1. Using differentiation (Application)
Outcome 3
Using differentiation (Application)
Finding the gradient for a polynomial
Increasing / Decreasing functions
Differentiating Brackets ( Type 1 )
Max / Min and inflexion Points
Differentiating Harder Terms (Type 2)
Curve Sketching
Differentiating with Leibniz Notation
Max & Min Values on closed Intervals
Equation of a Tangent Line ( Type 3 )
Optimization
Mind Map of Chapter

2. Gradients & Curves Demo Outcome 3
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. Gradients & Curves A B Outcome 3
Higher
Gradient of tangent = gradient of curve at A A B
Gradient of tangent = gradient of curve at B

4. To find the gradient at any point on a curve we need to modify the gradient formula
Gradients & Curves
Outcome 3
Higher
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

5. Gradients & Curves Outcome 3 ((x+h), f(x+h)) Approx gradient (x, f(x))
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.
((x+h), f(x+h))
Approx gradient
(x, f(x))
True gradient

6. So the points are even closer together.
Gradients & Curves
Outcome 3
Higher
We can improve upon this approximation by making the value of h even smaller.
So the points are even closer together.
((x+h), f(x+h))
Approx gradient
True gradient
(x, f(x))

7. Derivative
Outcome 3
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.
Finding the GRADIENT
Differentiating
The process of finding the gradient is called
Finding the rate of change
DIFFERENTIATING or
FINDING THE DERIVATIVE (Gradient)

8. Derivative
Outcome 3
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
Have guessed the rule yet !
5x10 50x9
6x x6
x x2
x x4
x x98

9. Derivative If f(x) = axn n n -1 ax then f '(x) =
Rule for Differentiating
Outcome 3
Higher
It can be given by this simple flow diagram ...
multiply by the power
reduce the power by 1
If f(x) = axn n n -1 ax
then f '(x) =
NB: the following terms & expressions mean the same
GRADIENT,
DERIVATIVE,
RATE OF CHANGE,
f '(x)

10. Derivative Rule for Differentiating Outcome 3
Higher
To be able to differentiate
it is VERY IMPORTANT that you are
comfortable using indices rules

11. Special Points Outcome 3 (I) f(x) = ax (Straight line function)
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 !!

12. Special Points Outcome 3 (II) f(x) = a, (Horizontal Line) Index Laws
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 !

13. Differentiation techniques=
Gradient
Differentiation =
Rate of change
Name :

14. Calculus Revision
Differentiate

15. Calculus Revision
Differentiate

16. Calculus Revision
Differentiate

17. Derivative Outcome 3 Example 1 A curve has equation f(x) = 3x4
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 = the gradient is
f '(-4) = 6 X -4 =
-24

18. If g(x) = 5x4 - 4x5 then find g '(2) .
Derivative
Outcome 3
Higher
Example 3
If g(x) = 5x4 - 4x5 then find g '(2) .
g '(x) = 20x3 - 20x4
g '(2) = 20 X X 24 =
= -160

19. Derivative Outcome 3 Example 4 h(x) = 5x2 - 3x + 19
Higher
Example 4
h(x) = 5x2 - 3x + 19
so h '(x) = 10x - 3
and h '(-4) = 10 X (-4) - 3
= = -43
Example 5
k(x) = 5x4 - 2x3 + 19x - 8, find k '(10) .
k '(x) = 20x3 - 6x
So k '(10) = 20 X X =

20. Derivative Outcome 3 Example 6 : Find the points on the curve
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 x2 - 6x + 2 = 2
f(0) = 7
or x2 - 6x = 0
ie x(x - 2) = 0
f(2) =
ie x = 0 or x - 2 = 0
= 7
so x = 0 or x = 2
Points are (0,7) & (2,7)

21. Calculus Revision
Differentiate

22. Calculus Revision
Differentiate
Straight line form
Differentiate

23. Calculus Revision
Differentiate
Straight line form
Differentiate

24. Calculus Revision
Differentiate
Straight line form
Chain Rule
Simplify

25. Calculus Revision
Differentiate
Straight line form
Differentiate

26. Calculus Revision
Differentiate
Straight line form
Differentiate

27. Calculus Revision
Differentiate
Straight line form
Differentiate

28. Basic Rule: Break brackets before you differentiate !
Outcome 3
Higher
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

29. Calculus Revision
Differentiate
Multiply out
Differentiate

30. Calculus Revision
Differentiate
multiply out
differentiate

31. Calculus Revision Differentiate Straight line form multiply out

32. Calculus Revision
Differentiate
multiply out
Differentiate

33. Calculus Revision Differentiate multiply out Simplify
Straight line form
Differentiate

34. Calculus Revision Differentiate Multiply out Straight line form

35. Fractions Outcome 3 Reversing the above we get the following “rule” !
Higher
Reversing the above we get the following “rule” !
This can be used as follows …..

36. Fractions f(x) = 3x3 - x + 2 x2 = 3x3 - x + 2 x2 x2 x2
Outcome 3
Higher
Example
f(x) = 3x3 - x + 2 x2
= x3 - x x x x2
= 3x - x x-2
f '(x) = x x-3
= x2 x3

37. Calculus Revision Differentiate Split up Straight line form

38. Leibniz Notation Outcome 3
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
Q = 9R R3
Find dQ/dR
NB: Q = 9R2 - 15R-3
So dQ/dR = 18R + 45R-4
= 18R R4

39. Leibniz Notation = 60 - (-16) = 76 Outcome 3 Example 20
Higher
Example 20
A curve has equation y = 5x3 - 4x
Find the gradient where x = ( differentiate ! )
gradient = dy/dx = 15x2 - 8x
if x = -2 then
gradient = 15 X (-2)2 - 8 X (-2)
= (-16) = 76

40. Real Life Example Physics Outcome 3 Newton’s 2ndLaw of Motion
Higher
Newton’s 2ndLaw of Motion
s = ut + 1/2at 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

41. Equation of Tangents y = mx +c Outcome 3 y = f(x) A(a,b) tangent
Higher
y = f(x)
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)

42. Straight line so we need a point plus the gradient then we can use the formula y - b = m(x - a) .
Equation of Tangents
Outcome 3
Higher
Example 21
Find the equation of the tangent line to the curve
y = x3 - 2x at the point where x = -1.
Point: if x = -1 then y = (-1)3 - (2 X -1) + 1
= (-2) + 1
= 2 point is (-1,2)
Gradient: dy/dx = 3x2 - 2
when x = -1 dy/dx = 3 X (-1)2 - 2
= = 1
m = 1

43. Equation of Tangents Outcome 3 Now using y - b = m(x - a)
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

44. Equation of Tangents m = 1 Outcome 3 Example 22
Higher
Example 22
Find the equation of the tangent to the curve y = x2 at the point where x = (x  0)
Also find where the tangent cuts the X-axis and Y-axis.
Point: when x = then y = (-2)2
= 4/4 = 1
point is (-2, 1)
Gradient: y = 4x so dy/dx = -8x-3
= x3
when x = -2 then dy/dx = (-2)3
= -8/-8 = 1
m = 1

45. Equation of Tangents Outcome 3 Now using y - b = m(x - a)
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 = = 3
at point (0, 3)
Tangent cuts X-axis when y = 0
so 0 = x + 3 or x = -3
at point (-3, 0)

46. Equation of Tangents Outcome 3 Example 23 - (other way round)
Higher
Example (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 =
= 45
Point is (10,45)

47. Increasing & Decreasing Functions and Stationary Points
Outcome 3
Higher
Consider the following graph of y = f(x) …..
y = f(x) + + - + + a b c - d e f X +

48. Increasing & Decreasing Functions and Stationary Points
Outcome 3
Higher
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.

49. Increasing & Decreasing Functions and Stationary Points
Outcome 3
Higher
Example 24
For the function f(x) = 4x2 - 24x 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
Check: f  (2) = 8 X 2 – 24 = -8
x < 3
f(x) increasing when f  (x) > 0
so 8x - 24 > 0
8x > 24
Check: f  (4) = 8 X 4 – 24 = 8
x > 3

50. Increasing & Decreasing Functions and Stationary Points
Outcome 3
Higher
Example 25
For the curve y = 6x – 5/x2
Determine if it is increasing or decreasing when x = 10.
y = 6x x2
= 6x - 5x-2
so dy/dx = x-3
= x3
when x = dy/dx = /1000
= 6.01
Since dy/dx > 0 then the function is increasing.

51. Increasing & Decreasing Functions and Stationary Points
Outcome 3
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.

52. Increasing & Decreasing Functions and Stationary Points
Outcome 3
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

53. Increasing & Decreasing Functions and Stationary Points
Outcome 3
Higher
We now use a special table of factors to 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 < x < 2

54. Stationary Points and Their Nature
Outcome 3
Higher
y = f(x)
Consider this graph of y = f(x) again + - + + + a - b c X +

55. Stationary Points and Their Nature
Outcome 3
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.

56. Stationary Points and Their Nature
Outcome 3
Higher
Maximum Turning point
Minimum Turning Point x a x b
f(x)
f(x)

57. Stationary Points and Their Nature
Outcome 3
Higher
Rising Point of inflexion
Other possible type of inflexion x c x d
f(x)
f(x)

58. Stationary Points and Their Nature
Outcome 3
Higher
Example 28
Find the co-ordinates of the stationary point on the curve y = 4x 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

59. + + Stationary Points and Their Nature Outcome 3 Nature Table x dy/dx
Higher
Nature Table x + +
dy/dx
dy/dx = 12x2
So (0,1) is a rising point of inflexion.

60. Stationary Points and Their Nature
Outcome 3
Higher
Example 29
Find the co-ordinates of the stationary points on the curve y = 3x4 - 16x and determine their nature.
SP occurs when dy/dx = 0
Using y = 3x4 - 16x3 + 24
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)

61. Stationary Points and Their Nature
Outcome 3
Higher
Nature Table x 4
dy/dx
dy/dx=12x3 - 48x2
So (0,24) is a Point of inflexion
and (4,-232) is a minimum Turning Point

62. Stationary Points and Their Nature
Outcome 3
Higher
Example 30
Find the co-ordinates of the stationary points on the curve y = 1/2x4 - 4x and determine their nature.
SP occurs when dy/dx = 0
Using y = 1/2x4 - 4x2 + 2
if x = 0 then y = 2
So x3 - 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
SP’s at(-2,-6), (0,2) & (2,-6)

63. Stationary Points and Their Nature
Outcome 3
Higher
Nature Table x -2 2
dy/dx
So (-2,-6) and (2,-6) are Minimum Turning Points
and (0,2) is a Maximum Turning Points

64. Curve Sketching Outcome 3
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.

65. Suppose that f(x) = -2x3 + 6x2 + 56x - 99
Curve Sketching
Outcome 3
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
As x  + then y  -
Graph roughly
As x  - then y  +

66. Curve Sketching Outcome 3 Example 31
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 = 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)

67. is a Maximum Turning Point
Curve Sketching
Outcome 3
Higher
(b) Stationary Points
occur where dy/dx = 0
so -6x = 0
If x = 2
then y = = 27
6x = 12
x = 2
Stationary Point is (2,27)
Nature Table x 2
dy/dx
So (2,27)
is a Maximum Turning Point

68. Curve Sketching Summarising Outcome 3 (c) Large values
Higher
Summarising
(c) Large values
as x  + then y  -
using y = -3x2
as x  - then y  - Y
Sketching
Cuts x-axis at -1 and 5 -1 5
Cuts y-axis at 15 15
Max TP (2,27)
(2,27) X
y = -3x2 + 12x + 15

69. Graph cuts axes at (0,0) and (4,0) .
Curve Sketching
Outcome 3
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 x2 (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

70. Curve Sketching Outcome 3 -2x(3x - 8) = 0 -2x = 0 or (3x - 8) = 0
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
8/3 - + -
dy/dx

71. Curve Sketching Summarising Outcome 3 (c) Large values
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 4
Max TP’s at (8/3, 512/27)
(8/3, 512/27) X
y = -2x2 (x – 4)

72. Curve Sketching Outcome 3 Example 33
Higher
Example 33
Sketch the graph of y = 8 + 2x2 - x4
(a) Axes
If x = 0 then y = 8 (0,8)
If y = 0 then x2 - x4 = 0
Let u = x2 so u2 = x4
Equation is now u - 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)

73. Curve Sketching Outcome 3 (b) SPs SPs occur where dy/dx = 0
Higher
(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 = = 9 (-1,9)
when x = 1 then y = = 9 (1,9)

74. So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .
Curve Sketching
Outcome 3
Higher
nature x -1 1 + - + -
dy/dx
So (0,8) is a min TP while (-1,9) & (1,9) are max TPs .

75. Curve Sketching Summarising Outcome 3 (c) Large values Using y = - x4
Higher
Summarising
(c) Large values
Using y = - x4
Sketch is
as x  + then y  - Y
as x  - then y  -
Cuts x – axis at -2 and 2 -2 2
Cuts y – axis at 8 8
Max TP’s at
(-1,9)
(-1,9)
(1,9)
(1,9) X
y = 8 + 2x2 - x4

76. Max & Min on Closed Intervals
Outcome 3
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.

77. Max & Min on Closed Intervals
Outcome 3
Higher
y =f(x)
(b, f(b))
max = f(b) end point
(a, f(a))
min = f(a) end point X
a b

78. Max & Min on Closed Intervals
Outcome 3
Higher
(c, f(c))
max = f(c ) max TP
y =f(x)
(b, f(b))
(a, f(a))
min = f(a) end point x
a b c
NB: a < c < b

79. Max & Min on Closed Intervals
Outcome 3
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 < c < b
a b c

80. Max & Min on Closed Intervals
Outcome 3
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 = = -11
If x = 2 then y = = -20

81. Max & Min on Closed Intervals
Outcome 3
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
Hence for -1  x  2 , max = 0 & min = -20

82. Extra bit Max & Min on Closed Intervals
Outcome 3
Higher
Extra bit
Using function notation we can say that
Domain = {xR: -1  x  2 }
Range = {yR: -20  y  0 }

83. Note: Optimum basically means the best possible.
Optimization
Outcome 3
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.

84. Optimization Q. What is the maximum volume
We can have for the given dimensions
Optimization
Outcome 3
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 > 0 but 2x < 10 or x < 5
ie 0 < x < 5
This gives us a particular interval to consider !

85. Optimization Outcome 3 x cm (10 - 2x) cm (16 - 2x) cm
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( x + 4x2)
= 4x3 - 52x2 +160x
We now try to maximize V(x) between 0 and 5

86. Considering the interval 0 < x < 5
Optimization
Outcome 3
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) = 12x x + 160
= 4(3x2 - 26x + 40)
= 4(3x - 20)(x - 2)

87. We now check gradient near x = 2
Optimization
Outcome 3
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

88. So max possible volume = 144cm3
Optimization
Outcome 3
Higher
Nature x 2 + -
V '(x)
Hence max TP when x = 2
So max possible volume = 144cm3

89. Optimization S(5) = 2/5 – 4/25 = 6/25 Outcome 3 Example 36
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.

90. There is no upper limit but as x   S(x)  0.
Optimization
Outcome 3
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

91. We now check the gradients either side of 4
Optimization
Outcome 3
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

92. And max share of market = S(4)
Optimization
Outcome 3
Higher
Nature
S (3.9 ) = …
x  
S (4.1) = … + -
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

93. Equation of tangent line
Nature Table
Equation of tangent line
Leibniz Notation -1 2 5 + - x
f’(x)
Max
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 f’x) = anxn-1