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8/12/2019 9) C2 Differentiation
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8/12/2019 9) C2 Differentiation
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Introduction
We have seen Differentiation in C1
In C2 we will be looking at solving moretypes of problem
We are also going to be applying theprocess to worded practical problems
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8/12/2019 9) C2 Differentiation
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Differentiation
You need to know thedifference between Increasing
and Decreasing Functions
An increasing function is one with a
positive gradient.
A decreasing function is one with anegative gradient.
9A
x
x
y
y
This functionis increasingfor all values
of x
This functionis decreasingfor all values
of x
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Differentiation
You need to know thedifference between Increasing
and Decreasing Functions
An increasing function is one with a
positive gradient.
A decreasing function is one with anegative gradient.
Some functions are increasing in one
interval and decreasing in another.
9A
x
y
This function isdecreasing for x > 0,
and increasing for x < 0
At x = 0, the gradient is0. This is known as a
stationary point.
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Differentiation
You need to know thedifference between Increasing
and Decreasing Functions
An increasing function is one with a
positive gradient.
A decreasing function is one with anegative gradient.
Some functions are increasing in one
interval and decreasing in another.
You need to be able to work outranges of values where a function is
increasing or decreasing..
9A
Example Question
Show that the function ;3( ) 24 3f x x x
is an increasing function.
3( ) 24 3f x x x
2'( ) 3f x x 24
Differentiate toget the gradient
function
Since x2has to be positive, 3x2+ 24will be as well
So the gradient will always bepositive, hence an increasing function
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Differentiation
You need to know thedifference between Increasing
and Decreasing Functions
An increasing function is one with a
positive gradient.
A decreasing function is one with anegative gradient.
Some functions are increasing in one
interval and decreasing in another.
You need to be able to work outranges of values where a function is
increasing or decreasing..
9A
Example Question
Find the range of values where:3 2( ) 3 9f x x x x
is an decreasing function.
3 2( ) 3 9f x x x x
2'( ) 3f x x 6x 923 6 9 0x x
23( 2 3) 0x x
3( 3)( 1) 0x x
1x 3x OR
3 1x
Differentiate for thegradient function
We want the gradientto be below 0
Factorise
Factorise again
Normally x = -3 and1
BUT, we want valuesthat will make the
function negative
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Differentiation
You need to know thedifference between Increasing
and Decreasing Functions
9A
Example Question
Find the range of values where:3 2( ) 3 9f x x x x
is an decreasing function.
3 2( ) 3 9f x x x x
2'( ) 3f x x 6x 923 6 9 0x x
23( 2 3) 0x x
3( 3)( 1) 0x x
1x 3x OR
3 1x
Differentiate for thegradient function
We want the gradientto be below 0
Factorise
Factorise again
Normally x = -3 and1
BUT, we want valuesthat will make the
function negative
x
y
-3 1
Decreasing Function range
f(x)
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Differentiation
You need to be able to calculatethe co-ordinates of Stationary
points, and determine their nature
A point where f(x) stops increasing and
starts decreasing is called a maximumpoint
A point where f(x) stops decreasing andstarts increasing is called a minimum point
A point of inflexion is where the gradientis locally a maximum or minimum (the
gradient does not have to change frompositive to negative, for example)
These are all known as turning points, and
occur where f(x) = 0 (for now at least!)9B
y
x
Localmaximum
Local
minimum
Point ofinflexion
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Differentiation
You need to be able to calculatethe co-ordinates of Stationary
points, and determine their nature
To find the coordinates of these points,
you need to:
1) Differentiate f(x) to get the GradientFunction
2) Solve f(x) by setting it equal to 0 (as
this represents the gradient being 0)
3) Substitute the value(s) of x into theoriginal equation to find thecorresponding y-coordinate
9B
y
x
Localmaximum
Local
minimum
Point ofinflexion
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Differentiation
You need to be able to calculatethe co-ordinates of Stationary
points, and determine their nature
To find the coordinates of these points,
you need to:
1) Differentiate f(x) to get the GradientFunction
2) Solve f(x) by setting it equal to 0 (as
this represents the gradient being 0)
3) Substitute the value(s) of x into theoriginal equation to find thecorresponding y-coordinate
9B
Example Question
Find the coordinates of the turning point on thecurve y = x432x, and state whether it is a
minimum or maximum.
4
32y x x 34 32
dyx
dx
34 32 0x
34 32x 2x
4 32y x x 4(2) 32(2)y
48y
Differentiate
Set equal to 0
Add 32
Divide by 4, then cube root
Sub 2 into the originalequation
Work out the y-coordinate
The stationary point is at(2, -48)
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Differentiation
You need to be able to calculatethe co-ordinates of Stationary
points, and determine their nature
To find the coordinates of these points, you
need to:
1) Differentiate f(x) to get the GradientFunction
2) Solve f(x) by setting it equal to 0 (as thisrepresents the gradient being 0)
3) Substitute the value(s) of x into the originalequation to find the corresponding y-coordinate
4) To determine whether the point is a minimumor a maximum, you need to work out f(x)
(differentiate again!)
9B
Example Question
Find the coordinates of the turning point on thecurve y = x432x, and state whether it is a
minimum or maximum.
4
32y x x 34 32
dyx
dx
The stationarypoint is at (2, -48)
22
2 12
d yx
dx
212x
212(2)
48
Differentiate again
Sub in the xcoordinate
Positive = Minimum
Negative = Maximum
So the stationary
point is a MINIMUMin this case!
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Differentiation
You need to be able to calculatethe co-ordinates of Stationary
points, and determine their nature
To find the coordinates of these points, you
need to:
1) Differentiate f(x) to get the GradientFunction
2) Solve f(x) by setting it equal to 0 (as thisrepresents the gradient being 0)
3) Substitute the value(s) of x into the originalequation to find the corresponding y-coordinate
4) To determine whether the point is a minimumor a maximum, you need to work out f(x)
(differentiate again!)
9B
Example Question
Find the stationary points on the curve:y = 2x315x2+ 24x + 6, and state whether they
are minima, maxima or points of inflexion
3 2
2 15 24 6y x x x 2'( ) 6f x x 30x 24
26 30 24 0x x
26( 5 4) 0x x
6( 4)( 1) 0x x
4x 1x OR
Substituting into the original formula will givethe following coordinates as stationary points:
(1, 17) and (4, -10)
Differentiate
Set equal to 0
Factorise
Factorise again
Write thesolutions
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Differentiation
You need to be able to calculatethe co-ordinates of Stationary
points, and determine their nature
To find the coordinates of these points, you
need to:
1) Differentiate f(x) to get the GradientFunction
2) Solve f(x) by setting it equal to 0 (as thisrepresents the gradient being 0)
3) Substitute the value(s) of x into the originalequation to find the corresponding y-coordinate
4) To determine whether the point is a minimumor a maximum, you need to work out f(x)
(differentiate again!)
9B
Example Question
Find the stationary points on the curve:y = 2x315x2+ 24x + 6, and state whether they
are minima, maxima or points of inflexion
3 2
2 15 24 6y x x x 2'( ) 6f x x 30x 24
Stationary points at:(1, 17) and (4, -10)
Differentiateagain
''( ) 12 30f x x
''( ) 12 30f x x
''(1) 12(1) 30f
''( ) 12 30f x x
''(4) 12(4) 30f
''(1) 18f ''(4) 18f
Sub in x = 1 Sub in x = 4
So (1,17) is
a MaximumSo (4,-10) is
a Minimum
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Differentiation
You need to be able to calculatethe co-ordinates of Stationary
points, and determine their nature
To find the coordinates of these points, youneed to:
1) Differentiate f(x) to get the GradientFunction
2) Solve f(x) by setting it equal to 0 (as thisrepresents the gradient being 0)
3) Substitute the value(s) of x into the originalequation to find the corresponding y-coordinate
4) To determine whether the point is a minimumor a maximum, you need to work out f(x)
(differentiate again!)
9B
Example Question
Find the maximum possible value for y in theformula y = 6x x2. State the range of the
function.
2
6y x x
6 2dy
xdx
6 2 0x
3x
26y x x 26(3) (3)y
9y
9y
Differentiate
Set equal to 0
Solve
Sub x into theoriginal equation
Solve
9 is the maximum, so the range
is less than but including 9
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Differentiation
You need to be able to recognisepractical problems that can be solved by
using the idea of maxima and minima
Whenever you see a question asking aboutthe maximum value or minimum value of aquantity, you will most likely need to use
differentiation at some point.
Most questions will involve creating aformula, for example for Volume or Area,and then calculating the maximum value of
it.
A practical application would be If I havea certain amount of material to make abox, how can I make the one with the
largest volume? (maximum)
9C
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Differentiation
You need to be able to recognisepractical problems that can be solved by
using the idea of maxima and minima
9C
Example Question
A large tank (shown) is to be made from 54m2ofsheet metal. It has no top.
Show that the Volume of the tank will be givenby:
32183
V x x xx
y
2
V x y Formula for the Volume
22 3SA x xy
1) Try to make formulae using theinformation you have
Formula for theSurface Area (no
top)
254 2 3x xy
2) Find a way to remove a constant, in thiscase y. We can rewrite the Surface Areaformula in terms of y.
254 2 3x xy 254 2 3x xy
254 2
3
xy
x
3) Substitute the SA formula into theVolume formula, to replace y.
22 54 2
3
xV x
x
2V x y
2 454 2
3
x x
V x
2 4
54 23 3
x xVx x
32183
V x x
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Differentiation
You need to be able to recognisepractical problems that can be solved by
using the idea of maxima and minima
9C
Example Question
A large tank (shown) is to be made from 54m2ofsheet metal. It has no top.
Show that the Volume of the tank will be givenby:
32
18 3V x x xx
y
b) Calculate the values of x that will givethe largest volume possible, and what this
Volume is.
32183
V x x
218 2dV
xdx
218 2 0x
218 2x
3x
254 2 3x xy
32183
V x x
32
18(3) (3)3V 336V m
Differentiate
Set equal to 0
Rearrange
Solve
Sub the x value in
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Differentiation
You need to be able to recognisepractical problems that can be solved by
using the idea of maxima and minima
9C
Example Question
A wire of length 2m is bent into the shapeshown, made up of a Rectangle and a Semi-circle.
x
y
y a) Find an expressionfor y in terms of x.
1) Find the length of the semi-circle,as this makes up part of the length.
2 2y x 2
xx2
2 22
xx y
12 4
x x y
Rearrangeto get y
alone
Divide by 2
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Differentiation
You need to be able to recognisepractical problems that can be solved by
using the idea of maxima and minima
9C
Example Question
A wire of length 2m is bent into the shapeshown, made up of a Rectangle and a Semi-circle.
x
y
y a) Find an expressionfor y in terms of x.
1) Work out the areas of theRectangle and Semi-circle separately.
b) Show that the Area is:
(8 4 )8
xA x x
xy
2
2
2
x
Rectangle Semi Circle
2 2r
2
24
x
2
8
x
1
2 4
x xy
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Differentiation
You need to be able to recognisepractical problems that can be solved by
using the idea of maxima and minima
9C
Example Question
A wire of length 2m is bent into the shapeshown, made up of a Rectangle and a Semi-circle.
x
y
y a) Find an expressionfor y in terms of x.
1) Work out the areas of theRectangle and Semi-circle separately.
b) Show that the Area is:
(8 4 )8
xA x x
xy
Rectangle Semi Circle2
8
x
1
2 4
x xy
A xy2
8
x
A 12 4
x xx
2
8
x
A
2 2
2 4
x x
x
2
8
x
A 2 2
2 8
x xx
(8 4 )8
xA x x
Replacey
ExpandFactorise
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Differentiation
You need to be able to recognisepractical problems that can be solved by
using the idea of maxima and minima
9C
Example Question
A wire of length 2m is bent into the shapeshown, made up of a Rectangle and a Semi-circle.
x
y
y a) Find an expressionfor y in terms of x.
1) Use the formula we have for theArea
b) Show that the Area is:
(8 4 )8
xA x x
1
2 4
x xy
c) Find the maximum
possible Area
(8 4 )8
xA x x
2 2
2 8
x xA x
14
dA xxdx
21 0
8
xx
8 8 2 0x x
4 4 0x x
4 4x x
4 4x
0.56 x20.28A m
Expand
Differentiate
Set equal to 0
Multiply by 8
Divide by 2
Factorise
Divide by (4 + )
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Summary
We have expanded our knowledge ofDifferentiation to include stationarypoints
We have looked at using maxima andminima to solve more practical problems