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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)
