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Finding maxima and Minima Using Second Derivative Test (with Example) Video Lecture | Mathematics (Maths) Class 12 - JEE
FAQs on Finding maxima and Minima Using Second Derivative Test (with Example) Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. What is the second derivative test for finding maxima and minima? |  |
The second derivative test is a method used to determine whether a critical point of a function corresponds to a maximum, minimum, or neither. It involves evaluating the second derivative at the critical point. If the second derivative is positive, the critical point corresponds to a local minimum. If the second derivative is negative, the critical point corresponds to a local maximum. If the second derivative is zero, the test is inconclusive and further analysis is needed.
| 2. How can the second derivative test be applied to find maxima and minima? | |
To apply the second derivative test, follow these steps:
1. Find the critical points of the function by setting its first derivative equal to zero.
2. Evaluate the second derivative of the function at each critical point.
3. If the second derivative is positive at a critical point, it corresponds to a local minimum.
4. If the second derivative is negative at a critical point, it corresponds to a local maximum.
5. If the second derivative is zero at a critical point, the test is inconclusive, and further analysis is needed using other methods.
| 3. Can the second derivative test be used to find absolute maxima and minima? | |
No, the second derivative test can only determine whether a critical point corresponds to a local maximum or minimum. It does not provide information about absolute maxima or minima. To find absolute maxima and minima, other methods such as the first derivative test or evaluating endpoints of a closed interval need to be used.
| 4. What should be done if the second derivative is zero at a critical point? | |
If the second derivative is zero at a critical point, the second derivative test is inconclusive. In such cases, further analysis is needed. One possible method is to use the first derivative test. By examining the sign changes of the first derivative around the critical point, it is possible to determine whether it corresponds to a maximum, minimum, or neither.
| 5. Are there any limitations or conditions for applying the second derivative test? | |
Yes, there are some limitations and conditions for applying the second derivative test. The function must be twice differentiable, meaning that its first and second derivatives exist and are continuous. Additionally, the critical points must be isolated, meaning that there are no other critical points nearby. If these conditions are not met, the second derivative test may not provide accurate results, and alternative methods should be used for finding maxima and minima.
Text Transcript from Video
in this video we're gonna use the second
derivative test to determine if there's
any relative extrema in the function if
there's a relative maximum or a relative
minimum so let's talk about a relative
maximum or a local maximum in order to
get a local max the critical number has
to be found so the first derivative has
to equal zero at some point C so C is
the point where the local max will be
located along the x-axis so f prime of C
has to be equal to zero and the second
derivative at C has to be negative or
less than zero whenever the second
derivative is negative at a critical
number it's going to be concave down so
you could have this shape and you could
see clearly from the graph that there is
a maximum at this point and that's gonna
be the C value that we need that's where
we have the horizontal tangent which is
typical of a local maximum or relative
maximum now to use the second derivative
test to identify the local minimum which
will occur at a critical number F prime
of C must be 0 so that's how we could
find the critical number and the second
derivative at that critical number has
to be greater than 0 or has to be
positive which means that it has to be
concave up so we're gonna have that
shape and so we can see that we have a
minimum at the bottom
now let's work on an example problem so
let's say that f of X is Q X cube minus
12 x squared so go ahead and use the
second derivative test to determine any
relative extrema in this particular
function so feel free to try it now
let's determine the first derivative so
we can identify any critical numbers the
derivative of x cube is 3 x squared
times 2 well let's do it one step at a
time the derivative of x squared is 2x
and so 2 times 3 is 6 12 times 2 is 24
now to find the critical numbers we need
to set the first derivative equal to
zero next let's factor out the GCF which
in this example is going to be 6x so 6x
squared divided by 6x is X negative 24x
divided by 6x is negative 4 and then set
each factor equal to zero so if we set
6x equal to zero and then divide by 6 we
can see that the first critical number
is at x equals zero and if we set x
minus 4 equal to zero and then add 4 to
both sides the second critical number is
at x equal 4 so I'm going to write that
at the top over here so I can erase the
stuff on the bottom
so we have two critical numbers now
let's find the second derivative to
determine the concavity so we said the
first derivative was 6x squared minus
24x now for the second derivative it's
going to be 6 times the derivative of x
squared which is 2x and 24 times the
derivative of X which is 1 and so that's
going to be 12x minus 24
so now let's set the second derivative
equal to zero and let's take out the GCF
which is 12 so we have a potential
inflection point at x equals 2 so let's
create a sign chart now if we plug in a
test point greater than 2 let's say 3 &
2 the second derivative 3 minus 2 is
positive if we plug in a number less
than 2 let's say 1 & 1 minus 2 is
negative so it's concave down between
negative infinity and 2 and concave up
between 2 and infinity so therefore the
second derivative at 0 we could say is
less than 0 it's negative because 0 is
between negative infinity and 2 now
granted you could just take this number
and plug it into this expression
actually not that one but you could plug
it into 12x minus 24 you really don't
need to create a sign chart so just keep
that in mind so for the second example
instead of using a sign chart I can
evaluate f double prime of 4 so that's
going to be 12 times 4 minus 24 12 times
4 is 48 48 minus 24 is 24 so that's
positive and as you can see 4 is between
2 and infinity so that's going to be
positive as well
so f double prime of four is going to be
greater than zero
when the second derivative is less than
zero it's going to have a negative sign
and when it's greater than zero that
means it's positive when the second
derivative is positive we have a graph
that is concave up and when the second
derivative is negative its concave down
so you can see that a concave up graph
is associated with a minimum and a
concave down graph is associated with a
maximum so at x equals 0 we have a
maximum and x equals 4 we have a minimum
and that's how you can use the second
derivative tests to determine the
relative extrema in the function you
need to find the critical numbers and
then find out what the second derivative
is at those critical numbers but now
let's confirm our answer using the first
derivative test so let's put the
critical numbers on this number line now
we need to rewrite the first derivative
in its factored form so it was 6x
squared minus 24x and if we take out 6x
it's gonna be 6x times X minus 4 so
let's say if we plug in a test point
that is greater than 4 like 5 6 times 5
is positive 5 minus 4 is positive if you
multiply two positive numbers that will
give you a positive result now let's
plug in a number between 0 & 4 let's try
2 6 times 2 is positive 2 minus 4 is
negative a positive and a negative
equals a negative now if we plug a
negative 1 6 times negative 1 is
negative negative 1 minus 4 is negative
a negative times a negative is a
positive number so going from positive
to negative the function is increasing
and then it's decreasing so this
indicates that we have a maximum at 0
which indeed we do and going from
negative to positive is to decrease in
and then increase in so that's a minimum
at 4 which we do have and so that's how
you can use the first derivative test to
confirm the results of the second
derivative
now it's your turn let's say if we have
the function f of X is equal to 4x cubed
minus 6x squared minus 24x plus 1 use
the second derivative test to determine
the relative extrema of the function so
first let's find the first derivative to
identify all the critical numbers so
it's going to be 12x squared and the
derivative of 6x squared that's 12x and
for 24x it's just going to become 24 now
let's set it equal to zero and let's
take out a 12 so it's going to be x
squared minus X minus 2 and so we need
to factor this expression two numbers
that multiply to negative 2 but add to
negative 1 that's going to be negative 2
and positive 1 so if the factor it's
going to be X minus 2 times X plus 1 so
the critical numbers are positive 2 and
negative 1 if you set each factor equal
to 0 so now that we have those critical
numbers let's determine the sign of the
second derivative at those points so
first we need to find a second
derivative the derivative of 12x squared
is 24x and the derivative of 12x is just
12 but we have a negative sign in front
so the second derivative at 2 that's
gonna be 24 times 2 minus 12 24 times 2
is 48 minus 12 that's positive 36 now
let's do the same for the other critical
number negative 1 so it's 24 times
negative 1 which is negative 24 minus 12
that's a negative 36
so at 2:00 it's positive therefore it's
concave up and negative 1 it's negative
so if the second derivative is negative
then it's concave down and any time it's
concave up its associated with a minimum
when it's concave down its associated
with a maximum this is a max and this is
a minimum so we can write the answers
over here so at 2 we have a minimum and
that negative 1 we have a maximum so
let's confirm the results with the first
derivative test so let's make a sign
chart so first let me get rid of this so
the critical numbers are negative 1 & 2
so using the first derivative and its
factored form let's plug in some test
points so I'm going to try 3 3 minus 2
is positive 3 plus 1 is positive next
I'm going to try 0 0 minus 2 is negative
0 plus 1 is positive so that will give
us a negative result and then negative 3
minus 2 is negative negative 3 plus 1 is
negative so we're going to get a
positive result so 4 negative 1 it's
increasing and then it's decreasing so
this is a maximum which we do have a
negative 1 and a 2 its decrease and then
increase in so we have a minimum at you
and don't let this negative 2 confuse
you a 2 we do indeed have a minimum so
this answer is confirmed and so that's
it now you know how to use the second
derivative test to determine the
relative extrema of the function and you
can confirm it with the first derivative
test so make sure you find the critical
numbers and then see if the second
derivative is positive or negative at
those critical numbers if the second
derivative is positive that means that
it's concave up and so you have a
minimum if the second derivative is a
negative
that means it's concave down and you're
gonna have a maximum at that critical
number and so that's it for this video
hopefully you found it to be helpful
thanks for watching
derivative test to determine if there's
any relative extrema in the function if
there's a relative maximum or a relative
minimum so let's talk about a relative
maximum or a local maximum in order to
get a local max the critical number has
to be found so the first derivative has
to equal zero at some point C so C is
the point where the local max will be
located along the x-axis so f prime of C
has to be equal to zero and the second
derivative at C has to be negative or
less than zero whenever the second
derivative is negative at a critical
number it's going to be concave down so
you could have this shape and you could
see clearly from the graph that there is
a maximum at this point and that's gonna
be the C value that we need that's where
we have the horizontal tangent which is
typical of a local maximum or relative
maximum now to use the second derivative
test to identify the local minimum which
will occur at a critical number F prime
of C must be 0 so that's how we could
find the critical number and the second
derivative at that critical number has
to be greater than 0 or has to be
positive which means that it has to be
concave up so we're gonna have that
shape and so we can see that we have a
minimum at the bottom
now let's work on an example problem so
let's say that f of X is Q X cube minus
12 x squared so go ahead and use the
second derivative test to determine any
relative extrema in this particular
function so feel free to try it now
let's determine the first derivative so
we can identify any critical numbers the
derivative of x cube is 3 x squared
times 2 well let's do it one step at a
time the derivative of x squared is 2x
and so 2 times 3 is 6 12 times 2 is 24
now to find the critical numbers we need
to set the first derivative equal to
zero next let's factor out the GCF which
in this example is going to be 6x so 6x
squared divided by 6x is X negative 24x
divided by 6x is negative 4 and then set
each factor equal to zero so if we set
6x equal to zero and then divide by 6 we
can see that the first critical number
is at x equals zero and if we set x
minus 4 equal to zero and then add 4 to
both sides the second critical number is
at x equal 4 so I'm going to write that
at the top over here so I can erase the
stuff on the bottom
so we have two critical numbers now
let's find the second derivative to
determine the concavity so we said the
first derivative was 6x squared minus
24x now for the second derivative it's
going to be 6 times the derivative of x
squared which is 2x and 24 times the
derivative of X which is 1 and so that's
going to be 12x minus 24
so now let's set the second derivative
equal to zero and let's take out the GCF
which is 12 so we have a potential
inflection point at x equals 2 so let's
create a sign chart now if we plug in a
test point greater than 2 let's say 3 &
2 the second derivative 3 minus 2 is
positive if we plug in a number less
than 2 let's say 1 & 1 minus 2 is
negative so it's concave down between
negative infinity and 2 and concave up
between 2 and infinity so therefore the
second derivative at 0 we could say is
less than 0 it's negative because 0 is
between negative infinity and 2 now
granted you could just take this number
and plug it into this expression
actually not that one but you could plug
it into 12x minus 24 you really don't
need to create a sign chart so just keep
that in mind so for the second example
instead of using a sign chart I can
evaluate f double prime of 4 so that's
going to be 12 times 4 minus 24 12 times
4 is 48 48 minus 24 is 24 so that's
positive and as you can see 4 is between
2 and infinity so that's going to be
positive as well
so f double prime of four is going to be
greater than zero
when the second derivative is less than
zero it's going to have a negative sign
and when it's greater than zero that
means it's positive when the second
derivative is positive we have a graph
that is concave up and when the second
derivative is negative its concave down
so you can see that a concave up graph
is associated with a minimum and a
concave down graph is associated with a
maximum so at x equals 0 we have a
maximum and x equals 4 we have a minimum
and that's how you can use the second
derivative tests to determine the
relative extrema in the function you
need to find the critical numbers and
then find out what the second derivative
is at those critical numbers but now
let's confirm our answer using the first
derivative test so let's put the
critical numbers on this number line now
we need to rewrite the first derivative
in its factored form so it was 6x
squared minus 24x and if we take out 6x
it's gonna be 6x times X minus 4 so
let's say if we plug in a test point
that is greater than 4 like 5 6 times 5
is positive 5 minus 4 is positive if you
multiply two positive numbers that will
give you a positive result now let's
plug in a number between 0 & 4 let's try
2 6 times 2 is positive 2 minus 4 is
negative a positive and a negative
equals a negative now if we plug a
negative 1 6 times negative 1 is
negative negative 1 minus 4 is negative
a negative times a negative is a
positive number so going from positive
to negative the function is increasing
and then it's decreasing so this
indicates that we have a maximum at 0
which indeed we do and going from
negative to positive is to decrease in
and then increase in so that's a minimum
at 4 which we do have and so that's how
you can use the first derivative test to
confirm the results of the second
derivative
now it's your turn let's say if we have
the function f of X is equal to 4x cubed
minus 6x squared minus 24x plus 1 use
the second derivative test to determine
the relative extrema of the function so
first let's find the first derivative to
identify all the critical numbers so
it's going to be 12x squared and the
derivative of 6x squared that's 12x and
for 24x it's just going to become 24 now
let's set it equal to zero and let's
take out a 12 so it's going to be x
squared minus X minus 2 and so we need
to factor this expression two numbers
that multiply to negative 2 but add to
negative 1 that's going to be negative 2
and positive 1 so if the factor it's
going to be X minus 2 times X plus 1 so
the critical numbers are positive 2 and
negative 1 if you set each factor equal
to 0 so now that we have those critical
numbers let's determine the sign of the
second derivative at those points so
first we need to find a second
derivative the derivative of 12x squared
is 24x and the derivative of 12x is just
12 but we have a negative sign in front
so the second derivative at 2 that's
gonna be 24 times 2 minus 12 24 times 2
is 48 minus 12 that's positive 36 now
let's do the same for the other critical
number negative 1 so it's 24 times
negative 1 which is negative 24 minus 12
that's a negative 36
so at 2:00 it's positive therefore it's
concave up and negative 1 it's negative
so if the second derivative is negative
then it's concave down and any time it's
concave up its associated with a minimum
when it's concave down its associated
with a maximum this is a max and this is
a minimum so we can write the answers
over here so at 2 we have a minimum and
that negative 1 we have a maximum so
let's confirm the results with the first
derivative test so let's make a sign
chart so first let me get rid of this so
the critical numbers are negative 1 & 2
so using the first derivative and its
factored form let's plug in some test
points so I'm going to try 3 3 minus 2
is positive 3 plus 1 is positive next
I'm going to try 0 0 minus 2 is negative
0 plus 1 is positive so that will give
us a negative result and then negative 3
minus 2 is negative negative 3 plus 1 is
negative so we're going to get a
positive result so 4 negative 1 it's
increasing and then it's decreasing so
this is a maximum which we do have a
negative 1 and a 2 its decrease and then
increase in so we have a minimum at you
and don't let this negative 2 confuse
you a 2 we do indeed have a minimum so
this answer is confirmed and so that's
it now you know how to use the second
derivative test to determine the
relative extrema of the function and you
can confirm it with the first derivative
test so make sure you find the critical
numbers and then see if the second
derivative is positive or negative at
those critical numbers if the second
derivative is positive that means that
it's concave up and so you have a
minimum if the second derivative is a
negative
that means it's concave down and you're
gonna have a maximum at that critical
number and so that's it for this video
hopefully you found it to be helpful
thanks for watching
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Oct 23, 2025
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