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Higher Unit 1Higher Unit 1
www.mathsrevision.comwww.mathsrevision.com
Finding the gradient for a polynomial
Differentiating Brackets ( Type 1 ) Differentiating Harder Terms (Type 2)Differentiating with Leibniz NotationEquation of a Tangent Line ( Type 3 )
Increasing / Decreasing functionsMax / Min and inflexion Points
Curve Sketching
Max & Min Values on closed IntervalsOptimization
Higher Outcome 3
Mind Map of Chapter
Using differentiation (Application)
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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).
B
S
Gradients & CurvesHigher Outcome 3
Demo
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A
Gradient of tangent = gradient of curve at A
B
Gradient of tangent = gradient of curve at B
Gradients & CurvesHigher Outcome 3
Demo
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Gradients & CurvesHigher Outcome 3
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, f(x))
((x+h), f(x+h))
True gradient
Approx gradient
To find the gradient at any point on a curve we need
to modify the gradient formula
2 1
2 1
--
y ym
x x
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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
smallerThis means the two points are closer together.
(x, f(x))
((x+h), f(x+h))
True gradient
Approx gradient
Gradients & CurvesHigher Outcome 3
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We can improve upon this approximation by making the value of h even smaller.
(x, f(x))
((x+h), f(x+h))True gradientApprox gradient
So the points are even closer together.
Gradients & CurvesHigher Outcome 3
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Higher Outcome 3
Derivative
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.
The process of finding the gradient is called
DIFFERENTIATING
or
FINDING THE DERIVATIVE (Gradient)
Differentiating
Finding the GRADIENT
Finding the rate of change
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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
DerivativeHigher Outcome 3
Have guessed the
rule yet !
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If f(x) = axn
n-1n
Rule for Differentiating
It can be given by this simple flow diagram ...
multiply by the power
reduce the power by 1
then f '(x) =
NB: the following terms & expressions mean the same
GRADIENT, DERIVATIVE, RATE OF CHANGE, f '(x)
Derivative
Higher Outcome 3
ax
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Rule for Differentiating
To be able to differentiate
it is VERY IMPORTANT that you are
comfortable using indices rules
Derivative
Higher Outcome 3
0
1
11
nn
mm n m n m n
n
nm n n mm
x xx
xx x x x
x
x x x
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Special Points
(I) f(x) = ax (Straight line function)
If f(x) = ax = ax1
then f '(x) = 1 X ax0
= a X 1 = a
Index Laws
x0 = 1
So if g(x) = 12x then g '(x) = 12
Also using y = mx + c
The line y = 12x has gradient 12,
and derivative = gradient !!
Higher Outcome 3
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If f(x) = a = a X 1 = ax0
then f '(x) = 0 X ax-1 = 0
Index Laws
x0 = 1
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 !
Special PointsHigher Outcome 3
Name :
3( ) 4f x x 2
2( )
3f x
x( ) 3f x x 3 2
1( )
2f x
x
3
2( )
3
xf x
x
( ) ( 2)(3 1)f x x x 2 3
( )x
f xx
41
( ) 5f x xx
Differentiation techniquesGradien
t
=Rate of change=
Differentiation
Differentiation
2'( ) 12f x x
1
2( ) 3f x x1
23
'( )2
f x x
3'( )
2f x
x
22( )
3
xf x
34
'( )3
xf x
3
4'( )
3f x
x
2
3
( )2
xf x
3 5
1'( )
3f x
x
3( ) 23
xf x x
41'( ) 6
3f x x
4
1 6'( )
3f x
x
1 1
2 4( ) 5f x x x
3 3
2 41 5
'( )2 4
f x x x
3 34
1 5'( )
2 4f x
x x
2( ) 3 5 2f x x x
'( ) 6 5f x x
2 3( )
xf x
x x
1 1
2 2( ) 2 3f x x x
3
1 3'( )
2f x
x x
Calculus Revision
Differentiate 24 3 7x x
8 3x
Calculus Revision
Differentiate 3 22 7 4 4y x x x
26 14 4y x x
Calculus Revision
Differentiate 216( ) 240
3A x x x
32( ) 240
3A x x
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Example 1 A curve has equation f(x) = 3x4
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 = 2
Its gradient is f '(x) = 6x
At the point where x = -4 the gradient isf '(-4) = 6 X -4 =-24
DerivativeHigher Outcome 3
Find the formula for its gradient and find the gradient when x = -4
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g '(x) = 20x3 - 20x4 g '(2) = 20 X 23 - 20 X 24
= 160 - 320
= -160
DerivativeHigher Outcome 3
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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
DerivativeHigher Outcome 3
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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 f '(x) = 2
ie 3x2 - 6x + 2 = 2
or 3x2 - 6x = 0
ie 3x(x - 2) = 0
ie 3x = 0 or x - 2 = 0
so 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)
DerivativeHigher Outcome 3
Calculus Revision
Differentiate
3 1
2 22 5x x
1 3
2 23 12 2
2 5x x
1 3
2 25
23x x
Calculus Revision
Differentiate3
1 1
x x
1 3x x Straight line form
Differentiate2 4( 3)x x
2 43x x
Calculus Revision
Differentiate 2 8003 r
r
2 13 800r r Straight line form
Differentiate26 ( 1)800r r
Calculus Revision
Differentiate 2 43200( )A x x
x
Chain Rule
Simplify
Straight line form2 1( ) 43200A x x x
2( ) 2 43200A x x x
2
43200( ) 2A x x
x
Calculus Revision
Differentiate2
3 1xx
1 1
2 23 2 1x x
Straight line form
Differentiate
1 3
2 21
2
32
2x x
1 3
2 23
2x x
Calculus Revision
Differentiate 2
2( )f x x
x
Differentiate
Straight line form1
22( ) 2f x x x
1321
2( ) 4f x x x
Calculus Revision
Differentiate16
, 0y x xx
Differentiate
Straight line form1
216y x x
3
21 8dy
xdx
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Brackets
Basic Rule: Break brackets before you differentiate !
Example h(x) = 2x(x + 3)(x -3)
= 2x(x2 - 9)
= 2x3 - 18x
So h'(x) = 6x2 -18
Higher Outcome 3
Calculus Revision
Differentiate (3 5)( 2)x x
23 6 5 10x x x Multiply out
Differentiate 6 1x
23 10x x
Calculus Revision
Differentiate2( 2 )x x x
multiply out3 22x x
differentiate23 4x x
Calculus Revision
Differentiate 2x x x
1
22x x xStraight line form
multiply out
Differentiate
5 3
2 2x x3 1
2 25 3
2 2x x
Calculus Revision
Differentiate3
(8 )4
A a a
multiply out
Differentiate
236
4A a a
36
2A a
Calculus Revision
Differentiate 2 4x x
multiply out
Simplify
4 8 2x x x
Differentiate
6 8x x
Straight line form1
26 8x x 1
23 1x
Calculus Revision
Differentiate 23 3 16( )
2A x x
x
Straight line form
Multiply out 23 3 3 3 16( )
2 2A x x
x
Differentiate
2 13 3( ) 24 3
2A x x x
2( ) 3 3 24 3A x x x
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Recall 17 7 7
Fractions
Reversing the above we get the following “rule” !
This can be used as follows …..
a + bc
a bc c
Higher Outcome 3
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Example
f(x) = 3x3 - x + 2 x2
= 3x - x-1 + 2x-2
f '(x) = 3 + x-2 - 4x-3
Fractions
= 3x3 - x + 2 x2 x2 x2
= 3 + 1 - 4 x2 x3
Higher Outcome 3
Calculus Revision
Differentiate
3 26 3 9x x x
x
3 26 3 9x x x
x x x x
2 16 3 9x x x
Split up
Straight line form
22 6 9x x Differentiate
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If y is expressed in terms of x then the derivative is written as dy/dx .
Leibniz Notation
Leibniz Notation is an alternative way of expressing derivatives to f'(x) , g'(x) , etc.
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 = 18R + 45 R4
Q = 9R2 - 15 R3
Higher Outcome 3
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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
Leibniz NotationHigher Outcome 3
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Newton’s 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 Newton’s 1st Law of Motion
Real Life ExamplePhysics
Higher Outcome 3
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y = mx +c
y = f(x)
Equation of Tangents
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)
Higher Outcome 3
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Higher Outcome 3
Equation of Tangents
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
= 3 - 2 = 1 m = 1
Straight line so we need a point plus the gradient then we can use the formula y - b = m(x -
a) .
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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
Higher Outcome 3
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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 x = -2 then y = 4
(-2)2 = 4/4 =
1
point is (-2, 1)
Gradient: y = 4x-2 so dy/dx = -8x-3 = -8 x3
when x = -2 then dy/dx = -8 (-2)3
= -8/-8 = 1m = 1
Equation of TangentsHigher Outcome 3
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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 TangentsHigher Outcome 3
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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 = 2x - 6
so 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)
Equation of TangentsHigher Outcome 3
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Increasing & Decreasing Functions and Stationary Points
Consider the following graph of y = f(x) …..
X
y = f(x)
a b c d e f+
+
+
++
-
-
0
0
0
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 negativeand f (x) < 0 when b < x < d .
The function is stationary if the gradient is zeroand 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
Higher Outcome 3
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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
f(x) decreasing when f (x) < 0 so 8x - 24 < 0 8x < 24
x < 3
f(x) increasing when f (x) > 0 so 8x - 24 > 0
8x > 24x > 3
Check: f (2) = 8 X 2 – 24 = -8
Check: f (4) = 8 X 4 – 24 = 8
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
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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 > 0 then the function is increasing.
Increasing & Decreasing Functions and Stationary Points
y = 6x - 5 x2
= 6 + 10 x3
Higher Outcome 3
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Example 26Show 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
Since (x - 3)2 0 for all values of x
then g (x) can never be negative
so the function is never decreasing.
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.
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
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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
Higher Outcome 3
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determine when f (x) is positive & negative.
x -3 2
f’(x) +
Function increasing when f (x) > 0
ie x < -3 or x > 2
Function decreasing when f (x) < 0
ie -3 < x < 2
Increasing & Decreasing Functions and Stationary Points
Higher Outcome 3
-0 +0
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Stationary Points and Their Nature
Consider this graph of y = f(x) again
X
y = f(x)
a b c+
+
+
+
+-
-
0
0
0
Higher Outcome 3
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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
Higher Outcome 3
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Maximum Turning point
x af(x) + 0 -
Minimum Turning Point
x bf(x) - 0
+
Stationary Points and Their Nature
Higher Outcome 3
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of inflexion
x c
f(x) + 0 +
Other possible type of inflexion
x d
f(x) - 0 -
Stationary Points and Their Nature
Higher Outcome 3
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Example 28Find the co-ordinates of the stationary point on the curve y = 4x3 + 1 and determine its nature.
SP occurs when dy/dx = 0so 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
Higher Outcome 3
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x 0
dy/dx +
So (0,1) is a rising point of inflexion.
Stationary Points and Their Nature
dy/dx = 12x2
Higher Outcome 3
+
0
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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
Higher Outcome 3
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Nature Table
x 0 4
dy/dx - 0 - 0 +
So (0,24) is a Point of inflexion
and (4,-232) is a minimum Turning Point
Stationary Points and Their Nature
dy/dx=12x3 - 48x2
Higher Outcome 3
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Example 30Find 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
So 2x3 - 8x = 0
2x(x2 - 4) = 0
2x(x + 2)(x - 2) = 0
x = 0 or x = -2 or x = 2
Using y = 1/2x4 - 4x2 + 2if x = 0 then y = 2
if x = -2 then y = -6
SP’s at(-2,-6), (0,2) & (2,-6)
if x = 2 then y = -6
Stationary Points and Their Nature
Higher Outcome 3
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Nature Table
x 0
dy/dx
-2 2
- 0 + 0 - 0 +
So (-2,-6) and (2,-6) are Minimum Turning Points
and (0,2) is a Maximum Turning Points
Stationary Points and Their Nature
Higher Outcome 3
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Curve Sketching
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.
Higher Outcome 3
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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
As x + then y -
As x - then y
+
Graph roughly
Curve SketchingHigher Outcome 3
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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
( -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)
Curve SketchingHigher Outcome 3
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(b) Stationary Points occur where dy/dx = 0
so -6x + 12 = 0
6x = 12
x = 2
If x = 2
then y = -12 + 24 + 15 = 27
Nature Table
x 2
dy/dx + 0 -So (2,27)
is a Maximum Turning Point
Stationary Point is (2,27)
Curve SketchingHigher Outcome 3
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using y = -3x2
as x + then y -
as x - then y -
Sketching
X
Y
y = -3x2 + 12x + 15
Curve Sketching
Cuts x-axis at -1 and 5
Summarising
Cuts y-axis at 15-1 5
Max TP (2,27)(2,27)
15
Higher Outcome 3
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Example 32Sketch the graph of y = -2x2 (x - 4)
(a) Axes If x = 0 then y = 0 X (-4) = 0If y = 0 then -2x2 (x - 4) = 0
x = 0 or x = 4
Graph cuts axes at (0,0) and (4,0) .
-2x2 = 0 or (x - 4) = 0
(b) SPs
y = -2x2 (x - 4) = -2x3 + 8x2
SPs occur where dy/dx = 0
so -6x2 + 16x = 0
Curve SketchingHigher Outcome 3
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-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
naturex 0 8/3
dy/dx -
Curve SketchingHigher Outcome 3
-
+
0
0
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(c) Large valuesusing y = -2x3 as x + then y -
as x - then y +Sketch
Xy = -2x2 (x – 4)
Curve Sketching
Cuts x – axis at 0 and 40 4
Max TP’s at (8/3, 512/27) (8/3, 512/27)
Y
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Example 33
Sketch the graph of y = 8 + 2x2 - x4
(a) Axes If x = 0 then y = 8 (0,8)If y = 0 then 8 + 2x2 - x4 = 0
Graph cuts axes at (0,8) , (-2,0) and (2,0)
Let u = x2 so u2 = x4
Equation is now 8 + 2u - u2 = 0
(4 - u)(2 + u) = 0
(4 - x2)(2 + x2) = 0
or (2 + x) (2 - x)(2 + x2) = 0So x = -2 or x = 2 but x2 -2
Curve SketchingHigher Outcome 3
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(b) SPs SPs occur where dy/dx = 0
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)
Curve SketchingHigher Outcome 3
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x -1 0 1
dy/dx +
So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .
Curve SketchingHigher Outcome 3
+- -0 00
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Cuts y – axis at 8Cuts x – axis at -2 and 2
(c) Large valuesUsing y = -
x4
as x + then y -as x - then y -
Sketch is
X
Y
-2 28
(-1,9) (1,9)
y = 8 + 2x2 - x4
Max TP’s at (-1,9) (1,9)
Curve Sketching
Summarising
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Max & Min on Closed Intervals
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.
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y =f(x)
Xa b
(a, f(a))
(b, f(b)) max = f(b) end point
min = f(a) end point
Max & Min on Closed Intervals
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x
y =f(x)(b, f(b))
(a, f(a))
max = f(c ) max TP
min = f(a) end point
a b
(c, f(c))
c NB: a < c < b
Max & Min on Closed Intervals
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y =f(x)
xa b
c
(a, f(a))
(b, f(b))
(c, f(c))
max = f(b) end point
min = f(c) min TP
NB: a < c < b
Max & Min on Closed Intervals
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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
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Stationary pointsdy/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
in interval not in interval
If x = 0 then y = 0 - 0 = 0
Hence for -1 x 2 , max = 0 & min = -20
Max & Min on Closed Intervals
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Using function notation we can say that
Domain = {xR: -1 x 2 }
Range = {yR: -20 y 0 }
Max & Min on Closed Intervals
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Optimization
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.
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OptimizationHigher Outcome 3
Problem
Practical exercise on optimizing volume.
Graph
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Example 35Higher Outcome 3
OptimizationQ. What is the maximum
volume
We can have for the given dimensions
A rectangular sheet of foil measuring 16cm X 10 cm has four small squares each x cm cut from each
corner. 16cm
10cm
x cm
NB: x > 0 but 2x < 10 or x < 5ie 0 < x < 5
This gives us a particular interval to consider !
x cm
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(16 - 2x) cm
(10 - 2x) cmx 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
By folding up the four flaps we get a small cuboid
Higher Outcome 3
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5End 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)
OptimizationHigher Outcome 3
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ie 4(3x - 20)(x - 2) = 0
3x - 20 = 0 or x - 2 = 0
ie x = 20/3 or x = 2
not in intervalin interval
When x = 2 then
V(2) = 2 X 12 X 6 = 144
We now check gradient near x = 2
OptimizationHigher Outcome 3
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x 2
V '(x) +
Hence max TP when x = 2
So max possible volume = 144cm3
Nature
OptimizationHigher Outcome 3
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Example 36
When a company launches a new product its share of the market after x months is calculated by the formula
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.
(x 2)2
2 4( )S x
x x
OptimizationHigher Outcome 3
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End points S(2) = 1 – 1 = 0
There is no upper limit but as x S(x) 0.
SPs occur where S (x) = 0
3 2
8 2'( ) 0S x
x x
1 22
2 4( ) 2 4 S x x x
x x
1 2 2 32 3
2 8'( ) 2 4 2 8S x x x x x
x x
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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
3 2
8 2'( ) 0S x
x xrearrange
OptimizationHigher Outcome 3
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x 4
S (x)
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
Optimization
Nature
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+ -0
Differentiationof Polynomials
f(x) = axn
then f’x) = anxn-1
Derivative
= gradient
= rate of change
Graphs
f’(x)=0
54
2( )
3f x
x
5
42( )
3
xf x
1
4
4
552'( )
3 6
xf x
x
1
2( ) 2 1f x x x 3 1
2 2( ) 2f x x x 1 1
2 21
'( ) 32
f x x x
1
21
'( ) 32
f x xx
f’(x)=0
Stationary Pts
Max. / Mini Pts
Inflection Pt
Nature Table-1 2 5+ 0 -
xf’(x)
MaxGradient at a point
Equation of tangent line
Straight Line
Theory
Leibniz Notation
'( )dy
f xdx
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Are you on Target !
• Update you log book
• Make sure you complete and correct
ALL of the Differentiation 1
questions in the past paper booklet.
Outcome 3
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