Differentiation (Maxima and Minima)

Summary

This 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

00:00

📚 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.

05:01

🔍 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.

Keywords

💡Differentiation

Differentiation in the context of the video refers to the mathematical process of finding the derivative of a function, which represents the rate of change of the function with respect to one of its variables. It is central to the theme of the video, as it is used to analyze the behavior of functions and identify points of maxima and minima. For example, the script discusses how the derivative (dy/dx) indicates whether a function is increasing or decreasing.

💡Maxima and Minima

Maxima and minima are the highest and lowest points, respectively, that a function can reach. In the video, these terms are used to describe the points on a curve where the function switches from increasing to decreasing (maxima) or decreasing to increasing (minima). The script explains how to find these points by setting the first derivative equal to zero and then determining their nature using the second derivative.

💡Stationary Points

Stationary points are points on a curve where the derivative (gradient) is zero, indicating no change in the rate of increase or decrease of the function. The video script uses this term to describe the specific points where the curve changes direction, such as the top or bottom of a curve, and provides a method to find these points by solving the equation dy/dx = 0.

💡Gradient

The gradient, in the context of the video, is synonymous with the derivative of a function and represents the slope of the tangent line to the curve at a given point. The script explains how a horizontal tangent (zero gradient) at the top or bottom of a curve indicates a stationary point.

💡Tangent

A tangent to a curve at a given point is a straight line that touches the curve at that point without crossing it. The video uses the concept of a tangent to illustrate how the slope of the tangent can be used to determine the gradient at specific points, such as at the top or bottom of a curve where the tangent is horizontal.

💡Quadratic Equation

A quadratic equation is a polynomial equation of degree two, typically in the form ax^2 + bx + c = 0. The video script mentions solving a quadratic equation to find the x-coordinates of stationary points, illustrating the process of factoring and solving for x.

💡Factorize

To factorize a quadratic expression means to express it as a product of its factors. In the script, the process of factorizing is used to simplify the equation 3x^2 - 12x + 9 = 0, leading to the solutions x = 1 and x = 3, which are the x-coordinates of the stationary points.

💡First Derivative

The first derivative of a function, denoted as dy/dx, is the rate of change of the function with respect to its variable. The video script discusses finding the first derivative to identify the gradient of the function and to locate stationary points where the gradient is zero.

💡Second Derivative

The second derivative, denoted as d^2y/dx^2, is the derivative of the first derivative. It represents the rate of change of the gradient with respect to the variable. The video explains how the second derivative can be used to determine whether a stationary point is a maximum or minimum by checking if it is positive or negative, respectively.

💡Local Maximum and Minimum

A local maximum or minimum is a point on a curve that is higher or lower, respectively, than all other points in its immediate vicinity. The video script describes how to identify these points without graphing by using the first and second derivatives, with a local maximum having a second derivative less than zero and a local minimum having a second derivative greater than zero.

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