Title: Higher Unit 1 Applications 1.4
1Higher 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
2Gradients Curves
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 halfpipe are very steep(S) but
it is not very steep near the base(B).
S
Demo
B
3Gradients Curves
Higher
Gradient of tangent gradient of curve at A
A
B
Gradient of tangent gradient of curve at B
4To find the gradient at any point on a curve we
need to modify the gradient formula
Gradients Curves
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))
Approx gradient
(x, f(x))
True gradient
5Gradients Curves
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))
Approx gradient
(x, f(x))
True gradient
6Gradients Curves
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))
Approx gradient
True gradient
(x, f(x))
7Derivative
Higher
We have seen that on curves the gradient changes
continually and is dependant on the position on
the curve. ie the xvalue of the given point.
Finding the GRADIENT
Differentiating
The process of finding the gradient is called
Finding the rate of change
DIFFERENTIATING or FINDING THE DERIVATIVE
(Gradient)
8Derivative
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
9Derivative
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
GRADIENT,
DERIVATIVE,
RATE OF CHANGE,
f '(x)
10Derivative
Rule for Differentiating
Higher
To be able to differentiate it is VERY IMPORTANT
that you are comfortable using Fractions and
Surds Indices rules
11xm . xn xmn
Surds Indices
12Do YOU need extra help or revision then do
Surds Indices HHM page 340 Ex8 HHM page 342
Ex9
13Special 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,
and derivative gradient !!
14Special 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 ax1
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 !
15Differentiation techniques
Differentiation
Gradient
Rate of change
Differentiation
Name
16Calculus
Revision
Differentiate
17Calculus
Revision
Differentiate
18Calculus
Revision
Differentiate
19Derivative
Higher
HHM Ex6D , Ex6E and Ex6F Even Numbers only
20Derivative
Higher
Example 1
A curve has equation f(x) 3x4
Find the formula for its gradient and find the
gradient when x 2
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
gradient when x 4
Its gradient is f '(x) 6x
At the point where x 4 the gradient is
f '(4) 6 X 4
24
21Derivative
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
22Derivative
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
23Derivative
Higher
Example 6 Find the points on the curve f(x)
x3  3x2 2x 7 where the gradient is 2.
NB gradient derivative f '(x)
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
24Calculus
Revision
Differentiate
25Calculus
Revision
Differentiate
Straight line form
Differentiate
26Calculus
Revision
Differentiate
Straight line form
Differentiate
27Calculus
Revision
Differentiate
Straight line form
Chain Rule
Simplify
28Calculus
Revision
Differentiate
Straight line form
Differentiate
29Calculus
Revision
Differentiate
Straight line form
Differentiate
30Calculus
Revision
Differentiate
Straight line form
Differentiate
31Differentiation techniques
Differentiation
Gradient
Rate of change
Differentiation
Name
32Calculus
Revision
Differentiate
Multiply out
Differentiate
33Calculus
Revision
Differentiate
multiply out
differentiate
34Calculus
Revision
Differentiate
Straight line form
multiply out
Differentiate
35Calculus
Revision
Differentiate
multiply out
Differentiate
36Calculus
Revision
Differentiate
multiply out
Simplify
Straight line form
Differentiate
37Calculus
Revision
Differentiate
Multiply out
Straight line form
Differentiate
38Calculus
Revision
Differentiate
Split up
Straight line form
Differentiate
39Leibniz 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  15R3
So dQ/dR 18R 45R4
40Leibniz Notation
Higher
Example 20
A curve has equation y 5x3  4x2 7 .
Find the gradient where x 2 (
differentiate ! )
gradient dy/dx 15x2  8x
if x 2 then
gradient 15 X (2)2  8 X (2)
60  (16) 76
41Derivative
Higher
HHM Ex6H Q1 Q3 HHM Ex6G Q1,4,7,10,13,16,19,22
,25
42Real 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
43Derivative
Higher
HHM Ex6H Q4 Q6
44Equation of Tangents
y mx c
Higher
y f(x)
Demo
A(a,b)
tangent
NB at A(a, b) gradient of line gradient of
curve
gradient of line m (from y mx c )
gradient of curve at (a, b) f? (a)
it follows that m f? (a)
45Straight 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)
Gradient dy/dx 3x2  2
when x 1 dy/dx 3 X (1)2  2
m 1
3  2 1
46Equation 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
47Equation 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 Xaxis and
Yaxis.
Point when x 2 then y 4
(2)2
4/4 1
point is (2, 1)
Gradient y 4x2 so dy/dx
8x3
8 x3
when x 2 then dy/dx 8 (2)3
8/8 1
m 1
48Equation 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 Yaxis when x 0
so y 0 3 3
at point (0, 3)
Tangent cuts Xaxis when y 0
so 0 x 3 or x 3
at point (3, 0)
49Equation 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.
gradient of tangent gradient of curve
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
50Derivative
Higher
HHM Ex6J
51Increasing 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
52Increasing 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 Xaxis.
53Increasing 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
54Increasing 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  5x2
so dy/dx 6 10x3
when x 10 dy/dx 6 10/1000
6.01
Since dy/dx gt 0 then the function is
increasing.
55Increasing 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.
56Increasing 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
57Increasing 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
58Derivative
Higher
HHM Ex6K and Ex6L
59Stationary Points and Their Nature
Higher
y f(x)
Consider this graph of y f(x) again
0
0


c
a
b
X
0
60Stationary 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.
61Stationary Points and Their Nature
Higher
Maximum Turning point
Minimum Turning Point
x
a
x
b
 0
f?(x)
f?(x)
0 
62Stationary Points and Their Nature
Higher
Rising Point of inflexion
Other possible type of inflexion
x
c
x
d
f?(x)
0
f?(x)
 0 
63Stationary Points and Their Nature
Higher
Example 28
Find the coordinates 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
64Stationary Points and Their Nature
Higher
Nature Table
x
0
dy/dx
0
dy/dx 12x2
So (0,1) is a rising point of inflexion.
65Stationary Points and Their Nature
Higher
Example 29
Find the coordinates 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)
66Stationary 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
67Stationary Points and Their Nature
Higher
Example 30
Find the coordinates 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)
68Stationary 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
69Curve 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 coordinate
axes.
for Yaxis put x 0
for Xaxis 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.
70Curve 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 ? ?
71Curve 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)
72Curve 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
73Curve Sketching
Higher
Summarising
as x ? ? then y ? ?
(c) Large values
as x ? ? then y ? ?
using y 3x2
Y
Sketching
5
Cuts xaxis at 1 and 5
1
Cuts yaxis at 15
15
Max TP (2,27)
(2,27)
X
y 3x2 12x 15
74Curve 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
75Curve 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
76Curve 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)
77Derivative
Higher
HHM Ex6M
78Max 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.
79Max 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
80Max 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
81Max 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
82Max 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
83Max 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
84Max 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
85Derivative Graphs
Higher
Demo
86Optimization
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.
87Optimization
Higher
Practical exercise on optimizing volume.
Graph
Problem
88Q. 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 !
89Optimization
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
90Optimization
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)
91Optimization
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
92Optimization
Higher
Nature
x
2

V '(x)
0
Hence max TP when x 2
So max possible volume 144cm3
93Optimization
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.
94Optimization
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
95Optimization
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
96Optimization
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
97Derivative
Higher
HHM Ex6Q and Ex6R
98Nature Table
Equation of tangent line
Leibniz Notation
Straight Line Theory
Gradient at a point
f(x)0 Stationary Pts Max. / Mini Pts Inflection
Pt
Graphs f(x)0
Derivative gradient rate of change
Differentiation of Polynomials
f(x) axn then fx) anxn1
99Are you on Target !
 Make sure you complete and correct
 ALL of the Differentiation 1
 questions in the past paper booklet.