Block 1

Stationary Points

What is to be learned?

• How to use differentiation to identify SPs• What SPs are!!!!!!• What the difference between a stationary

point and a stationary value• What SVs are• How to find SP/SVs• How to find the nature of SP/SVs

m ~ ½

m ~ 3

m ~ ½

m ~ 3

m ~ ½

m ~ ½

m ~ 3

m ~ ½

m ~ ½

m ~ 3

m ~ ½

m = 0

m ~ ½

m ~ 3

m ~ ½

m = 0

At Turning Points the gradient = 0

Turning points are known as stationary points (or values)(SPs or SVs)

So for SPs the derivative = 0

Ex y = x2 – 8x + 10 SPs?For SPs dy/dx = 0

Ex y = x2 – 8x + 10 SPs?

For SPs dy/dx = 0

Ex y = x2 – 8x + 10 SPs?

For SPs dy/dx = 0

dy/dx = 2x – 8

2x – 8 = 02x = 8 x = 4 y?

Ex y = x2 – 8x + 10 SPs?

For SPs dy/dx = 0

dy/dx = 2x – 8

2x – 8 = 02x = 8 x = 4 y?

y = 42 – 8(4) + 10 = - 6

SP is (4 , -6)

at x = 4

Stationary Points

• Max TPs, Min TPs or…….• For SPs gradient = 0• For SPs dy/dx = 0

TacticsFind derivativeSolve equation → xy → sub x into original equation

= 0

Ex y = x2 + 4x – 11 SPs?

For SPs dy/dx = 0

dy/dx = 2x + 4

2x + 4 = 02x = -4 x = -2 y?

y = (-2)2 + 4(-2) – 11 = -15

SP is (-2 , -15)

at x = -2

Ex y = 20x – 2x2 SP?

For SPs dy/dx = 0

dy/dx = 20 – 4x

20 – 4x = 0 x = 5 y?

y = 20(5) – 2(5)2 = 50

SP is (5 , 50)

at x = 5

Key Question

Ex y = 2x3 – 6x2 + 10 SPs?

For SPs dy/dx = 0

dy/dx = 6x2 – 12x

6x2 – 12x = 06x(x – 2) = 06x = 0 or x – 2 = 0 x = 0 or x = 2

Quadratic Equation

Factorise

x = 0 y = 10x = 2 y =2(2)3 – 6(2)2 + 10

= 2

(0 , 10) and (2 , 2)

Two SPs!!! → Need two y values

Ex y = 1/3x3 – 2x2 – 12x SPs?

For SPs dy/dx = 0

dy/dx = x2 – 4x – 12

x2 – 4x – 12 = 0(x – 6)(x + 2) = 0x–6= 0 or x+2 = 0 x = 6 or x = -2

Quadratic EquationFactorise

More Than One SP

Two SPs!!!→ Need two y values

Ex y = 1/3x3 – 2x2 – 12x SPs?

For SPs dy/dx = 0

dy/dx = x2 – 4x – 12

x2 – 4x – 12 = 0(x – 6)(x + 2) = 0x–6= 0 or x+2 = 0 x = 6 or x = -2

More Than One SP

x = 6

y = 1/3(6)3 – 2(6)2 – 12(6)

= - 72x = -2

y = 1/3(-2)3 – 2(-2)2 – 12(-2)

= 132/3

SPs are (-2 , 132/3) and (6 , -72)Two SPs!!!→ Need two y values

Find the SPs.1. y = 1/3x3 – 3x2 + 8x

2. y = 2x3 – 6x2

3. y = x3 – 3x2 – 24x + 2

4. y = x3 – 48x

(0 , 0) and (2 , -8)

(4 , -78) and (-2 , 30)

(4 , -128) and (-4 , 128)

Key Question

Ex y = 1/3x3 – 3x2 + 8x SPs?

For SPs dy/dx = 0

dy/dx = x2 – 6x + 8

x2 – 6x + 8 = 0(x – 4)(x – 2 ) = 0x–4= 0 or x–2 = 0 x = 4 or x = 2

x = 4

y = 1/3(4)3 – 3(4)2 + 8(4))

= 51/3x = 2

y = 1/3(2)3 – 3(2)2 + 8(2)

= 62/3

SPs are (4 , 51/3) and (2 , 62/3)

Key Question

What is to be learned?

• How to use differentiation to identify SPs• What SPs are!!!!!!• What the difference between a stationary

point and a stationary value• What SVs are• How to find SP/SVs• How to find the nature of SP/SVs

Finding the nature

m negative m positive m = 0

for nature need to know gradient just before and after SP

Use the derivative!

Making a nature table

Making a Nature Tabley = x2 – 8x + 10

For SPs dy/dx = 0

dy/dx = 2x – 8

2x – 8 = 02x = 8 x = 4 y?

at x = 4 y = 42 – 8(4) + 10 = - 6

SP is (4 , -6)

Making a Nature Tabley = x2 – 8x + 10 dy/dx = 2x – 8

SP is (4 , -6)

x 4dydy//dxdx = 2x – 8 = 2x – 8 0

3 5

- +

Slope

Min TPat (4 , -6)

Making a Nature Tabley = 2x3 – 6x2 + 10dy/dx = 6x2 – 12x

SPs (0 , 10) and (2 , 2)

x 0dydy//dxdx = 6x = 6x22 – 12x – 12x 0

-1 1

+ -

Slope

Max TPat (0 , 10)

2 3

+0

Min TPat (2 , 2)

Nature Table

• Used to find nature of SPs• Show gradient just before and after SPs• Use derivative

Making a Nature Tabley = 1/3x3 –– 2x2 –– 12xdy/dx = x2 – 4x - 12

SPs (-2, 132/3) and (6 , -72)

x -2dydy//dxdx = x = x22 – 4x – 4x –– 12 12 0

-3 0

+ -

Slope

Max TPat (-2, 132/3)

6 7

+0

Min TPat (6 , -72)

Making a Nature Tabley = 1/3x3 –– 2x2 –– 12xdy/dx = x2 – 4x – 12

SPs (-2, 132/3) and (6 , -72)

x -2dydy//dxdx = x = x22 – 4x – 12 – 4x – 12 0

-3 0

+ -

Slope

Max TPat (-2, 132/3)

6 7

+0

Min TPat (6 , -72)

Making a Nature Tabley = 1/3x3 –– 2x2 –– 12xdy/dx = x2 – 4x –– 12

SPs (-2, 132/3) and (6 , -72)

x -2dydy//dxdx = (x = (x –– 6)(x + 2) 6)(x + 2) 0

-3 0

= +

Slope

Max TPat (-2, 132/3)

6 7

0

Min TPat (6 , -72)

= - = +- X - - X + + X +

Making a Nature Tabley = 1/3x3 - 2x2 – 12xdy/dx = x2 – 4x – 12

SPs (-2, 132/3) and (6 , -72)

x -2dydy//dxdx = x = x22 – 4x – 4x –– 12 12 0

-3 0

+ -

Slope

Max TPat (-2, 132/3)

6 7

+0

Min TPat (6 , -72)

*

*could use factorised derivative

(( (x (x –– 6)(x + 2) 6)(x + 2) ))

Find Stationary Point and Nature fory = 12x – 3x2

For SPs dy/dx = 0

dy/dx = 12 – 6x

12 – 6x = 0 x = 2

y = 12(2) – 3(2)2 = 12

SP is (2 , 12)

at x = 2

Key Question

x 2

dydy//dxdx = 12 – 6x = 12 – 6x 0

1 3

+ -

Slope Max TPat (2 , 12)