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

also be written as ( ) < 0 for the interval

Now, a rough sketch of can be drawn:

-5 0

(- 4 ; 8)

(0 ; 1)

𝑓

MATHEMATICS GRADE 12 DIFFERENTIAL CALCULUS PART 2

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ACTIVITY 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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