MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2 1 D I F F E R E N T I A L C A L C U L U S THE GRAPH OF THE CUBIC FUNCTION Turning Points (also called ‘Stationary Points’ or ‘Critical Points’) When we determine () we are dealing with the gradient of which can be increasing, decreasing or equal to zero. Minimum turning points Maximum turning points ()= 3 + 2 + + + + + + + + + + + + + + + - - - - - - - m = positive ∴ is increasing m = 0 ∴Turning Point m = positive ∴ is increasing m =negative ∴ decreasing m = 0 ∴Turning Point ()= + + + ()=0 ()<0 ()>0 At a minimum turning point the sign of the gradient changes from negative to positive. ()<0 ()>0 ()=0 At a maximum turning point the sign of the gradient changes from positive to negative.
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MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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D I F F E R E N T I A L C A L C U L U S
THE GRAPH OF THE CUBIC FUNCTION
Turning Points (also called ‘Stationary Points’ or ‘Critical Points’)
When we determine ( ) we are dealing with the gradient of which can be
increasing, decreasing or equal to zero.
Minimum turning points
Maximum turning points
𝑓(𝑥) = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑
+ +
+ +
+ +
+ + +
+ +
+ +
- - -
- - - -
m = positive
∴ 𝒇 is increasing
m = 0 ∴Turning Point
m = positive
∴ 𝒇 is increasing
m =negative
∴ 𝒇 decreasing
m = 0 ∴Turning Point
𝒇(𝒙) = 𝒂𝒙𝟑 + 𝒃𝒙𝟐 + 𝒄𝒙 + 𝒅
𝑓 (𝑥) = 0
𝑓 (𝑥) < 0 𝑓 (𝑥) > 0 At a minimum turning point the sign
of the gradient changes from negative
to positive.
𝑓 (𝑥) < 0 𝑓 (𝑥) > 0
𝑓 (𝑥) = 0
At a maximum turning point the sign
of the gradient changes from positive
to negative.
MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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TO SKETCH THE CUBIC FUNCTION
To sketch the graph of ( ) = 3 + 2 + + , first determine the following:
1. SHAPE
If > 0 (positive), then
If < 0 (negative), then
2. TURNING POINTS
The x – coordinates: AND The y – coordinates:
Let ( ) = 0 Calculate ( ) and ( )
2 + + = 0
∴ = =
The turning points are ( ( )) and ( ( )).
3. THE x – INTERCEPT
*Let = 0, then factorize if possible (e.g. take out a common factor).
Otherwise….
* Let = 0
* “Guess” the first factor, e.g.
If f (2) = 0 then (x – 2) is a factor of f
If f (–3) = 0 then (x + 3) is a factor of f
* long division / synthetic division
* factorize to get ( )( )( ) = 0
* solve for to get the – intercepts
MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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4. THE y – INTERCEPT (d – value)
Let = 0, then calculate y.
5. Point of INFLECTION
The x – coordinate: The y – coordinate:
*Let ( ) = 0 *Substitute the x – value into ( )
*Solve for x *Simplify
OR
* Determine x = ( ) ( )
2
CONCAVITY & THE POINT OF INFLECTION
MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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EXAMPLES:
MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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ACTIVITY 1
MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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FINDING THE EQUATION OF A CUBIC CURVE
The graphs of ( ) vs the graph of ( )
MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2
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MAKE SURE YOU UNDERSTAND…..
Zeros at –5
‘Zeros’ = x – intercepts. In this case only one, at =
Can also be given as ( ) = 0 or ( 0)
Stationary points at x = – 4 and x = 0
‘Stationary points’ = Turning points
Can also be given as ( ) = 0 (0) = 0
f (–4) = 8 and f (0) = 1
tell us that for the function ……
= at = and
= at = 0
Can also be given as ( ) (0 )
Increasing on the interval (– ; – 4) and (0 ; )
’increasing’ means that the gradient will be positive for these intervals
Can also be written as ( ) > 0 for these intervals
Decreasing on the interval (– 4 ; 0) ’decreasing’ means that the gradient will be negative for the interval Can