The Laws Of Surds.
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Transcript The Laws Of Surds.
Higher Unit 1
Outcome 3
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Higher
Finding the gradient for a polynomial
Increasing / Decreasing functions
Differentiating a polynomial
Max / Min and Infection Points
Differentiating Negative Indices
Curve Sketching
Differentiating Roots
Max & Min Values on closed Intervals
Differentiating Brackets
Optimization
Differentiating Fraction Terms
Mind Map of Chapter
Differentiating with Leibniz Notation
Equation of a Tangent Line
Exam Type Questions
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Gradients & Curves
Outcome 3
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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.
S
B
The sides of the half-pipe
are very steep(S) but it is
not very steep near the
base(B).
Gradients & Curves
Outcome 3
Higher
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Gradient of tangent = gradient of curve at A
A
B
Gradient of tangent = gradient of curve at B
To find the gradient at any
y 2 - y1
point on a curve we need to m
x 2 - x1
modify the gradient Outcome
formula 3
Gradients & Curves
Higher
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For the function y = f(x) we do this by taking the point (x, f(x))
and another “very close point” ((x+h), f(x+h)).
Then we find the gradient between the two.
((x+h), f(x+h))
Approx gradient
(x, f(x))
True gradient
Gradients & Curves
Higher
Outcome 3
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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.
((x+h), f(x+h))
Approx gradient
(x, f(x))
True gradient
Gradients & Curves
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Higher
Outcome 3
We can improve upon this approximation by making the
value of h even smaller.
So the points are even closer together.
Approx gradient
(x, f(x))
((x+h), f(x+h))
True gradient
Derivative
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Higher
Outcome 3Finding the
GRADIENT
Differentiating
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.
We can use the formula for the curve to produce a
formula for the gradient.
Finding the rate
ofThis
change
process is called DIFFERENTIATING
or FINDING THE DERIVATIVE
Derivative
Higher
Outcome 3
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If the formula/equation of the curve is given by f(x)
Then the derivative is called f '(x)
There is a simple way
of finding f '(x) from f(x).
-
“f dash x”
f(x)
f '(x)
2x2
4x2
5x10
6x7
x3
x5
x99
4x
8x
50x9
42x6
3x2
5x4
99x98
Derivative
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Higher
Or
Rules for Differentiating
Outcome 3
These can be given by the simple flow diagram ...
multiply by
the power
If
reduce the
power by 1
f(x)
= axn
then f '(x) = naxn-1
NB: the following terms & expressions mean the same
GRADIENT, DERIVATIVE, RATE OF CHANGE, f '(x)
Derivative
Outcome 3
Higher
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Example 1
A curve has equation
Its gradient is
f(x) = 3x4
f '(x) = 12x3
At the point where x = 2 the gradient is
f '(2) = 12 X 23 =
Example 2
12 X 8 =
96
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
Derivative
Outcome 3
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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
Special Points
Outcome 3
Higher
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(I) f(x) = ax (Straight line function)
If f(x) = ax
=
Index Laws
ax1
then f '(x) = 1 X ax0
x0 = 1
= aX1 = a
So if g(x) = 12x then
Also using
g '(x) = 12
y = mx + c
The line y = 12x has gradient 12,
and derivative = gradient !!
Special Points
Outcome 3
Higher
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(II) f(x) = a, (Horizontal Line)
If f(x) = a
= a X 1 = ax0
then f '(x) = 0 X ax-1
So if g(x) = -2 then
Also using formula
Index Laws
x0 = 1
= 0
g '(x) = 0
y = c , (see outcome 1 !)
The line y = -2 is horizontal so has gradient 0 !
Derivative
Outcome 3
Higher
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Example 4
h(x) = 5x2 - 3x + 19
and
so
h '(-4) = 10 X (-4) - 3
h '(x) = 10x - 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
Derivative
Outcome 3
Higher
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Example 6 : Find the points on the curve
f(x) = x3 - 3x2 + 2x + 7 where the gradient is 2.
NB: gradient = derivative = f '(x)
We need
ie
f '(x) = 2
3x2 - 6x + 2 = 2
or
3x2 - 6x = 0
ie
3x(x - 2) = 0
ie
so
3x = 0 or
x-2=0
x = 0 or x = 2
Now using original
formula
f(0) = 7
f(2) = 8 -12 + 4 + 7
=7
Points are (0,7) & (2,7)
Negative Indices
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Higher
Outcome 3
Index Law
1 = x-m
xm
NB: Before we can differentiate a term it must be
in the form axn
Bottom line terms get negative powers !!
YOU need to be good with
FRACTIONS and INDICES
Negative Indices
Outcome 3
Higher
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Example 7
f(x) = 1 = x -2
x2
So
f '(x) =
-2x-3
-2
= 3
x
Example 8
g(x) = -3
x4
So
g '(x) =
12x-5
= -3x-4
=
12
x5
Negative Indices
Outcome 3
Higher
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Example 9
h(y) = 4
=
3y3
So
h '(y) =
4/
3
-12/
3
y-4
=
3/
4
y-3
= -4
y4
Example 10
k(t) = 3
4t2/3
So
k '(t) =
-6/12
t-5/3
t-2/3
=
-1
2t5/3
Negative Indices
Outcome 3
Higher
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Example 11
The equation of a curve is f(x) = 8
x4
(x 0)
Find the gradient at the point where x = -2 .
f(x) = 8 = 8x-4
x4
so
Required gradient =
f '(x) = -32x-5 =-32
x5
f '(-2) = -325
(-2)
= -32
-32
=1
Differentiating Roots
Outcome 3
Higher
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Fractional Indices
From the credit or Intermediate 2 course
nx
nxm
or
=
1/n
x
n
m
( x)
=
m/n
x
nxm
or (nx)m
= xm/n
Outcome 3
Higher
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Check the following using the power button on your calc.
641/2 = 64 =
641/3 =
=
364
8
163/4 = (416)3 = 23 = 8
4
1255/3 = (3125)5 = 55 = 3125
Fractional Powers
25-1/2 = 1
25
= 1 or 0.2
5
top line - power
bottom line - root
Differentiating Roots
Outcome 3
Higher
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Example 12
f(x) = x = x1/2
so f '(x) = 1/2 x-1/2 = 1
2x
Example 13
g(t) = t5 = t5/2
so g'(t) = 5/2 t3/2 = 5t3
2
Example 14
h(y) = 1 = y-7/3
3y7
so h'(y) = -7/3 y-10/3 = -7
33y10
Differentiating Roots
Outcome 3
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Higher
Example 15
Find the rate of change at the point where x = 4 on the
curve with equation g(x) = 4 .
x
g(x) = 4
= 4x-1/2
x
NB: rate of change = gradient = g'(x) .
g'(x) = -2x-3/2 =
so g'(4) =
-2
(x)3
-2
-2/ = -1/
=
8
4
(4)3
Brackets
Outcome 3
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Higher
Basic Rule: Break brackets before you differentiate !
Example 16
h(x) = 2x(x + 3)(x -3)
= 2x(x2 - 9)
= 2x3 - 18x
So h'(x) = 6x2 -18
Fractions
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Higher
Outcome 3
3 4 7
Recall
1
7 7 7
Reversing the above we get the following “rule” !
a+b a b
c
c c
This can be used as follows …..
Fractions
Outcome 3
Higher
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Example 17
f(x) = 3x3 - x + 2
x2
=
3x3 - x + 2
x2
x2 x2
=
3x - x-1 + 2x-2
f '(x) = 3 + x-2 - 4x-3 = 3 + 1 - 4
x2 x3
Fractions
Outcome 3
Higher
Change to
powers √=½
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Example 18 (tricky)
g(y) = (y + y)(y + 1)
yy
= (y + y1/2)(y + 1)
y X y1/2
Brackets
= y2 + y3/2 + y + y1/2
Single
fractions
Correct form
y3/2
Indices
= y2 + y3/2 + y + y1/2
y3/2 y3/2
y3/2 y3/2
= y1/2 + 1 + y-1/2 + y-1
Fractions
Outcome 3
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Higher
g'(y) = 1/2y-1/2 - 1/2y-3/2 - y-2
=
Also
g'(4) =
=
1
- 1 2y1/2
2y3/2
1
y2
1
1
2 X 41/2
2 X 43/2
1
2 X 4
-
1
2 X (4)3
1
42
-
1
16
= 1/4 - 1/16 - 1/16 = 1/8
Leibniz Notation
Outcome 3
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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
Example 19
NB:
So
so
dy/
dx
= 6x - 7 .
Q = 9R2 - 15
R3
Find
dQ/
dR
Q = 9R2 - 15R-3
dQ/
dR
= 18R + 45R-4 = 18R + 45
R4
Leibniz Notation
Outcome 3
Higher
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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
Real Life Example
Physics
Outcome 3
Higher
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Newton’s 2ndLaw of Motion
s = ut + 1/2at2
Finding
ds/
dt
where s = distance & t = time.
means “diff in dist”
ie
so
but
ds/
dt
ds/
dt
and this is
“diff in time”
speed or velocity
= u + at
= v so we get
v = u + at
Newton’s 1st Law of Motion
Equation of Tangents
y = mx +c
Outcome 3
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Higher
A(a,b)
y = f(x)
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)
Straight line so
need a point
Equation
ofweTangents
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Higher
plus the gradient then we can use
the formula y - b = m(x - a) .
Outcome 3
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
Gradient:
when x = -1
point is (-1,2)
dy/
dx
= 3x2 - 2
dy/
dx
= 3 X (-1)2 - 2
= 3-2 = 1
m=1
Equation of Tangents
Outcome 3
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Higher
Now using
y - b = m(x - a)
we get
y - 2 = 1( x + 1)
or
y-2=x+1
or
y=x+3
point is (-1,2)
m=1
Equation of Tangents
Outcome 3
Higher
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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
Gradient:
x = -2
y = 4x-2
when x = -2 then
then y = 4 =
(-2)2
so
dy/
dx
dy/
dx
=
-8
=
(-2)3
4/
4
= 1
point is (-2, 1)
= -8x-3 =
-8/
-8
=1
-8
x3
m=1
Equation of Tangents
Outcome 3
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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)
Equation of Tangents
Outcome 3
Higher
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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
so
= 2x - 6
2x - 6 = 14
2x = 20
x = 10
Put x = 10 into y = x2 - 6x + 5
Giving
y = 100 - 60 + 5 = 45
Point is (10,45)
Increasing & Decreasing Functions
and Stationary Points
Outcome 3
Higher
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Consider the following graph of y = f(x) …..
y = f(x)
0
+
+ a
0
b
c
+
-
d
0
e
+
f
+
X
Increasing & Decreasing Functions
and Stationary Points
Higher
Outcome 3
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In the graph of
y = f(x)
The function is increasing if the gradient is positive
i.e. f (x) > 0 when x < b or d < x < f or x > f .
The function is decreasing if the gradient is negative
and f (x) < 0 when b < x < 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.
Increasing & Decreasing Functions
and Stationary Points
Outcome 3
Higher
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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 < 0
8x < 24
x<3
2 – 24 = -8
f(x) decreasing when f (x) < 0
Check: f (2) = 8
X
f(x) increasing when f (x) > 0
Check: f (4) = 8 X 4 – 24 = 8
so 8x - 24 > 0
8x > 24
x>3
Increasing & Decreasing Functions
and Stationary Points
Outcome 3
Higher
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Example 25
For the curve y = 6x – 5/x2
Determine if it is increasing or decreasing when x = 10.
y = 6x - 5 = 6x - 5x-2
x2
so
dy/
dx
= 6 + 10x-3 = 6 + 10
x3
when x = 10
Since
dy/
dx
dy/
dx
= 6 + 10/1000 = 6.01
> 0 then the function is increasing.
Increasing & Decreasing Functions
and Stationary Points
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Higher
Outcome 3
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.
Increasing & Decreasing Functions
and Stationary Points
Outcome 3
Higher
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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
= 6(x2 + x - 6)
= 6(x + 3)(x - 2)
Function is stationary
when f (x) = 0
ie
6(x + 3)(x - 2) = 0
ie
x = -3 or
x=2
Increasing & Decreasing Functions
and Stationary Points
Outcome 3
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Higher
We now use a special table of factors to
determine when f (x) is positive & negative.
x
f’(x)
+
-3
0
-
2
0
+
Function increasing when f (x) > 0
ie
x < -3 or x > 2
Function decreasing when f (x) < 0
ie
-3 < x < 2
Stationary Points
and Their Nature
Outcome 3
Higher
y = f(x)
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Consider this graph of y = f(x) again
0
-
+
+
0
a
+
+
-
b
0
+
c
X
Stationary Points
and Their Nature
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Higher
Outcome 3
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.
Stationary Points
and Their Nature
Outcome 3
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Higher
Maximum Turning point
x
f(x)
a
+
0
Minimum Turning Point
x
-
f(x)
b
-
0
+
Stationary Points
and Their Nature
Outcome 3
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Higher
Rising Point
of Inflection
Other possible
type of infection
x
x
f(x)
c
+
0
+
f(x)
d
-
0
-
Stationary Points
and Their Nature
Outcome 3
Higher
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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
so 12x2 = 0
x2 = 0
x= 0
Using
y = 4x3 + 1
if x = 0 then y = 1
SP is at (0,1)
Stationary Points
and Their Nature
Outcome 3
Higher
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Nature Table
x
dy/
dx
0
+
0
+
dy/
dx
= 12x2
So (0,1) is a rising point of inflexion.
Stationary Points
and Their Nature
Outcome 3
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Higher
Example 29
Find the co-ordinates of the stationary points on the
curve y = 3x4 - 16x3 + 24 and determine their nature.
SP occurs when
dy/
dx
= 0
So 12x3 - 48x2 = 0
12x2(x - 4) = 0
12x2 = 0 or (x - 4) = 0
x = 0 or x = 4
Using y = 3x4 - 16x3 + 24
if x = 0 then y = 24
if x = 4 then y = -232
SPs at (0,24) & (4,-232)
Stationary Points
and Their Nature
Outcome 3
Higher
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Nature Table
x
dy/
dx
4
0
-
0
-
0
+
So (0,24) is a Point of Infection
and (4,-232) is a minimum Turning Point
dy/
dx=12x
3
- 48x2
Stationary Points
and Their Nature
Outcome 3
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Higher
Example 30
Find the co-ordinates of the stationary points on the
curve y = 1/2x4 - 4x2 + 2 and determine their nature.
= 0
Using y = 1/2x4 - 4x2 + 2
2x3 - 8x = 0
if x = 0 then y = 2
2x(x2 - 4) = 0
if x = -2 then y = -6
SP occurs when
So
dy/
dx
2x(x + 2)(x - 2) = 0
if x = 2 then y = -6
x = 0 or x = -2 or x = 2 SP’s at(-2,-6), (0,2) & (2,-6)
Stationary Points
and Their Nature
Outcome 3
Higher
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Nature Table
x
dy/
dx
-2
-
0
0
+
0
2
-
0
+
So (-2,-6) and (2,-6) are Minimum Turning Points
and (0,2) is a Maximum Turning Points
Curve Sketching
Outcome 3
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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.
Curve Sketching
Higher
Suppose that
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Outcome 3
Dominant Terms
As
x +/-
f(x) = -2x3 + 6x2 + 56x - 99
(ie
for large positive/negative values)
The formula is approximately the same as f(x) = -2x3
As
As
x +
x -
then
y
-
then
y
+
Graph roughly
Curve Sketching
Outcome 3
Higher
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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
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)
( -3)
Curve Sketching
Outcome 3
Higher
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(b) Stationary Points
occur where
so
If x = 2
6x = 12
x = 2
Nature Table
x
So (2,27)
is a Maximum Turning Point
= 0
-6x + 12 = 0
then y = -12 + 24 + 15 = 27
Stationary Point is (2,27)
dy/
dx
dy/
dx
2
+
0
-
Curve Sketching
Outcome 3
Higher
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Summarising
(c) Large values
as
x +
then
y -
using y = -3x2
as
x -
then
y -
Sketching
Y
Cuts x-axis at -1 and 5
Cuts y-axis at 15
Max TP (2,27)
X
y = -3x2 + 12x + 15
Curve Sketching
Outcome 3
Higher
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Example 32
(a) Axes
(b) SPs
Sketch the graph of
If x = 0 then
If y = 0 then
-2x2 = 0
y = -2x2 (x - 4)
y = 0 X (-4) = 0
-2x2 (x - 4) = 0
or (x - 4) = 0
x = 0 or x = 4
Graph cuts axes at (0,0) and (4,0) .
y = -2x2 (x - 4)
= -2x3 + 8x2
SPs occur where
so
dy/
dx
= 0
-6x2 + 16x = 0
Curve Sketching
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Higher
Outcome 3
-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
dy/
dx
-
0
0
8/
3
+ 0
-
Curve Sketching
Outcome 3
Higher
www.mathsrevision.com
Summarising
(c) Large values
using y = -2x3
Y
Sketch
as
x +
then
y
as
x -
then
y
-
+
Cuts x – axis at 0 and 4
512/ )
Max TP’s at (88/33, 512
27
27
X
y = -2x2 (x – 4)
Curve Sketching
Outcome 3
Higher
www.mathsrevision.com
Example 33 Sketch the graph of
(a) Axes
y = 8 + 2x2 - x4
If x = 0 then y = 8
If y = 0 then 8 + 2x2 - x4 = 0
Let u = x2 so u2 = x4
(0,8)
Equation is now 8 + 2u - u2 = 0
(4 - u)(2 + u) = 0
(4 - x2)(2 + x2) = 0
or (2 + x) (2 - x)(2 + x2) = 0
So x = -2 or x = 2 but x2 -2
Graph cuts axes at (0,8) , (-2,0) and (2,0)
Curve Sketching
Outcome 3
Higher
www.mathsrevision.com
(b) SPs
SPs occur where
dy/
dx
So 4x - 4x3 = 0
4x(1 - x2) = 0
4x(1 - x)(1 + x) = 0
x = 0 or x =1 or x = -1
Using y = 8 + 2x2 - x4
when x = 0 then y = 8
when x = -1 then y = 8 + 2 - 1 = 9 (-1,9)
when x = 1 then y = 8 + 2 - 1 = 9 (1,9)
= 0
Curve Sketching
Outcome 3
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Higher
nature
x
dy/
dx
-1
0
1
+ 0 - 0 + 0 -
So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .
Curve Sketching
Outcome 3
Higher
www.mathsrevision.com
Summarising
Sketch is
Y
(c) Large values
Using y = - x4
as
as
x +
x -
then
y
then
y
-
-
Cuts x – axis at -2 and 2
Cuts y – axis at 8
Max TP’s at (-1,9) (1,9)
X
y = 8 + 2x2 - x4
Max & Min on Closed Intervals
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Higher
Outcome 3
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.
Max & Min on Closed Intervals
Outcome 3
Higher
www.mathsrevision.com
y =f(x)
(b, f(b))
max = f(b) end point
(a, f(a))
a
min = f(a)
b
X
end point
Max & Min on Closed Intervals
Outcome 3
Higher
www.mathsrevision.com
(c, f(c))
max = f(c ) max TP
(b, f(b))
y =f(x)
min = f(a) end point
(a, f(a))
a
c
b
x
NB:
a<c<b
Max & Min on Closed Intervals
Outcome 3
y =f(x)
max = f(b) end point
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Higher
(b, f(b))
(a, f(a))
(c, f(c))
a
c
b
min = f(c) min TP
x
NB:
a<c<b
Max & Min on Closed Intervals
Outcome 3
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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
Max & Min on Closed Intervals
Outcome 3
Higher
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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
Max & Min on Closed Intervals
Outcome 3
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Higher
Extra bit
Using function notation we can say that
Domain = {xR: -1 x 2 }
Range = {yR: -20 y 0 }
Optimization
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Higher
Outcome 3
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.
Q. What is the maximum volume
Optimization
We can have for the given
www.mathsrevision.com
Higher
Example 35
dimensions
Outcome
3
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 > 0 but 2x < 10 or x < 5
ie 0 < x < 5
This gives us a particular interval to consider !
Optimization
Outcome 3
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Higher
x cm
By folding up the four flaps we
get a small cuboid
(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
Optimization
Outcome 3
www.mathsrevision.com
Higher
End Points
Considering the interval 0 < x < 5
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)
Optimization
Outcome 3
Higher
www.mathsrevision.com
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
When x = 2 then
in interval
V(2) = 2 X 12 X 6 = 144
We now check gradient near x = 2
Optimization
Outcome 3
Higher
www.mathsrevision.com
Nature
x
V '(x)
+
2
0
-
Hence max TP when x = 2
So max possible volume = 144cm3
Optimization
www.mathsrevision.com
Higher
Example 36
Outcome 3
When a company launches a new product its share of
the market after x months is calculated by the formula
2 4
S ( x) 2
x x
(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.
Optimization
Outcome 3
Higher
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End points
S(2) = 1 – 1 = 0
There is no upper limit but as x S(x) 0.
2 4
S ( x) 2 2 x 1 4 x 2
x x
S '( x) 2 x 1 4 x 2 2 x 2 8 x 3
2 8
3
2
x
x
SPs occur where S (x) = 0
8
2
S '(x ) 3 2 0
x
x
Optimization
Outcome 3
www.mathsrevision.com
Higher
S '(x )
8
2
0
3
2
x
x
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
Optimization
Outcome 3
Higher
www.mathsrevision.com
Nature
x
S (x)
+
4
0
-
S (3.9 ) = 0.00337…
S (4.1) = -0.0029…
Hence max TP at x = 4
And max share of market = S(4)
= 2/4 – 4/16
= 1/2 – 1/4
= 1/ 4
Nature Table
x -1 2 5
+ 0 Max
Leibniz
Notation
dy
f '( x)
dx
f’(x)=0
Graphs
Stationary Pts
f’(x)=0
f’(x)
Max. / Mini Pts
Equation of
tangent line
Straight Line
Gradient
at a point
Derivative
= gradient
Differentiation
of Polynomials
Inflection Pt
1
2
f ( x) 2 x x
1
2
1
2
1
f '( x) 3x x
2
1
1
f '( x) 3x 2
2 x
1
2
= rate of change
f ( x)
f ( x) x 2 x 1
3
2
Theory
f(x) = axn
then f’x) = anxn-1
2
3 4 x5
f ( x)
1
5
4
2x
3
5
x4
5
f '( x) 2
4
3
6 x
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Differentiation
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Calculus
Revision
Differentiate
4 x 3x 7
2
8x 3
Back
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Calculus
Revision
Differentiate
x3 6 x 2 3x 9
x
3
Split up
Straight line form
Differentiate
Back
2
x 6 x 3x 9
x
x
x x
x 6x 3 9x
2
2x 6 9x
Quit
1
2
Next
Calculus
Revision
Differentiate
3
2
2 x 5x
3
2x
2
1
2
1
2
1
2
1
5x
2
5
2
3x x
Back
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3
2
3
2
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Calculus
Revision
Differentiate
2
3 x
1
x
1
2
Straight line form
Differentiate
Back
3x 2 x
3
x
2
1
2
1
2
1
1
2x
2
Quit
3
2
3
3 12
x x 2
2
Next
Calculus
Revision
Differentiate
Multiply out
Differentiate
Back
(3 x 5)( x 2)
3x 6 x 5 x 10
2
3x 2 x 10
6x 1
Quit
Next
Calculus
Revision
Differentiate
Straight line form
Differentiate
Back
800
3 r
r
2
3 r 800r
2
1
6 r (1)800r
Quit
2
Next
Calculus
Revision
Differentiate
x( x 2 x)
multiply out
x 2x
differentiate
3x 4 x
Back
2
3
2
2
Quit
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Calculus
Revision
Differentiate
1 1
3
x x
x 1 x 3
Straight line form
Differentiate
2
x (3) x
4
x 2 3 x 4
Back
Quit
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Calculus
Revision
Differentiate
x x x
2
Straight line form
multiply out
Differentiate
Back
x
1
2
5
2
x
x x
5
2
3
2
2
x
3
2
3
2
x x
1
2
Quit
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Calculus
Revision
Differentiate
multiply out
Simplify
Straight line form
Differentiate
Back
x 2 4 x
4 x x 8 2 x
6 x x 8
1
2
6x x 8
3x
1
2
1
Quit
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Calculus
Revision
Differentiate
Chain rule
Simplify
Back
( x 2)
5
5( x 2) 1
4
5( x 2)
Quit
4
Next
Calculus
Revision
Differentiate
(5 x 1)3
Chain Rule
3(5 x 1)2 5
Simplify
15(5 x 1)
Back
Quit
2
Next
Calculus
Revision
Differentiate
Chain Rule
Back
(5 x 3x 2)
2
4
4(5x2 3x 2)3 10 x 3
Quit
Next
Calculus
Revision
Differentiate
Chain Rule
Simplify
Back
(7 x 1)
5
2
5
(7 x 1)
2
3
2
7
35
(7 x 1)
2
Quit
3
2
Next
Calculus
Revision
Differentiate
Chain Rule
Simplify
Back
(2 x 5)
1
2
1
(2 x 5)
2
(2 x 5)
3
2
2
3
2
Quit
Next
Calculus
Revision
Differentiate
3x 1
3x 1
Straight line form
Chain Rule
Simplify
Back
1
2
3x 1
3
2
1
2
3x 1
Quit
1
2
3
1
2
Next
Calculus
Revision
Differentiate
x
Straight line form
Chain Rule
Simplify
Back
5
2
3
4
5( x 2 3)4
20( x 2 3)5 2 x
40 x( x 2 3)5
Quit
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Calculus
Revision
Differentiate
Straight line form
Chain Rule
Simplify
Back
1
2x 1
(2 x 1)
1
(2 x 1)
2
1
2
3
2
(2 x 1)
Quit
2
3
2
Next
Calculus
Revision
Differentiate
2
2 cos x sin x
3
2
2sin x cos x
3
Back
Quit
Next
Calculus
Revision
Differentiate
3cos x
3sin x
Back
Quit
Next
Calculus
Revision
Differentiate
cos 4x 2sin 2x
4sin 4x 4cos 2x
Back
Quit
Next
Calculus
Revision
cos 4 x
Differentiate
4 cos3 x sin x
Back
Quit
Next
Calculus
Revision
Differentiate
Straight line form
Chain Rule
Simplify
Back
sin x
(sin x)
1
(sin x)
2
1
2
1
2
cos x
1
cos x(sin x)
2
Quit
1
2
1
2
cos x
sin x
Next
Calculus
Revision
Differentiate
Chain Rule
Simplify
Back
sin(3 2 x)
cos(3 2 x) (2)
2 cos(3 2 x)
Quit
Next
Calculus
Revision
Differentiate
Straight line form
Chain Rule
Simplify
Back
cos 5 x
2
1
cos 5x
2
1
sin 5 x 5
2
5
sin 5x
2
Quit
Next
Calculus
Revision
Differentiate
f ( x) cos(2 x) 3sin(4 x)
f ( x) 2sin(2 x) 12 cos(4 x)
Back
Quit
Next
Calculus
Revision
Differentiate
43200
A( x) x
x
2
Straight line form
A( x) x 2 43200 x 1
Chain Rule
A( x) 2 x 43200 x 2
Simplify
Back
A( x) 2 x
Quit
43200
x2
Next
Calculus
Revision
Differentiate
2
f ( x) x 2
x
Straight line form
Differentiate
Back
1
2
f ( x) x 2 x2
f ( x) x
1
2
Quit
1
2
4 x 3
Next
Calculus
Revision
Differentiate
y 2 x3 7 x 2 4 x 4
y 6 x 2 14 x 4
Back
Quit
Next
Calculus
Revision
Differentiate
y 2 sin x
6
Chain Rule
dy
2 cos x 1
6
dx
Simplify
dy
2 cos x
6
dx
Back
Quit
Next
Calculus
Revision
Differentiate
A
3
a (8 a )
4
multiply out
3 2
A 6a a
4
Differentiate
3
A 6 a
2
Back
Quit
Next
Calculus
Revision
Differentiate
Chain Rule
Simplify
Back
1
3 2
f ( x) (8 x )
1
2
1
f ( x) (8 x3 ) (2 x 2 )
2
f ( x) x 2 (8 x3 )
Quit
1
2
Next
Calculus
Revision
Differentiate
16
y x
, x0
x
1
2
Straight line form
y x 16 x
Differentiate
3
dy
1 8x 2
dx
Back
Quit
Next
Calculus
Revision
Differentiate
Multiply out
3 3 2 16
A( x)
x
2
x
A( x)
3 3 2 3 3 16
x
2
2 x
3 3 2
x 24 3x 1
2
Straight line form
A( x)
Differentiate
A( x) 3 3x 24 3x 2
Back
Quit
Next
Calculus
Revision
Differentiate
f ( x) 5 x 4
Chain Rule
f ( x)
Simplify
f ( x)
Back
1
2
1
2
5x 4
5
2
5x 4
Quit
1
2
5
1
2
Next
Calculus
Revision
Differentiate
A( x) 240 x
16 2
x
3
32
A( x) 240
x
3
Back
Quit
Next
Calculus
Revision
Differentiate
f ( x) 3x 2 (2 x 1)
Multiply out
f ( x) 6 x 3 3 x 2
Differentiate
f ( x) 18 x 2 6 x
Back
Quit
Next
Calculus
Revision
Differentiate
Chain Rule
Simplify
Back
f ( x) cos2 x sin 2 x
f ( x) 2 cos x sin x 2 sin x cos x
f ( x) 4 cos x sin x
Quit
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