Class 12 Exam > Class 12 Videos > Examples Second Derivative Test (Part - 23) - Application of Derivatives, Maths, Class 12
Examples Second Derivative Test (Part - 23) - Application of Derivatives, Maths, Class 12 Video Lecture
FAQs on Examples Second Derivative Test (Part - 23) - Application of Derivatives, Maths, Class 12 Video Lecture
| 1. What is the second derivative test in calculus? |  |
Ans. The second derivative test is a method in calculus used to determine the nature of critical points of a function. It involves finding the second derivative of the function and analyzing its sign to determine whether a critical point is a local maximum, local minimum, or a point of inflection.
| 2. How is the second derivative test applied in real-world scenarios? | |
Ans. The second derivative test can be applied in various real-world scenarios involving optimization problems. For example, it can be used to find the maximum or minimum values of a cost function in economics, the maximum or minimum speed of an object in physics, or the maximum or minimum profit of a business.
| 3. Can the second derivative test be used to determine the global maximum or minimum of a function? | |
Ans. No, the second derivative test can only determine the nature of critical points. It cannot determine whether a critical point is a global maximum or minimum. To find the global maximum or minimum of a function, additional analysis such as checking the behavior of the function at the endpoints of the domain is required.
| 4. Are there any limitations to the second derivative test? | |
Ans. Yes, the second derivative test has certain limitations. It can only be applied to continuous functions that have a defined second derivative. Additionally, the test may fail to provide conclusive results when the second derivative is zero at a critical point or when the function is not differentiable at a critical point.
| 5. What other methods can be used alongside the second derivative test for analyzing critical points? | |
Ans. Alongside the second derivative test, other methods such as the first derivative test and the use of intervals can be employed to analyze critical points. The first derivative test involves analyzing the sign of the first derivative to determine if a critical point is a local maximum or minimum. Intervals can be used to determine the behavior of the function between critical points.
Text Transcript from Video
hello friends this video an application
of derivatives part 23 has brought you
by exam Viacom no more fear from exam
before watching this video please make
sure that you have washed part 1 2 part
when you chew say for example we have to
prove that dysfunctions does not have
Maxima or C for a function to have
maximum anymore
right they are freaking select or honest
it can have Maxima or minima at the
endpoints it can have Maxima or minima
at F dash X is equal to 0 turning points
or non differentiable points correct so
let us say case 1 FX is going to eat Y
over X there is no turning point given
there is no end point given because it
is open interval so in so this part you
can know in fact in all the squeeze the
is the open interval so I can ignore the
8 points altogether
the second is f dash X equal to 0 and
third is non differentiable also since
it is differentiable so this part of
degree you know the only part is place
where F dash X is equal to 0 is the
place where this guy will be
differentiable so let's find em des
exposition F does this is only what if
you won X itself and this will never be
0 you take any value of x it will never
visual you take 0 this becomes 1 you
take 1 it becomes me it will never be 0
so I can see that since it is never be 0
it will never have maximal let's take
this guy GX is equal to log X here also
endpoint not given it is differentiable
everywhere so we'll have the only case
where G dash X is equal to 0 G dash this
is nothing but 1 by X this guy will also
never be 0 you put X value 0 this
becomes infinite
if you pirate X going inside it becomes
0 so at infinity it is 0 but that is not
consider
so this guy's also never zero since it
is never zero it doesn't have maximum
anymore
let's take third scenario HX is equal to
XQ plus X square plus X plus 1 now if
you see here also endpoints we are not
talking about endpoint is open interval
correct it is continuous function so it
is differentiable the only place is f
dash X equal to 0 is my place where it
can have a maximum on so let's find H
dash X here that is nothing but 3 X
square plus 2x plus 1 and if this is
greater 0 let's find the value of x X is
nothing but minus B patient in the form
ax square plus BX plus C X is nothing
but minus B plus minus B Square minus 2e
v square by 2 B Square minus 4 AC by
doing yeah so minus B is what minus 2
plus minus root v square is 4 minus 4
into AC a is 3 C is 1 correct by 2 into
a a is 3 this is nothing but minus 2
plus minus this becomes 4 minus 12 this
is - a true there's I 2 biases so you
see this is the imaginary number so on
an imaginary number at this extra little
on a real number F desk or SDS X is
nonzero plus I can see that as this X is
not 0 for X here so I can see that this
also function doesn't have a mind small
so in if you see in all this question
the only thing I prove that the
differentiable of these functions are
not equal to 0 thus it doesn't have
local Maxima thank you
visit exam fair com2
watch free educational videos try free
online deaths get the best quality study
materials study from the best tutors and
mentors and much more thanks again
of derivatives part 23 has brought you
by exam Viacom no more fear from exam
before watching this video please make
sure that you have washed part 1 2 part
when you chew say for example we have to
prove that dysfunctions does not have
Maxima or C for a function to have
maximum anymore
right they are freaking select or honest
it can have Maxima or minima at the
endpoints it can have Maxima or minima
at F dash X is equal to 0 turning points
or non differentiable points correct so
let us say case 1 FX is going to eat Y
over X there is no turning point given
there is no end point given because it
is open interval so in so this part you
can know in fact in all the squeeze the
is the open interval so I can ignore the
8 points altogether
the second is f dash X equal to 0 and
third is non differentiable also since
it is differentiable so this part of
degree you know the only part is place
where F dash X is equal to 0 is the
place where this guy will be
differentiable so let's find em des
exposition F does this is only what if
you won X itself and this will never be
0 you take any value of x it will never
visual you take 0 this becomes 1 you
take 1 it becomes me it will never be 0
so I can see that since it is never be 0
it will never have maximal let's take
this guy GX is equal to log X here also
endpoint not given it is differentiable
everywhere so we'll have the only case
where G dash X is equal to 0 G dash this
is nothing but 1 by X this guy will also
never be 0 you put X value 0 this
becomes infinite
if you pirate X going inside it becomes
0 so at infinity it is 0 but that is not
consider
so this guy's also never zero since it
is never zero it doesn't have maximum
anymore
let's take third scenario HX is equal to
XQ plus X square plus X plus 1 now if
you see here also endpoints we are not
talking about endpoint is open interval
correct it is continuous function so it
is differentiable the only place is f
dash X equal to 0 is my place where it
can have a maximum on so let's find H
dash X here that is nothing but 3 X
square plus 2x plus 1 and if this is
greater 0 let's find the value of x X is
nothing but minus B patient in the form
ax square plus BX plus C X is nothing
but minus B plus minus B Square minus 2e
v square by 2 B Square minus 4 AC by
doing yeah so minus B is what minus 2
plus minus root v square is 4 minus 4
into AC a is 3 C is 1 correct by 2 into
a a is 3 this is nothing but minus 2
plus minus this becomes 4 minus 12 this
is - a true there's I 2 biases so you
see this is the imaginary number so on
an imaginary number at this extra little
on a real number F desk or SDS X is
nonzero plus I can see that as this X is
not 0 for X here so I can see that this
also function doesn't have a mind small
so in if you see in all this question
the only thing I prove that the
differentiable of these functions are
not equal to 0 thus it doesn't have
local Maxima thank you
visit exam fair com2
watch free educational videos try free
online deaths get the best quality study
materials study from the best tutors and
mentors and much more thanks again
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