# Higher Unit 1 Applications 1.4 - PowerPoint PPT Presentation

PPT – Higher Unit 1 Applications 1.4 PowerPoint presentation | free to download - id: 7a1636-ZjQxN

The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
Title:

## Higher Unit 1 Applications 1.4

Description:

### Higher Unit 1 Applications 1.4 Higher Finding the gradient for a polynomial Increasing / Decreasing functions Differentiating Easy Functions Max / Min and inflexion ... – PowerPoint PPT presentation

Number of Views:123
Avg rating:3.0/5.0
Slides: 100
Provided by: AlanP198
Category:
Tags:
Transcript and Presenter's Notes

Title: Higher Unit 1 Applications 1.4

1
Higher Unit 1Applications 1.4
Higher
Finding the gradient for a polynomial
Increasing / Decreasing functions
Differentiating Easy Functions
Max / Min and inflexion Points
Differentiating Harder Functions
Curve Sketching
Differentiating with Leibniz Notation
Max Min Values on closed Intervals
Equation of a Tangent Line
Optimization
Mind Map of Chapter
2
Higher
On a straight line the gradient remains constant,
however with curves the gradient changes
continually, and the gradient at any point is in
fact the same as the gradient of the tangent at
that point.

The sides of the half-pipe are very steep(S) but
it is not very steep near the base(B).
S
Demo
B
3
Higher
A
B
4
To find the gradient at any point on a curve we
need to modify the gradient formula
Higher
For the function y f(x) we do this by taking
the point (x, f(x)) and another very close
point ((xh), f(xh)).
Then we find the gradient between the two.
((xh), f(xh))
(x, f(x))
5
Higher
The gradient is not exactly the same but is
quite close to the actual value
We can improve the approximation by making the
value of h smaller
This means the two points are closer together.
((xh), f(xh))
(x, f(x))
6
Higher
We can improve upon this approximation by
making the value of h even smaller.
So the points are even closer together.
Demo
((xh), f(xh))
(x, f(x))
7
Derivative
Higher
We have seen that on curves the gradient changes
continually and is dependant on the position on
the curve. ie the x-value of the given point.
Differentiating
The process of finding the gradient is called
Finding the rate of change
DIFFERENTIATING or FINDING THE DERIVATIVE
8
Derivative
Higher
If the formula/equation of the curve is given by
f(x)
Then the derivative is called f '(x) - f
dash x
There is a simple way of finding f '(x) from
f(x).
f(x) f '(x)
2x2 4x
4x2 8x
5x10 50x9
6x7 42x6
x3 3x2
x5 5x4
x99 99x98
9
Derivative
Rule for Differentiating
Higher
It can be given by this simple flow diagram ...
multiply by the power
reduce the power by 1
n
n
-1
ax
then f '(x)
NB the following terms expressions mean the
same

DERIVATIVE,
RATE OF CHANGE,
f '(x)
10
Derivative
Rule for Differentiating
Higher
To be able to differentiate it is VERY IMPORTANT
that you are comfortable using Fractions and
Surds Indices rules
11
xm . xn xmn
Surds Indices
12
Do YOU need extra help or revision then do
Surds Indices HHM page 340 Ex8 HHM page 342
Ex9
13
Special Points
Higher
(I) f(x) ax (Straight line function)
Index Laws x0 1
If f(x) ax
ax1
then f '(x) 1 X ax0
a X 1 a
So if g(x) 12x then g '(x) 12
Also using y mx c
The line y 12x has gradient 12,
14
Special Points
Higher
(II) f(x) a, (Horizontal Line)
Index Laws x0 1
If f(x) a
a X 1 ax0
then f '(x) 0 X ax-1
0
So if g(x) -2 then g '(x) 0
Also using formula y c , (see outcome 1
!)
The line y -2 is horizontal so has gradient
0 !
15
Differentiation techniques
Differentiation

Rate of change

Differentiation
Name
16
Calculus

Revision
Differentiate
17
Calculus

Revision
Differentiate
18
Calculus

Revision
Differentiate
19
Derivative
Higher
HHM Ex6D , Ex6E and Ex6F Even Numbers only
20
Derivative
Higher
Example 1
A curve has equation f(x) 3x4
Find the formula for its gradient and find the
Its gradient is f '(x) 12x3
f '(2) 12 X 23
12 X 8
96
Example 2
A curve has equation f(x) 3x2
Find the formula for its gradient and find the
Its gradient is f '(x) 6x
At the point where x -4 the gradient is
f '(-4) 6 X -4
-24
21
Derivative
Higher
Example 3
If g(x) 5x4 - 4x5 then find g '(2) .
g '(x) 20x3 - 20x4
g '(2) 20 X 23 - 20 X 24
160 - 320
-160
22
Derivative
Higher
Example 4
h(x) 5x2 - 3x 19
so h '(x) 10x - 3
and h '(-4) 10 X (-4) - 3
-40 - 3 -43
Example 5
k(x) 5x4 - 2x3 19x - 8, find k '(10) .
k '(x) 20x3 - 6x2 19
So k '(10) 20 X 1000 - 6 X 100 19
19419
23
Derivative
Higher
Example 6 Find the points on the curve f(x)
x3 - 3x2 2x 7 where the gradient is 2.
Now using original formula
We need f '(x) 2
ie 3x2 - 6x 2 2
f(0) 7
or 3x2 - 6x 0
ie 3x(x - 2) 0
f(2) 8 -12 4 7
ie 3x 0 or x - 2 0
7
Points are (0,7) (2,7)
so x 0 or x 2
24
Calculus

Revision
Differentiate
25
Calculus

Revision
Differentiate
Straight line form
Differentiate
26
Calculus

Revision
Differentiate
Straight line form
Differentiate
27
Calculus

Revision
Differentiate
Straight line form
Chain Rule
Simplify
28
Calculus

Revision
Differentiate
Straight line form
Differentiate
29
Calculus

Revision
Differentiate
Straight line form
Differentiate
30
Calculus

Revision
Differentiate
Straight line form
Differentiate
31
Differentiation techniques
Differentiation

Rate of change

Differentiation
Name
32
Calculus

Revision
Differentiate
Multiply out
Differentiate
33
Calculus

Revision
Differentiate
multiply out
differentiate
34
Calculus

Revision
Differentiate
Straight line form
multiply out
Differentiate
35
Calculus

Revision
Differentiate
multiply out
Differentiate
36
Calculus

Revision
Differentiate
multiply out
Simplify
Straight line form
Differentiate
37
Calculus

Revision
Differentiate
Multiply out
Straight line form
Differentiate
38
Calculus

Revision
Differentiate
Split up
Straight line form
Differentiate
39
Leibniz Notation
Higher
Leibniz Notation is an alternative way of
expressing derivatives to f'(x) , g'(x) , etc.
If y is expressed in terms of x then the
derivative is written as dy/dx .
eg y 3x2 - 7x
so dy/dx 6x - 7 .
Example 19
Find dQ/dR
NB Q 9R2 - 15R-3
So dQ/dR 18R 45R-4
40
Leibniz Notation
Higher
Example 20
A curve has equation y 5x3 - 4x2 7 .
Find the gradient where x -2 (
differentiate ! )
if x -2 then
gradient 15 X (-2)2 - 8 X (-2)
60 - (-16) 76
41
Derivative
Higher
HHM Ex6H Q1 Q3 HHM Ex6G Q1,4,7,10,13,16,19,22
,25
42
Real Life Example Physics
Higher
Newtons 2ndLaw of Motion
s ut 1/2at2 where s distance t
time.
Finding ds/dt means diff in dist ? diff in
time
ie speed or velocity
so ds/dt u at
but ds/dt v so we get
v u at
and this is Newtons 1st Law of Motion
43
Derivative
Higher
HHM Ex6H Q4 Q6
44
Equation of Tangents
y mx c
Higher
y f(x)
Demo
A(a,b)
tangent
curve
gradient of line m (from y mx c )
gradient of curve at (a, b) f? (a)
it follows that m f? (a)
45
Straight line so we need a point plus the
gradient then we can use the formula y - b
m(x - a) .
Equation of Tangents
Higher
Demo
Example 21
Find the equation of the tangent line to the
curve y x3 - 2x 1 at the point where x
-1.
Point if x -1 then y (-1)3 - (2 X -1)
1
-1 - (-2) 1
2 point is (-1,2)
when x -1 dy/dx 3 X (-1)2 - 2
m 1
3 - 2 1
46
Equation of Tangents
Higher
Now using y - b m(x - a)
point is (-1,2)
m 1
we get y - 2 1( x 1)
or y - 2 x 1
or y x 3
47
Equation of Tangents
Higher
Example 22
Find the equation of the tangent to the curve y
4 x2 at the point where x -2. (x
? 0)
Also find where the tangent cuts the X-axis and
Y-axis.
Point when x -2 then y 4
(-2)2
4/4 1

point is (-2, 1)
-8x-3
-8 x3
when x -2 then dy/dx -8 (-2)3
-8/-8 1
m 1
48
Equation of Tangents
Higher
Now using y - b m(x - a)
we get y - 1 1( x 2)
or y - 1 x 2
or y x 3
Axes
Tangent cuts Y-axis when x 0
so y 0 3 3
at point (0, 3)
Tangent cuts X-axis when y 0
so 0 x 3 or x -3
at point (-3, 0)
49
Equation of Tangents
Higher
Example 23 - (other way round)
Find the point on the curve y x2 - 6x 5
where the gradient of the tangent is 14.
dy/dx
2x - 6
so 2x - 6 14
2x 20
x 10
Put x 10 into y x2 - 6x 5
Point is (10,45)
Giving y 100 - 60 5
45
50
Derivative
Higher
HHM Ex6J
51
Increasing Decreasing Functions and
Stationary Points
Higher
Consider the following graph of y f(x) ..
y f(x)

0
0

-

a
b
c
d
e
f

-
X

0
52
Increasing Decreasing Functions and
Stationary Points
Higher
In the graph of y f(x)
The function is increasing if the gradient is
positive
i.e. f ? (x) gt 0 when x lt b or d lt x lt f
or x gt f .
The function is decreasing if the gradient is
negative
and f ? (x) lt 0 when b lt x lt d .
The function is stationary if the gradient is zero
and f ? (x) 0 when x b or x d
or x f .
These are called STATIONARY POINTS.
At x a, x c and x e the curve is
simply crossing the X-axis.
53
Increasing Decreasing Functions and
Stationary Points
Higher
Example 24
For the function f(x) 4x2 - 24x 19
determine the intervals when the function is
decreasing and increasing.
f ? (x) 8x - 24
so 8x - 24 lt 0
f(x) decreasing when f ? (x) lt 0
8x lt 24
x lt 3
Check f ? (2) 8 X 2 24 -8
f(x) increasing when f ? (x) gt 0
so 8x - 24 gt 0

8x gt 24
Check f ? (4) 8 X 4 24 8
x gt 3
54
Increasing Decreasing Functions and
Stationary Points
Higher
Example 25
For the curve y 6x 5/x2
Determine if it is increasing or decreasing when
x 10.
6x - 5x-2
so dy/dx 6 10x-3
when x 10 dy/dx 6 10/1000
6.01
Since dy/dx gt 0 then the function is
increasing.
55
Increasing Decreasing Functions and
Stationary Points
Higher
Example 26
Show that the function g(x) 1/3x3 -3x2 9x
-10 is never decreasing.
g ?(x) x2 - 6x 9
(x - 3)(x - 3)
(x - 3)2
Squaring a negative or a positive value produces
a positive value, while 02 0. So you will
never obtain a negative by squaring any real
number.
Since (x - 3)2 ? 0 for all values of x then
g ?(x) can never be negative so the function
is never decreasing.
56
Increasing Decreasing Functions and
Stationary Points
Higher
Example 27
Determine the intervals when the function f(x)
2x3 3x2 - 36x 41 is (a) Stationary (b)
Increasing (c) Decreasing.
f ?(x) 6x2 6x - 36
Function is stationary when f ?(x) 0
6(x2 x - 6)
ie 6(x 3)(x - 2) 0
6(x 3)(x - 2)
ie x -3 or x 2
57
Increasing Decreasing Functions and
Stationary Points
Higher
We now use a special table of factors to
determine when f ?(x) is positive negative.
x
-3
2
-

0
0
f(x)

Function increasing when f ?(x) gt 0
ie x lt -3 or x gt 2
Function decreasing when f ?(x) lt 0
ie -3 lt x lt 2
58
Derivative
Higher
HHM Ex6K and Ex6L
59
Stationary Points and Their Nature
Higher
y f(x)
Consider this graph of y f(x) again

0
0
-

-
c
a
b

X

0
60
Stationary Points and Their Nature
Higher
This curve y f(x) has three types of
stationary point.
When x a we have a maximum turning point
(max TP)
When x b we have a minimum turning point
(min TP)
When x c we have a point of inflexion (PI)
Each type of stationary point is determined by
the gradient ( f?(x) ) at either side of the
stationary value.
61
Stationary Points and Their Nature
Higher
Maximum Turning point
Minimum Turning Point
x
a
x
b
- 0
f?(x)
f?(x)
0 -
62
Stationary Points and Their Nature
Higher
Rising Point of inflexion
Other possible type of inflexion
x
c
x
d
f?(x)
0
f?(x)
- 0 -
63
Stationary Points and Their Nature
Higher
Example 28
Find the co-ordinates of the stationary point on
the curve y 4x3 1 and determine its nature.
SP occurs when dy/dx 0
Using y 4x3 1
so 12x2 0
if x 0 then y 1
x2 0
SP is at (0,1)
x 0
64
Stationary Points and Their Nature
Higher
Nature Table
x
0

dy/dx
0
dy/dx 12x2
So (0,1) is a rising point of inflexion.
65
Stationary Points and Their Nature
Higher
Example 29
Find the co-ordinates of the stationary points on
the curve y 3x4 - 16x3 24 and determine
their nature.
Using y 3x4 - 16x3 24
SP occurs when dy/dx 0
So 12x3 - 48x2 0
if x 0 then y 24
12x2(x - 4) 0
if x 4 then y -232
12x2 0 or (x - 4) 0
x 0 or x 4
SPs at (0,24) (4,-232)
66
Stationary Points and Their Nature
Higher
Nature Table
x
0
4
dy/dx
- 0 - 0
dy/dx12x3 - 48x2
So (0,24) is a Point of inflexion and
(4,-232) is a minimum Turning Point
67
Stationary Points and Their Nature
Higher
Example 30
Find the co-ordinates of the stationary points on
the curve y 1/2x4 - 4x2 2 and determine
their nature.
SP occurs when dy/dx 0
Using y 1/2x4 - 4x2 2
if x 0 then y 2
So 2x3 - 8x 0
if x -2 then y -6
2x(x2 - 4) 0
if x 2 then y -6
2x(x 2)(x - 2) 0
x 0 or x -2 or x 2
SPs at(-2,-6), (0,2) (2,-6)
68
Stationary Points and Their Nature
Higher
Nature Table
x
0
-2
2
dy/dx
- 0 0 - 0

So (-2,-6) and (2,-6) are Minimum Turning
Points and (0,2) is a Maximum Turning Points
69
Curve Sketching
Higher
Note A sketch is a rough drawing which
includes important details. It is not an
accurate scale drawing.
Process
(a) Find where the curve cuts the co-ordinate
axes.
for Y-axis put x 0
for X-axis put y 0 then solve.
(b) Find the stationary points determine
their nature as done in previous section.
(c) Check what happens as x ? /- ? .
This comes automatically if (a) (b) are correct.
70
Curve Sketching
Higher
Dominant Terms
Suppose that f(x) -2x3 6x2 56x - 99
As x ? /- ? (ie for large
positive/negative values)
The formula is approximately the same as f(x)
-2x3
Graph roughly
As x ? ? then y ? -?
As x ? -? then y ? ?
71
Curve Sketching
Higher
Example 31
Sketch the graph of y -3x2 12x 15
(a) Axes
If x 0 then y 15
If y 0 then -3x2 12x 15 0
(? -3)
x2 - 4x - 5 0
(x 1)(x - 5) 0
x -1 or x 5
Graph cuts axes at (0,15) , (-1,0) and (5,0)
72
Curve Sketching
Higher
(b) Stationary Points
occur where dy/dx 0
so -6x 12 0
If x 2 then y -12 24 15 27
6x 12
x 2
Stationary Point is (2,27)
Nature Table
x
2
dy/dx
0 -
So (2,27) is a Maximum Turning Point
73
Curve Sketching
Higher
Summarising
as x ? ? then y ? -?
(c) Large values
as x ? -? then y ? -?
using y -3x2
Y
Sketching
5
Cuts x-axis at -1 and 5
-1
Cuts y-axis at 15
15
Max TP (2,27)
(2,27)
X
y -3x2 12x 15
74
Curve Sketching
Higher
Example 32
Sketch the graph of y -2x2 (x - 4)
(a) Axes
If x 0 then y 0 X (-4) 0
If y 0 then -2x2 (x - 4) 0
-2x2 0 or (x - 4) 0
x 0 or x 4
(b) SPs
Graph cuts axes at (0,0) and (4,0) .
y -2x2 (x - 4)
-2x3 8x2
SPs occur where dy/dx 0
so -6x2 16x 0
75
Curve Sketching
Higher
-2x(3x - 8) 0
-2x 0 or (3x - 8) 0
x 0 or x 8/3
If x 0 then y 0 (see part (a) )
If x 8/3 then y -2 X (8/3)2 X (8/3 -4)
512/27
nature
x
0
8/3
-

-
0
0
dy/dx
76
Curve Sketching
Higher
Summarising
(c) Large values
as x ? ? then y ? -?
using y -2x3
as x ? -? then y ? ?
Y
Sketch
Cuts x axis at 0 and 4
0
4
Max TPs at (8/3, 512/27)
(8/3, 512/27)
X
y -2x2 (x 4)
77
Derivative
Higher
HHM Ex6M
78
Max Min on Closed Intervals
Higher
In the previous section on curve sketching we
dealt with the entire graph.
In this section we shall concentrate on the
important details to be found in a small section
of graph.
Suppose we consider any graph between the points
where x a and x b (i.e. a ? x ? b)
then the following graphs illustrate where we
would expect to find the maximum minimum values.
79
Max Min on Closed Intervals
Higher
y f(x)
(b, f(b))
max f(b) end point
(a, f(a))
min f(a) end point
X
a b
80
Max Min on Closed Intervals
Higher
max f(c ) max TP
(c, f(c))
y f(x)
(b, f(b))
(a, f(a))
min f(a) end point
x
a b
c
NB a lt c lt b
81
Max Min on Closed Intervals
Higher
y f(x)
max f(b) end point
(b, f(b))
(a, f(a))
(c, f(c))
min f(c) min TP
x
NB a lt c lt b
c
a b
82
Max Min on Closed Intervals
Higher
From the previous three diagrams we should be
able to see that the maximum and minimum values
of f(x) on the closed interval a ? x ? b can be
found either at the end points or at a stationary
point between the two end points
Example 34
Find the max min values of y 2x3 - 9x2 in
the interval where -1 ? x ? 2.
End points
If x -1 then y -2 - 9 -11
If x 2 then y 16 - 36 -20
83
Max Min on Closed Intervals
Higher
Stationary points
dy/dx 6x2 - 18x
6x(x - 3)
SPs occur where dy/dx 0
6x(x - 3) 0
6x 0 or x - 3 0
x 0 or x 3
not in interval
in interval
If x 0 then y 0 - 0 0
Hence for -1 ? x ? 2 , max 0 min -20
84
Max Min on Closed Intervals
Higher
Extra bit
Using function notation we can say that
Domain x?R -1 ? x ? 2
Range y?R -20 ? y ? 0
85
Derivative Graphs
Higher
Demo
86
Optimization
Higher
Note Optimum basically means the best possible.
In commerce or industry production costs and
profits can often be given by a mathematical
formula.
Optimum profit is as high as possible so we would
look for a max value or max TP.
Optimum production cost is as low as possible so
we would look for a min value or min TP.
87
Optimization
Higher
Practical exercise on optimizing volume.
Graph
Problem
88
Q. What is the maximum volume We can have for the
given dimensions
Optimization
Higher
Example 35
A rectangular sheet of foil measuring 16cm X 10
cm has four small squares each x cm cut from each
corner.
16cm
x cm
10cm
x cm
NB x gt 0 but 2x lt 10 or x lt 5
ie 0 lt x lt 5
This gives us a particular interval to consider !
89
Optimization
Higher
By folding up the four flaps we get a small
cuboid
x cm
(10 - 2x) cm
(16 - 2x) cm
The volume is now determined by the value of x so
we can write
V(x) x(16 - 2x)(10 - 2x)
x(160 - 52x 4x2)
4x3 - 52x2 160x
We now try to maximize V(x) between 0 and 5
90
Optimization
Higher
Considering the interval 0 lt x lt 5
End Points
V(0) 0 X 16 X 10 0
V(5) 5 X 6 X 0 0
SPs
V '(x) 12x2 - 104x 160
4(3x2 - 26x 40)
4(3x - 20)(x - 2)
91
Optimization
Higher
SPs occur when V '(x) 0
ie 4(3x - 20)(x - 2) 0
3x - 20 0 or x - 2 0
ie x 20/3 or x 2
not in interval
in interval
When x 2 then
V(2) 2 X 12 X 6 144
We now check gradient near x 2
92
Optimization
Higher
Nature
x
2
-

V '(x)
0
Hence max TP when x 2
So max possible volume 144cm3
93
Optimization
Higher
Example 36
When a company launches a new product its share
of the market after x months is calculated by the
formula
(x ? 2)
So after 5 months the share is
S(5) 2/5 4/25
6/25
Find the maximum share of the market that the
company can achieve.
94
Optimization
Higher
End points
S(2) 1 1 0
There is no upper limit but as x ? ? S(x) ? 0.
SPs occur where S ?(x) 0
95
Optimization
Higher
rearrange
8x2 2x3
8x2 - 2x3 0
2x2(4 x) 0
x 0 or x 4
Out with interval
In interval
We now check the gradients either side of 4
96
Optimization
Higher
Nature
S ?(3.9 ) 0.00337
x ? 4 ?
S ?(4.1) -0.0029

-
0
S ?(x)
Hence max TP at x 4
And max share of market S(4)
2/4 4/16
1/2 1/4
1/4
97
Derivative
Higher
HHM Ex6Q and Ex6R
98
Nature Table
Equation of tangent line
Leibniz Notation
Straight Line Theory
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}