then we're going to graph these 3
values on the number line so we have
negative 1 that's our left most points
we'll say negative 1 here then 0 then
positive 1 so these are our three
potential critical points we graph them
from right to left according to their
values notice that plotting these three
potential critical points on a number
line divides the entire number line into
four intervals so we have this first
interval here from negative infinity to
negative 1 the second interval from
negative 1 to 0 the third interval from
zero to positive 1 and the fourth
interval from 1 to positive infinity we
have to test each of these intervals
which means that we have to pick a test
value inside each interval it doesn't
matter which value we pick as long as it
falls within the interval so in other
words for interval number 1 here we have
to pick a test value that's between
negative infinity and negative 1 pick
something easy so in this case we'll
pick negative 2 because negative 2 is
between negative infinity and negative 1
we'll be able to use the first
derivative test with negative 2 to
determine the function
behavior on the interval negative
infinity to negative 1 in other words
we'll be able to use this test value of
negative 2 to say whether or not the
function is increasing or decreasing on
this interval negative infinity to
negative 1 because these values negative
1 0 and 1 are the only potential
critical points that means they're the
only points where the function can
change direction so the function can't
possibly change direction inside any of
these intervals which is why we can just
pick one single value to represent the
behavior of the function over the entire
interval so we'll pick test values on
each interval picking a value between
negative 1 and 0 let's pick negative 1/2
between zero and one will pick positive
1/2 and between 1 and positive infinity
we'll pick positive 2 so these values in
green are going to be our test interval
and the reason again that it's called
the first derivative test is because
we're going to plug these test values
into our first derivative so I always
like to remind myself by writing F prime
the first derivative next to my number
line so I remember where I'm plugging
everything in so we'll start with this
leftmost test value negative 2 we're
going to plug that into our first
derivative so we're going to say F prime
of negative 2 is going to be equal to
plugging that into 4 X cubed minus 4 X
we get 4 times negative 2 cubed minus 4
times negative 2 that's going to give us
here negative 2 cubed is a negative 8
negative 8 times 4 is a negative 32 then
we have minus 4 times a negative 2 which
is minus a negative 8 or plus 8 so
negative 32 plus 8 is going to give us a
negative 24 we'll come back to this
value in a second let's plug in the
other 3 test values and then we'll look
at what conclusions we can draw so
plugging negative 1/2 into the first
derivative F prime if negative 1/2 gives
us 4 times quantity negative 1/2 cubed
minus 4 times negative 1/2 when we
simplify here negative 1/2 cubed is a
negative 1/8 negative 1/8 times 4 is a
negative 1/2 and then we have negative 4
times a negative 1/2 which is a positive
2 so negative 1
plus two is a positive three-halves a
positive value let's plug in positive
1/2 so f prime positive 1/2 is equal to
4 times 1/2 cubed minus 4 times 1/2 1/2
cubed is a positive 1 8 times 4 is 1/2 4
times 1/2 is 2 so we get minus 2 that's
going to give us a negative 3 halves and
then our last test value of positive 2
we get F prime a positive 2 is 4 times 2
cubed minus 4 times 2 which is going to
be 8 times 4 or 32 minus 8 that's going
to be a positive 24 so the resulting
values we get here the specific values
are not important what's important is
whether or not those values are positive
or negative so this value here of
negative 24 all that matters is that
it's negative so this is less than 0 we
get a negative value right here positive
3 halves all that matters is it's
greater than 0 that's a positive value
negative 3 halves that's less than 0
it's a negative value and positive 24
that's greater than 0 that's a positive
value so we want to take these
conclusions about negative and positive
values and plot them here on our number
line plugging in the test value of
negative 2 resulted in a negative value
so we want to go ahead and put negative
right there about the negative 2
plugging in negative 1/2 gave us a
positive value
plugging in positive 1/2 gave us a
negative value and plugging in positive
2 gave us a positive value so we get
positive right here
so what the first derivative test tells
us is that when we plug in a test value
and we get a negative result the
functions behavior over the interval
represented by that test value is
decreasing so if we get a negative
result the function is decreasing on
this interval
because we got a positive result for
this test value the test value that
represents the interval from negative 1
to 0 that means the function is
increasing over that interval so the
function is increasing over this
interval then the function is decreasing
over this interval and then increasing
again over this interval
and I like to draw in those arrows
because it makes the function's behavior
very visual I can see a visual
representation of the functions behavior
and because the function changes
direction at each of these values x
equals negative 1 0 and positive 1 I can
confirm that negative 1 0 and positive 1
all represent critical points of the
function so I have a critical point here
at x equals negative 1 I have a critical
point here at x equals 0 and a critical
point here at x equals positive 1 and I
can see that because to the left of
negative 1 my function is decreasing
then when it hits negative 1 it starts
increasing so if it was decreasing and
then starts increasing once it hits this
value I know my function change
direction there and therefore I have a
critical point same thing here at 0 my
function was increasing to the left of 0
and then once it got to 0 it started
decreasing so zero represents a critical
point and so does x equals 1 because I
had decreasing on the left and
increasing on the right so I do in fact
have three critical points now I have to
say whether or not these critical points
represent local and/or global Maxima or
minima you can only have extrema where
you have critical points I just have to
plug in these values negative 1 0 and
positive 1 to the original function to
see what the value of the function is at
those critical points so that I can draw
a conclusion about the extrema at these
three critical points so plugging
negative 1 into the original function
we're going to say F of negative 1 and
then plugging it into this function here
I have X to the 4 so negative 1 to the
4th minus 2 times x squared or negative
1 squared negative 1 to the 4th is a
positive 1 negative 1 squared is a
positive 1 negative 2 times positive 1
is a negative 2 so the value I get then
is negative 1 we'll come back to that in
a second first let's plug in x equals 0
and x equals 1 so f of 0 is going to
give me 0 to the 4th minus 2 times 0
squared so I'm going to get 0 minus 0 or
just 0 and then plugging in x equals 1
I'm going to have F of positive 1 which
is going to be 1 to the 4th minus 2
times 1 squared which is going to give
me 1 minus 2 or a negative
so when X was equal to negative 1 Y is
equal to negative 1 so I can say that
the coordinate point here is negative 1
negative 1 when X was equal to 0 Y was
equal to 0 so I can say the coordinate
point here is 0 0 and when X was equal
to positive 1 Y was negative 1 so the
coordinate point here is positive 1
negative 1 now what we can tell just by
looking at these coordinate points is
that the function has global minimum at
x equals negative 1 and x equals
positive 1 and a local maximum at x
equals 0 you'll get more and more
comfortable drawing that conclusion just
from the coordinate points as you do
more of these problems but for now let's
look at the graph of this function to
make it really clear if you can graph
the function it's always going to help
you double-check the results of your
first derivative test so if we look here
at the function here's what we see we
have this graph and it kind of mimics
the arrows that we drew which should
make sense so we can see that our
function is decreasing from negative
infinity to negative 1 and you can see
here this is x equals negative 1 this
line right here so we can see the
function coming in from left to right
decreasing until it hits x equals
negative 1 then we saw that the function
starts increasing from negative 1 to 0
so from negative 1 until x equals 0
right here the function is increasing
then it starts decreasing again from 0
to negative 1 and then at negative 1 all
the way to positive infinity it starts
increasing again so the results that we
found about where the function is
increasing and decreasing matches the
graph of the function here so then we
can go ahead and say that the value of
the function at x equals negative 1 is
negative 1 so we can say this point here
is negative 1 negative 1 right here we
can say that this value we found we just
found this point was 0 0 and that this
point right here that we just found was
positive 1 negative 1 so what this shows
us then is that the function is never
going to reach a value lower than y
equals negative 1 right these two points
right here along this line
that we found these two critical points
right here the function is never going
to go below these two points these are
the lowest values the function will ever
obtain because we're going up here to
the left and up here to the right and
there's never going to be a point where
the function starts coming back down if
there were we would have found another
critical point so we're going to
increase up here to the left and
increase up here to the right these are
the lowest values the function will ever
obtain their form the points negative 1
negative 1 and positive 1 negative 1 are
the global minima of the function they
are the lowest points on the function so
we can call this point here a global min
we can call this point here a global
minimum and the reason we can have two
global minimum normally you can only
ever have one but if they both have the
exact same y-value then you can have two
global
minimums this point here is 0 0 we can
only call a local maximum because it's
the highest point in its neighborhood or
in its region it's the highest point of
the function around 0 0 because you can
see that the function slopes down to
either side of 0 0 so this is the
highest point in this area but obviously
the function reaches values that are
higher than 0 so all of these points up
here have higher Y values than our point
0 0 and same thing over here all of
these points have higher Y values than
the point 0 0 but 0 0 is the highest
point in this area so we call it a local
maximum so those are all the conclusions
that we can draw about the extrema of
this function we have to global minimums
and a global maximum and that's how you
use the first derivative test to find
critical points test the functions
behavior to see where it's increasing or
decreasing and then use what you know
about where the function increases and
decreases to draw conclusions about
local at global extrema