Differentiation (Maxima and Minima)
Summary
TLDRThis educational video delves into the concept of differentiation, focusing on finding and identifying stationary points, which are points where the gradient of a curve is zero. It guides viewers through the process of differentiating a given function to find these points and then uses the second derivative to determine whether they are maxima or minima. The video provides step-by-step examples, including solving quadratic equations and interpreting the results to understand the nature of the stationary points without graphing.
Takeaways
- 📚 The video discusses the concept of differentiation, specifically focusing on finding maxima and minima of functions.
- 📈 It explains that stationary points occur where the gradient of a function is zero, indicated by a horizontal tangent line.
- 🔍 The process of finding stationary points involves setting the first derivative of a function equal to zero and solving for the variable.
- 📝 The video provides a step-by-step example of differentiating a given function and solving for the x-values where the derivative equals zero.
- 📉 After finding the x-values, the corresponding y-values are calculated by substituting these x-values back into the original function to get the coordinates of the stationary points.
- 📊 The nature of stationary points as maxima or minima is determined by the second derivative test, which involves differentiating the first derivative again.
- 🔎 A positive second derivative at a stationary point indicates a minimum, while a negative second derivative indicates a maximum.
- 📚 The video includes an example of applying the second derivative test to determine the nature of a stationary point without graphing the function.
- 📝 The script walks through the differentiation of a second function, solving for the x-coordinate of a stationary point, and then finding the y-coordinate.
- 🔑 The video emphasizes the importance of understanding both the x and y coordinates of stationary points, not just the x values.
- 👍 It concludes with an encouragement to practice with the provided exam questions and to subscribe for future educational content.
Q & A
What are the characteristics of a function's increasing and decreasing sections?
-The increasing sections of a function have a positive gradient, meaning dy/dx is greater than zero. The decreasing sections have a negative gradient, with dy/dx less than zero.
What is the significance of a horizontal tangent on a curve?
-A horizontal tangent at a point on a curve indicates that the gradient at that point is zero, which means dy/dx equals zero at that specific point.
What are stationary points on a curve?
-Stationary points are points on a curve where the derivative (gradient) is zero, indicating no change in the slope of the curve at that point.
How can you find the coordinates of stationary points on a curve?
-To find the coordinates of stationary points, you set the derivative of the curve (dy/dx) equal to zero and solve for the variable x. Then, substitute the x values back into the original equation to find the corresponding y values.
What is the process to determine if a stationary point is a maximum or minimum without graphing?
-You can determine the nature of a stationary point by finding the second derivative (d^2y/dx^2). If the second derivative is less than zero, the point is a maximum; if it is greater than zero, the point is a minimum.
How does the gradient of a function change as you move past a maximum point?
-As you move past a maximum point, the gradient starts positive, becomes zero at the maximum, and then turns negative as the curve descends.
What does a positive second derivative (d^2y/dx^2) indicate about a stationary point?
-A positive second derivative at a stationary point indicates that the point is a minimum, as the gradient is increasing from negative to positive.
How does the gradient change as you move past a minimum point on a curve?
-As you move past a minimum point, the gradient starts negative, becomes zero at the minimum, and then turns positive as the curve ascends.
What is the purpose of finding the second derivative in the context of stationary points?
-The second derivative helps in identifying whether a stationary point is a maximum or minimum by analyzing the concavity of the curve at that point.
Can you provide an example of how to find the second derivative of a function?
-To find the second derivative, differentiate the first derivative function again. For example, if the first derivative is 8x - 1/x^2, the second derivative would be found by differentiating each term, resulting in 8 for the derivative of 8x and 2/x^3 for the derivative of -1/x^2.
Outlines
📚 Introduction to Differentiation Maxima and Minima
This paragraph introduces the topic of differentiation maxima and minima, explaining the concept of stationary points where the gradient of a function is zero, indicated by a horizontal tangent to the curve. The speaker provides a brief overview of increasing and decreasing sections of a function and their relation to the sign of the derivative. The audience is encouraged to practice with exam questions linked in the video description. The focus is on finding the coordinates of stationary points by setting the first derivative equal to zero and solving the resulting equation.
🔍 Identifying and Analyzing Stationary Points
The second paragraph delves into the process of identifying stationary points on a curve by differentiating the given function and solving for when the derivative equals zero. The example provided involves a cubic function, and the process includes finding the first derivative, setting it to zero, and solving the quadratic equation to find the x-coordinates of potential stationary points. The y-coordinates are then determined by substituting these x-values back into the original function. The paragraph also touches on the concept of local maxima and minima and introduces the method of using the second derivative to determine the nature of these stationary points without graphing.
Mindmap
Keywords
💡Differentiation
💡Maxima and Minima
💡Stationary Points
💡Gradient
💡Tangent
💡Quadratic Equation
💡Factorize
💡First Derivative
💡Second Derivative
💡Local Maximum and Minimum
Highlights
Differentiation maxima and minima are discussed, with exam questions provided in the video description.
Stationary points are identified where the gradient of a function is zero, indicated by a horizontal tangent.
The process of finding the coordinates of stationary points by setting the first derivative equal to zero is explained.
A quadratic equation is factored to find x-values for stationary points.
Substituting x-values back into the original equation to find corresponding y-values is demonstrated.
The coordinates of stationary points are calculated for the given function.
Differentiation is used to determine whether a stationary point is a local maximum or minimum without graphing.
The second derivative test is introduced to identify the nature of stationary points.
A negative second derivative indicates a local maximum, while a positive one indicates a local minimum.
The process of differentiating the gradient function to find the second derivative is shown.
A specific curve's equation is analyzed to find the coordinates and nature of its stationary point.
The second derivative is calculated by differentiating the first derivative of a given function.
Substitution of the x-coordinate of the stationary point into the second derivative is demonstrated.
A positive second derivative confirms the stationary point as a local minimum.
The video concludes with an invitation to check out exam questions and subscribe for future content.
Transcripts
[Music]
in this video we're going to look at
differentiation maxima and minima
as usual you can find some exam
questions in this video's description to
try afterwards
now in the previous video we looked at
this function here
we saw how the black sections were
increasing
therefore they had a positive gradient
and dy by dx was greater than zero
and also the red sections were
decreasing they had a negative gradient
and dy by dx was less than zero
but what about the points where it
switches from positive to negative
or from negative to positive
so right at the top of the curve here or
right at the bottom of the curve here
well if you draw a tangent to these the
tangent will be completely horizontal
therefore the gradient is zero
so dy by dx must equal zero
these points are given a special name we
call them stationary points
so let's take a curve
and in this question we're asked to find
the coordinates of the stationary points
so when we have stationary points we
know that dy by dx equals zero
so let's find dy by dx well for this one
dy by dx
we've got x cubed so if you
differentiate that you get three x
squared
differentiating negative six x squared
gives you negative 12x
differentiating plus 9x gives you plus 9
and the plus 1 constant differentiates
to give 0.
so we know this must equal 0. so we
write 3x squared minus 12x plus 9 equals
0 and then we solve this equation
this one's a quadratic there's a common
factor of 3 here so you can divide 3 by
3 on both sides if you divide on the
left you get x squared minus 4 x plus 3
and on the right 0 divided by 3 is just
0.
you can now factorize this one
it would be x take one x take three
equals zero and this gives you two
solutions for x x equals one and x
equals three
now we have to be careful it didn't ask
us for the x coordinate it asks us for
the coordinates so we need to get the y
value as well
to do this we'll substitute our x values
back into the equation of the curve so y
equals x cubed minus six x squared plus
nine x plus one
so we'll start with x equals one
when x equals one y equals x cubed so 1
cubed
minus 6 x squared so minus 6 lots of 1
squared plus 9 x so plus 9 lots of 1 and
then plus 1.
1 cubed is just 1
1 squared is 1 so negative six times one
is negative six
nine times one is just nine
and if you simplify all of this
you'll get five
now we'll substitute in the other point
when x was equal to three
so for this one we do three cubed take
away six lots of three squared
plus nine lots of three plus one
three cubed is twenty seven
three squared is nine and if you times
this by negative six you get negative
fifty four
nine times three is twenty seven
and if you simplify all of this you'll
find a y value of 1.
so we found that when x equals 1
y equals 5 so we have the coordinate of
station 3.1 which is 1 5
and then when x equals 3 y equals one so
we have the coordinate of the second
stationary point which is three one
this is a sketch of the curve y equals x
cubed minus six x squared plus nine x
plus one and you can see we have our two
stationary points this one here which
was coordinates one five and this one
here with coordinates three one
we would say that the point with
coordinates one five is a local maximum
point because it reaches a maximum and
then comes back down again
we would say that the point three one is
a local minimum point because it reaches
a minimum and then goes back up again
what we need to be able to do is
identify if these points are maxima or
minima but without drawing out the graph
now you can do this just using
differentiation
when you find d y by dx which we've
called the gradient function you're
actually finding what we call the first
derivative
now you can find d2y by dx squared which
is the second derivative to find d2y by
dx squared you differentiate the
gradient function so you differentiate a
second time
you could think of the gradient function
as the rate of change of y with respect
to x but you could think of d2y by dx
squared as the rate of change of the
gradient with respect to x
so how can this be used to tell if it's
a maximum or a minimum point well let's
take the point right in the bottom left
hand corner
and we're going to move right along the
graph and consider what's happening to
the gradient
so right now the gradient is a positive
value because it's sloping up and it's
also very steep so a very high number
as we begin to move along the curve the
gradient starts to decrease it's still
positive here but it's getting smaller
and smaller and smaller and smaller
until it reaches the top and now the
gradient is zero
as we move past this maximum point the
gradient turns negative and then it
becomes more and more negative as the
line becomes steeper and steeper
so as we go past the maximum point you
can see the gradient is actually
decreasing all of the time it starts
positive it becomes zero and then it
turns negative
this means that if we're at a maximum
point the value of d2y by dx squared
will be less than zero
so if d2y by dx squared is less than
zero we have a maximum point
now if we continue on on this journey at
the moment the gradient's negative and
it's becoming less and less steep so
it's actually getting closer and closer
to zero and then it reaches zero at the
bottom and as we go past this minimum
point the gradient switches to positive
and then becomes larger and larger and
larger as the graph gets steeper
so as we went past the minimum point the
gradient started negative
went to zero and then became positive so
the rate of change of gradient with
respect to x is actually positive
this means if d2y by dx squared is
greater than zero we have a minimum
point
let's see how we can apply this to a
question
so we have this curve here
and first of all we want to find the
coordinates of the stationary point and
then we want to determine its nature
when it uses the word nature in the
question it refers to maximum or minimum
so let's find the stationary point first
we know when we have a stationary point
d y by dx must equal zero
so let's differentiate
if we differentiate four x squared we
get eight x
and if we differentiate x to the power
negative one we get negative x to the
power negative two
which we may rewrite in a fraction form
now we know this must equal zero
to solve this one i'm going to times
both sides by x squared if you times x
squared on the left you get 8x cubed the
x squared will cancel with the x squared
on the bottom of the fraction so just
take away 1 and x squared times 0 is 0.
now if you add 1 to both sides and
divide by 8 you'll get x cubed equals 1
8.
and if you cube root 1 8 you'll find x
equals a half
now we have the x coordinate but we also
need the y coordinate
so we'll substitute that back into the
original equation of the curve y equals
4 x squared plus x to the power negative
1.
so at x equals one half
y would equal four lots of one half
squared
plus one over one half
one half squared is a quarter and times
that by four gives you one
and then one divided by a half is two so
we have one plus two which equals three
so the coordinates of the stationary
point are one half three
now we need to do part b we need to
determine its nature remember the nature
is just if it's a maximum or a minimum
to do this we're going to find the
second derivative d2y by dx squared
to find this you just differentiate the
first derivative
so on the left we have d y by dx equals
eight x minus one over x squared we just
need to differentiate this
to make this easier we'll rewrite one
over x squared as x to the power
negative two
so to differentiate eight x you get
eight and if you differentiate negative
x to the power negative two you do
negative two times negative one which is
positive two
and then reduce the power of x from
negative two to negative three
which again we'll write back in its
fraction form
now we need to substitute in the x
coordinate of our stationary point our
stationary point had coordinates one
half three so we'll substitute in x
equals one half
so at x equals one half
d two y by d x squared would equal eight
plus two over one half cubed
one-half cubed equals one-eighth for
0.125
and 2 divided by 0.125 gives you 16.
so we have 8 plus 16
which makes 24.
now this is positive so d2y by dx
squared is positive
when it's positive at a stationary point
we have a minimum point
if it were negative it would have been a
maximum point
thank you for watching this video i hope
you found it useful check out the exam
questions in the video's description
what i think you should watch next and
also subscribe so you don't miss out on
future videos
you
5.0 / 5 (0 votes)