IGCSE Further Maths/C1
Differentiation
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
www.drfrostmaths.com
Objectives:
Last modified: 3rd January 2016
Gradient of a curve
The gradient of the curve at a given point can be found by:
1. Drawing the tangent at that point.
2. Finding the gradient of that tangent.
How could we find the gradient?
We want to find the gradient at the point A.
We could pick
find a point close
to 𝐴 (𝐵) and just
use “change in 𝑦
over change in 𝑥”
δx represents a small change in x and δy
represents a small change in y.
How could we find the gradient?
Suppose we’re finding the gradient of 𝑦 = 𝑥 2 at the point A(3, 9).
Gradient = 6 ?
How could we find the gradient?
For the curve 𝑦 = 𝑥 2 , we find the gradient for these various points.
Can you spot the pattern?
Point
(1, 1)
(4, 16)
(2.5, 6.25)
(10, 100)
Gradient
2
8
5
20
?
For 𝑦 = 𝑥 2 , gradient function = 𝟐𝒙
Let’s prove it...
Proof that ‘gradient function’ of y = x2 is 2x
Suppose we add some tiny value, h, to x. Then:
(x + h, (x+h)
? 2)
(x, x?2)
δy
δx
The “lim” bit means
“what this expression
approaches as h tends
towards 0”
𝛿𝑦
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝛿𝑥
𝑥 + ℎ 2 − 𝑥2
= lim
ℎ→0
ℎ
2𝑥ℎ + ℎ?2
= lim
ℎ→0
ℎ
= lim 2𝑥 + ℎ
ℎ→0
= 2𝑥
The h disappears as h
tends towards 0.
Note: This proof will not be examined.
Terminology
Let’s reflect on what we’ve done.
Give the function 𝑦 = 𝑥 2 , we wanted to find a function (2𝑥) that
would give us the gradient for any point on the line. This is known
as the gradient function.
Original function
𝑦=𝑥
2
Gradient function
𝑑𝑦
= 2𝑥
𝑑𝑥
The process of finding the gradient function is known as
‘differentiation’. We’d say that “we differentiated 𝑥 2 to get 2𝑥”.
Differentiating 𝑥 𝑛
Can you spot the pattern?
𝒚
𝑥2
𝑥3
𝑥4
𝑥5
𝑥6
𝒅𝒚
𝒅𝒙
2𝑥
3𝑥 2
4𝑥 3
5𝑥 4
6𝑥 5
𝑑𝑦
= 𝑛𝑥 𝑛−1
?
𝑑𝑥
𝑑𝑦
= 𝑎𝑛𝑥?𝑛−1
𝑑𝑥
! If 𝑦 = 𝑥 𝑛 , then
If 𝑦 = 𝑎𝑥 𝑛
then
Quickfire Questions!
𝑦 = 𝑥4
Gradient function
𝑑𝑦
= 𝟒𝒙𝟑?
𝑑𝑥
Gradient when 𝑥 = 2
𝑚 = 𝟒 𝟐𝟑 ?= 𝟑𝟐
𝑑𝑦
(Bro Tip: Use 𝑑𝑥 for the
gradient function and 𝑚 for
specific gradients)
Quickfire Questions!
𝑦 = 5𝑥 3
Gradient function
𝑑𝑦
= 𝟏𝟓𝒙?𝟐
𝑑𝑥
Gradient when 𝑥 = −2
𝑚 = 𝟏𝟓 −𝟐?𝟐 = 𝟔𝟎
Quickfire Questions!
1 5
𝑦 = 𝑥 − 3𝑥 2
2
Gradient function
𝑑𝑦 𝟓 𝟒
= 𝒙 ?
− 𝟔𝒙
𝑑𝑥 𝟐
Gradient when 𝑥 = 1
𝟓
𝟕
𝑚 = −𝟔=
−
? 𝟐
𝟐
Differentiating 𝑐𝑥 and 𝑐
y
x
y=4
x
𝑦 = 5𝑥
𝑦=3
→
𝑑𝑦
=𝟓
𝑑𝑥
→
𝑑𝑦
=𝟎
𝑑𝑥
Looking at the graph, what is the gradient of the
line 𝒚 = 𝟑𝒙?
How could you show it using differentiation?
𝒚 = 𝟑𝒙 = 𝟑𝒙𝟏
𝒅𝒚 ?
∴
= 𝟑𝒙𝟎 = 𝟑
𝒅𝒙
Looking at the graph, what is the gradient of the
line 𝒚 = 𝟒?
How could you show it using differentiation?
𝒚 = 𝟒 = 𝟒𝒙𝟎
𝒅𝒚
? −𝟏 = 𝟎
∴
= 𝟎𝒙
𝒅𝒙
?
?
𝑦 = 10
𝑦 = −𝑥
→
𝑑𝑦
=𝟎
𝑑𝑥
→
𝑑𝑦
= −𝟏?
𝑑𝑥
?
Exercise 1
Function (𝒚)
1
2
3
4
5
6
7
8
𝑦 = 2𝑥 5
𝑦 = 7𝑥 3
𝑦 = 2𝑥 + 𝑥 2
𝑦 = 𝑥 3 + 𝑥 −1 + 3
𝑦 = 𝑥4
𝑦 = 𝑥3 + 4
𝑦 = 2𝜋𝑥
𝑦 = 𝑥 3 − 3𝑥 + 𝜋 2 𝑥 2
𝒅𝒚
𝒅𝒙
Gradient Function ( )
Point of interest
Gradient at this
point (𝒎)
𝑑𝑦
= 10𝑥 4
𝑑𝑥
𝑑𝑦
= 21𝑥 2
𝑑𝑥
𝑑𝑦
= 2 + 2𝑥
𝑑𝑥
𝑑𝑦
= 3𝑥 2 − 𝑥 −2
𝑑𝑥
𝑑𝑦
= 4𝑥 3
𝑑𝑥
𝑑𝑦
= 3𝑥 2
𝑑𝑥
𝑑𝑦
= 2𝜋
𝑑𝑥
?
2,64
160
?
3,189
189
?
4, 24
10
?
2, 11.5
11.75
?
2,16
32
?
3,31
27
?
10, 20𝜋
2𝜋
𝑑𝑦
= 3𝑥 2 − 3 + 2𝜋 2 𝑥
𝑑𝑥
1, 𝜋 2 − 2
2𝜋 2
?
?
?
?
?
?
?
?
?
9 A rectangle has length 6𝑥 and width 3𝑥. The area of the rectangle is 𝑦.
(i) Write down 𝑦 in terms of 𝑥.
𝑦 = 18𝑥 2 ?
𝑑𝑦
𝑑𝑦
(ii) Work out 𝑑𝑥
= 36𝑥 (this represents the
? rate of change of area)
𝑑𝑥
Recap
Edexcel C1 May 2012
1
𝑑𝑦
2
= 15𝑥 ?− 8𝑥 3 + 2
𝑑𝑥
Turning more complex expressions into polynomials
We know how to differentiate things in the form 𝑎𝑥 𝑛 . Where possible, put
expressions in this form.
𝑥→
1
𝑥 2?
1
1
−
= 𝑥 ?2
𝑥
3
𝑥2
= 𝑥 2?
𝑥
? 𝑥2
𝑥2 𝑥 − 1 = 𝑥3 −
1
−2?
=
𝑥
𝑥2
1
1
1+𝑥
−2
? 𝑥2
=𝑥 +
𝑥
5
𝑥 − 𝑥2
−2
=𝑥 −
𝑥 −1
?
3
𝑥
C1 Exercise 7E (Page 115)
1
Use standard results to differentiate.
?
?
a) y = 2 𝑥
1
c) y = 3𝑥 3
2
e) y = 𝑥 3 + 𝑥
f) y =
2
3
1
𝑑𝑦
= 𝑥 −2
𝑑𝑥
𝑑𝑦
= −𝑥 −4
𝑑𝑥
𝑑𝑦
1 1
= −6𝑥 −4 + 𝑥 −2
𝑑𝑥
2
𝑑𝑦 1 −2 1 −2
= 𝑥 3− 𝑥
𝑑𝑥 3
2
𝑑𝑦
= 3 + 6𝑥 −2
𝑑𝑥
𝑑𝑦
= 3𝑥 2 − 2𝑥 + 2
𝑑𝑥
𝑑𝑦
= 24𝑥 − 8 + 2𝑥 −2
𝑑𝑥
1
𝑥 + 2𝑥
h)
j)
3𝑥 2 −6
y= 𝑥
y = 𝑥 𝑥2
l)
y = 3𝑥 − 2
−𝑥+2
1
4𝑥 + 𝑥
?
?
?
?
?
Find the gradient of the curve with equation
𝑦 = 𝑓(𝑥) at the point A where:
c) 𝑓 𝑥 =
1
𝑥
and A is at
4
1
,2
4
d) 𝑓 𝑥 = 3𝑥 − 𝑥 2 and A is at 2,5
−4?
4?
Further Practice
1
[Jan 2013 Paper 1]
2𝑥 2 3𝑥 3 − 7𝑥
𝑦=
𝑥
𝒅𝒚
= 𝟐𝟒𝒙𝟑 −?𝟐𝟖𝒙
𝒅𝒙
2
[Jan 2013 Paper 2]
𝑦 = 𝑥3 − 1 2 + 𝑥 4
𝒅𝒚
= 𝟔𝒙𝟓 − 𝟑𝒙?𝟐 + 𝟐𝒙
𝒅𝒙
3
[June 2012 Paper 1]
𝑦 = 2𝑥 3 + 𝑎𝑥 where 𝑎 is a constant. The
𝑑𝑦
value of 𝑑𝑥 when 𝑥 = 2 is twice the
5
𝑑𝑦
value of 𝑑𝑥 when 𝑥 = −1. Determine 𝑎.
𝟏𝟐 + 𝒂 = 𝟐 𝟑 + 𝒂 → 𝒂 = 𝟔
?
4
[June 2013 Paper 2]
𝑑𝑦
𝑦 = 5𝑥 − 2 2 . Determine 𝑑𝑥 in the form
𝑎 𝑏𝑥 − 𝑐 .
𝒅𝒚
𝒅𝒙
= 𝟏𝟎? 𝟓𝒙 − 𝟐
6
[Set 1 Paper 1]
A right circular cone is being filled
with water. The volume of the
water is 𝑦 cm3 when the depth of
water is 3𝑥 cm and the radius of
the surface is 𝑥 cm.
(a) Show that 𝑦 = 𝜋𝑥 3
𝟏 𝟐
𝒚 = 𝝅𝒙 ?𝟑𝒙 = 𝝅𝒙𝟑
𝟑
𝑑𝑦
(b) Work out giving your
𝑑𝑥
answer in terms of 𝜋.
𝟑𝝅𝒙
?𝟐
(c) Work out the rate of change
of 𝑦 with respect to 𝑥 when
𝑥 = 5, in terms of 𝜋.
𝟕𝟓𝝅
?
1
2
7
2
[Set 3 Paper 2] 𝑦 = 𝑥 𝑥 − 𝑥
𝒅𝒚
=?
𝟒𝒙𝟑 − 𝟏
𝒅𝒙
1
2
Finding equations of tangents
Find the equation of the tangent to the curve
𝑦 = 𝑥 2 when 𝑥 = 3.
Gradient function:
𝑥=3
𝒅𝒚
𝒅𝒙
=
? 𝟐𝒙
Gradient when 𝑥 = 3:
𝒎?
=𝟔
𝑦-value when 𝑥 = 3:
𝒚=
?𝟗
So equation of tangent:
𝒚 − 𝟗 =?𝟔 𝒙 − 𝟑
! To find equation of tangent:
1. Find 𝑦 value using original function.
𝑑𝑦
2. Differentiate to find 𝑑𝑥 and use 𝑥 to find gradient 𝑚 at that point.
3. Use 𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1 for equation.
Finding equations of tangents
Find the equation of the normal to the curve
𝑦 = 𝑥 2 when 𝑥 = 3.
Equation of tangent (from earlier):
𝒚 − 𝟗 = 𝟔(𝒙 − 𝟑)
So equation of normal:
𝟏
𝒚 − 𝟗 = − 𝒙?− 𝟑
𝟔
𝑥=3
Test Your Understanding
June 2012 Paper 1
(Q1 on provided sheet)
𝒅𝒚
= 𝟑𝒙𝟐 + 𝟏𝟎𝒙
𝒅𝒙
𝒎 = 𝟑 −𝟏 𝟐 + 𝟏𝟎 ?
−𝟏 = 𝟑 − 𝟏𝟎 = −𝟕
At 𝒙 = −𝟏, 𝒚 = −𝟏 𝟑 + 𝟓 −𝟏 𝟐 + 𝟏
= −𝟏 + 𝟓 + 𝟏 = 𝟓
?
Therefore:
𝒚 − 𝟓 = −𝟕 𝒙 + 𝟏
(questions on
worksheet)
Exercises
1
[IGCSEFM June 2012 Paper 1 Q8] A curve
has equation 𝑦 = 𝑥 3 + 5𝑥 2 + 1
𝑑𝑦
(a) When 𝑥 = −1, show that the value of
𝑑𝑥
is -7.
𝒅𝒚
= 𝟑𝒙𝟐 + 𝟏𝟎𝒙
𝒅𝒙
𝒎 = 𝟑 −𝟏 𝟐 + 𝟏𝟎 −𝟏 = −𝟕
(b) Work out the equation of the tangent to
the curve 𝑦 = 𝑥 3 + 5𝑥 2 + 1 at the point
where 𝑥 = −1.
𝒚 = −𝟏 𝟑 + 𝟓 −𝟏 𝟐 + 𝟏 = 𝟓
𝒚 − 𝟓 = −𝟕(𝒙 + 𝟏)
?
?
2
[IGCSEFM June 2013 Paper Q8] A curve has
equation 𝑦 = 𝑥 4 − 5𝑥 2 + 9
𝑑𝑦 𝒅𝒚
(a) Work out .
= 𝟒𝒙𝟑 − 𝟏𝟎𝒙
𝑑𝑥 𝒅𝒙
(b) Work out the equation of the tangent to
the curve at the point where 𝑥 = 2
Give your answer in the form 𝑦 = 𝑚𝑥 + 𝑐
𝒚 = 𝟐𝟒 − 𝟓 𝟐 𝟐 + 𝟗 = 𝟓
𝒎 = 𝟒 𝟐 𝟑 − 𝟏𝟎 𝟐 = 𝟏𝟐
𝒚 − 𝟓 = 𝟏𝟐 𝒙 − 𝟐
𝒚 = 𝟏𝟐𝒙 − 𝟏𝟗
?
?
that the
3 [IGCSEFM Set Paper 1 Q11] Show
3
2
tangents to the curve 𝑦 = 𝑥 + 3𝑥 +
3𝑥 + 1 at 𝑥 = 1 and 𝑥 = −3 are parallel.
𝒅𝒚
= 𝟑𝒙𝟐 + 𝟔𝒙 + 𝟑
𝒅𝒙
𝒅𝒚
At 𝒙 = 𝟏, = 𝟑 + 𝟔 + 𝟑 = 𝟏𝟐
𝒅𝒙
𝒅𝒚
−𝟑,
𝒅𝒙
?
At 𝒙 =
= 𝟐𝟕 − 𝟏𝟖 + 𝟑 = 𝟏𝟐
Same gradient therefore parallel.
4 [IGCSEFM Set 1 Paper 2 Q17] Work out the
equation of the normal to the curve 𝑦 =
2𝑥 3 − 𝑥 2 + 1 at the point (1, 2). Give your
answer in the form 𝑦 = 𝑚𝑥 + 𝑐.
𝒅𝒚
= 𝟔𝒙𝟐 − 𝟐𝒙
𝒅𝒙
𝒎=𝟔−𝟐=𝟒
𝟏
𝒎𝑻 = −
𝟒
𝟏
𝒚−𝟐=− 𝒙−𝟏
𝟒
𝟏
𝟗
𝒚=− 𝒙+
𝟒
𝟒
?
(questions on
worksheet)
Exercises
5
[IGCSEFM Set 2 Paper 1 Q15] The graph shows
a sketch of 𝑦 = (𝑥 − 2)(𝑥 − 3). The curve
intersects the 𝑥-axis at 𝑃 and 𝑄.
Show that the tangents at 𝑃 and 𝑄 are
perpendicular.
𝒚 = 𝒙𝟐 − 𝟓𝒙 + 𝟔
𝒅𝒚
= 𝟐𝒙 − 𝟓
𝒅𝒙
𝒅𝒚
At 𝒙 = 𝟐, = 𝟒 − 𝟓 = −𝟏
𝒅𝒙
𝒅𝒚
𝒅𝒙
?
At 𝒙 = 𝟑, = 𝟔 − 𝟓 = 𝟏
−𝟏 × 𝟏 = −𝟏 therefore the lines are
perpendicular.
6
[IGCSEFM Set 4 Paper 2 Q20] A sketch of
the curve 𝑦 = (𝑥 + 1)(2 − 𝑥) is shown.
𝐴 0,2 , 𝑃(2,0) and 𝑄 are points on the
curve.
(a) Write down the coordinates of point 𝑄.
(b) Show that the normal to the curve at 𝐴
intersects the curve again at 𝑃.
(a) 𝑸 −𝟏, 𝟎 𝑷 𝟐, 𝟎
(b) 𝒚 = −𝒙𝟐 + 𝒙 + 𝟐
𝒅𝒚
= −𝟐𝒙 + 𝟏
𝒅𝒙
At 𝒙 = 𝟎, 𝒎 = 𝟏 ∴ 𝒎𝑵 = −𝟏
𝒚 − 𝟐 = −𝟏 𝒙 − 𝟎
𝒚 = −𝒙 + 𝟐
At 𝒙 = 𝟐, 𝟎 = −𝟐 + 𝟐
𝟎=𝟎
Thus 𝑷 is on the line 𝒚 = −𝒙 + 𝟐
?
Exercises
7
(questions on
worksheet)
[IGCSEFM Specimen Paper 2 Q22] The diagram
shows the graph of 𝑦 = 𝑥 2 − 4𝑥 + 3
The curve cuts the 𝑥-axis at the points 𝐴 and 𝐵.
The tangent to the curve at the point (5,8) cuts
the 𝑥-axis at the point 𝐶.
Show that 𝐴𝐵 = 3𝐵𝐶.
𝒚= 𝒙−𝟏 𝒙−𝟑
∴ 𝑨 𝟏, 𝟎 𝑩(𝟑, 𝟎)
𝒅𝒚
= 𝟐𝒙 − 𝟒
𝒅𝒙
𝒅𝒚
At 𝟓, 𝟖 , 𝒅𝒙 = 𝟐(𝟓) − 𝟒 = 𝟔
Equation of tangent: 𝒚 − 𝟖 = 𝟔(𝒙 − 𝟓)
When 𝒚 = 𝟎: −𝟖 = 𝟔 𝒙?− 𝟓
𝟒
𝟏𝟏
𝒙=− +𝟓=
𝟑
𝟑
𝑨𝑩 = 𝟑 − 𝟏 = 𝟐
𝟏𝟏
𝟐
𝑩𝑪 =
−𝟑=
𝟑
𝟑
𝟐
𝟑𝑩𝑪 = 𝟑 × = 𝟐 = 𝑨𝑩
𝟑
Exercises
(questions on
worksheet)
Edexcel C1 Jan 2013
1
𝑑𝑦
−
= 2?− 4𝑥 2
𝑑𝑥
𝑦 = −6𝑥
? +3
(9, ?
-1)
Recap: If a line has gradient m and goes through 𝑥1 , 𝑦1 , then it has equation:
𝑦 − 𝑦1 = 𝑚 𝑥 − 𝑥1
Exercises
Expanding gives 𝑦 = 𝑥 3 + 7𝑥 2 + 15𝑥 + 9
𝑑𝑦
Thus is as given.
?
𝑑𝑥
𝑚 = 3 −5 2 + 14 −5 + 15 = 20
So 𝑦 + 16 = 20 𝑥 + 5
→
𝑦1 = −5 + 1 −5 + 3
𝑦 = 20𝑥 + 84
?
2
= −16
We’re interested where the gradient is 20.
3𝑥 2 + 14𝑥 + 15 = 20
3𝑥 2 + 14𝑥 − 5 = 0
3𝑥 − 1 𝑥 + 5 = 0
1
So 𝑥 = −5 (as before) or 𝑥 =
?
3
Exercises
Edexcel C1 Jan 2012
1
?, 0
2
17
−8,?
8
Exercises
Edexcel C1 Jan 2011
𝑑𝑦 3 2 27 1
= 𝑥 −? 𝑥 2 − 8𝑥 −2
𝑑𝑥 2
2
3
1
8
3
4 − 9 4 2 + + 30
2
4
= 32 − 72 + 2 + 30 = −8
?
𝑚𝑡𝑎𝑛𝑔 =
Thus 𝑚⊥ =
2
9
3
4
2
2
−
27
8
1
9
4 − 2 = 24 − 27 − = −
2
4
2
2
?2 𝑥 − 4
𝑦+8=
9
9𝑦 + 8 = 2𝑥 − 8
2𝑥 − 9𝑦 − 16 = 0
Increasing Functions
How could we use
differentiation to tell us if this is
a strictly increasing function?
...if the gradient is always
positive, i.e. f’(x)?> 0 for all x.
f(x2)
f(x1)
x1
x2
! A function is increasing if for any
two values of x, x1 and x2 where x2 >
x1, then f(x2) ≥ f(x1)
! A function is strictly increasing if
f(x2) > f(x1)
Increasing/Decreasing in an Interval
This is a decreasing
function in the
interval (𝑎, 𝑏)
i.e. where 𝑎 < 𝑥 < 𝑏
𝑎
𝑏
1
Find the values of x for which the function
𝑓 𝑥 = 𝑥 3 + 3𝑥 2 − 9𝑥 is a decreasing function.
2
Find the values of x for which the function
25
𝑓 𝑥 = 𝑥 + is a decreasing function.
𝑥
3x2 + 6x – 9 < 0
Thus -3 < x ?
<1
𝟐𝟓
𝟏− 𝟐 <𝟎
𝒙?
Thus −𝟓 < 𝒙 < 𝟓
Example Questions
A curve has the equation 𝑦 = 3𝑥 2 − 8𝑥.
𝑑𝑦
(a) Find 𝑑𝑥
(b) Hence determine the range of values of 𝑥 for which the function is increasing.
𝑑𝑦 a ?
= 6𝑥 − 8
𝑑𝑥
6𝑥 − 8 ≥ 0
4b ?
𝑥≥
3
A curve has the equation 𝑓 𝑥 = 𝑥 3 + 3𝑥 2 + 5𝑥.
(a) Find 𝑓′(𝑥) putting your answer in the form 𝑎 𝑥 + 𝑏
(b) Hence show that 𝑓(𝑥) is an increasing function.
𝑓 ′ 𝑥 = 3𝑥 2 + 6𝑥 + 5
5
= 3 𝑥 2 + 2𝑥 +
3
5
= 3 𝑥+1 2−1+
3
= 3 𝑥+1 2+2
𝑥 + 1 2 ≥ 0 for all 𝑥
∴ 𝑓 ′ 𝑥 > 0 for all 𝑥 so function is increasing.
a?
b?
2
+𝑐
! To show that a function is ALWAYS
increasing, complete the square to
show the minimum value of
expression is positive.
Test Your Understanding
1 3 5 2
𝑥 − 𝑥 + 6𝑥 + 3
3
2
Determine the range of values of 𝑥 for which 𝑦 is a decreasing function.
𝑦=
𝑑𝑦
= 𝑥 2 − 5𝑥 + 6
𝑑𝑥
𝑥 2 − 5𝑥 + 6? ≤ 0
𝑥−3 𝑥−2 ≤0
(Via a suitable sketch) 2 ≤ 𝑥 ≤ 3
Show that 𝑓 𝑥 = 𝑥 3 + 24𝑥 + 3 (𝑥 ∈ ℝ) is an increasing function.
𝒇′ 𝒙 = 𝟑𝒙𝟐 + 𝟐𝟒
?
𝟐
𝟑𝒙 is always positive for all 𝒙. Thus 𝒇′ 𝒙 > 𝟎
Exercises
1
Work out the values of 𝑥 for which the
following functions are increasing.
(i) 𝑦 = 𝑥 2 + 4
𝒙 ≥ 𝟎?
(ii) 𝑦 = 2𝑥 − 3
for all ?
𝒙
(iii) 𝑦 = 𝑥 2 + 2𝑥 − 5
𝒙 ≥ −𝟏
?𝟐
(iv) 𝑦 = 3𝑥 2 + 4𝑥 + 7
𝒙 ≥ −?
𝟑
1
3 Prove that 𝑦 = 3 𝑥 3 + 2𝑥 2 + 7𝑥 + 1 is
an increasing function for all values of 𝑥.
𝒅𝒚
= 𝒙𝟐 + 𝟒𝒙 + 𝟕
𝒅𝒙
= 𝒙+𝟐 𝟐+𝟑
Since 𝒙 + 𝟐
values of 𝒙.
1
(v) 𝑦 = 3 𝑥 3 − 2𝑥 2
𝒙 < 𝟎 or?𝒙 > 𝟒
(vi) 𝑦 = 𝑥 3 + 6𝑥 2 − 15𝑥 𝒙 < −𝟓 ?
or 𝒙 > 𝟏
2
Work out the values of 𝑥 for which the
following functions are decreasing.
(i) 𝑦 = 4𝑥 2
𝒙 ≤?𝟎
(ii) 𝑦 = 𝑥 2 − 6𝑥 + 2
𝒙 ≤ ?𝟑
(iii) 𝑦 = 12 − 𝑥
for all 𝒙
?
1 3
2
(iv) 𝑦 = 3 𝑥 + 𝑥
−𝟐 < 𝒙
<𝟎
?
(v) 𝑦 = 2𝑥 3 − 3𝑥 2 − 72𝑥 −𝟑 <?𝒙 < 𝟒
(vi) 𝑦 = 27𝑥 − 𝑥 3
𝒙 < −𝟑 or 𝒙 > 𝟑
?
4
𝟐
?
≥ 𝟎,
𝐝𝐲
𝐝𝐱
> 𝟎 for all
Prove that 𝑦 = 𝑥 3 − 6𝑥 2 + 27𝑥 − 4 is
an increasing function for all values of 𝑥.
𝒅𝒚
= 𝟑𝒙𝟐 − 𝟏𝟐𝒙 + 𝟐𝟕
𝒅𝒙
= 𝟑 𝒙𝟐 − 𝟒𝒙 + 𝟗
=𝟑
𝒙−𝟐
?𝟐
𝟐
+𝟓
= 𝟑 𝒙 − 𝟐 + 𝟏𝟓
𝒅𝒚
Since 𝒙 − 𝟐 𝟐 ≥ 𝟎, > 𝟎 for all
𝒅𝒙
values of 𝒙.
3
5 Prove that 𝑦 = 12 − 2𝑥 − 𝑥 is a
decreasing function for all values of 𝑥.
Stationary Points
Features you’ve previously
used to sketch graphs?
𝑓′ 𝑥 = 0
Maximum point
Stationary points are
those for which 𝑓 ′ 𝑥 = 0
Minimum point
𝑓′ 𝑥 = 0
Maximum/minimum points are known as ‘turning points’.
Examples
Find the turning point of the line with equation 𝑦 = 𝑥 2 + 4𝑥 − 3
𝑑𝑦
= 2𝑥 + 4 = 0
𝑑𝑥
𝑥 = −2
𝑦 = −2 2 + 4 −2 − 3
= 4 − 8 − 3 = −7
?
𝑑𝑦
Set gradient (𝑑𝑥 ) to be 0.
Use 𝑥 we’ve found to find 𝑦.
Turning point is −2, −7
Edexcel C2 May 2013 (R)
𝑦 = 2𝑥 + 3 + 8𝑥 −2
𝑑𝑦
= 2 − 16𝑥 −3 = 0
𝑑𝑥
16
2− 3 =0
𝑥
3
2𝑥 − 16 = 0
3
𝑥= 8=2
8
𝑦 =2 2 +3+ 2 =9
2
Stationary point is 2,9 . But is this a minimum or maximum point?
?
Test Your Understanding
Work out the turning points of the line with equation 𝑦 = 𝑥 3 − 12𝑥 + 3.
𝒅𝒚
= 𝟑𝒙𝟐 − 𝟏𝟐 = 𝟎
𝒅𝒙
𝟑𝒙𝟐 = 𝟏𝟐
𝒙𝟐 = 𝟒
?
𝒙 = ±𝟐
If 𝒙 = −𝟐: 𝒚 = −𝟐 𝟑 − 𝟏𝟐 −𝟐 + 𝟑 = 𝟏𝟗 → −𝟐, 𝟏𝟗
If 𝒙 = 𝟐: 𝒚 = 𝟐𝟑 − 𝟏𝟐 𝟐 + 𝟑 = −𝟏𝟑
→ 𝟐, −𝟏𝟑
𝑑𝑦
Given that 𝑦 = 𝑥 3 − 9𝑥 2 + 27𝑥 + 2, find 𝑑𝑥 . Put your answer in the form
𝑎 𝑥 + 𝑏 2 . Hence or otherwise, determine the turning points.
𝒅𝒚
= 𝟑𝒙𝟐 − 𝟏𝟖𝒙 + 𝟐𝟕
𝒅𝒙
= 𝟑 𝒙𝟐 − 𝟔𝒙 + 𝟗
= 𝟑 𝒙 − 𝟑?𝟐
𝟑 𝒙−𝟑 𝟐 =𝟎
∴ 𝒙 = 𝟑 → 𝒚 = 𝟐𝟕 − 𝟐𝟕 + 𝟖𝟏 + 𝟐 = 𝟖𝟑
What type of stationary point?
What are the 4 different things that could happen if the gradient is 0?
−
?
+
Type: minimum
?
Gradient just before
turning point:
Negative
?
Gradient just after
turning point:
Positive
?
?
?
Type: maximum
Gradient just before
turning point:
Positive ?
Gradient just after
turning point:
Negative
?
?
?
Type:
‘point of inflection’
?
Type:
‘point of inflection’
?
Gradient just before Gradient just before
turning point:
turning point:
Negative
Positive
?
Gradient just after
turning point:
Positive ?
?
Gradient just after
turning point:
Negative
?
Do we have a minimum or maximum point?
Find the coordinates of the turning point on the curve with
equation y = x4 – 32x. Determine whether this is a minimum or
maximum point.
Value of x
x<2
e.g. x = 1.9
x=2
x>2
e.g. x = 2.1
Gradient
e.g. -4.56
?
?0
e.g. 5.04
?
?
Shape
We can see from this
shape that this is a
minimum point.
?
?
(2, -48)
𝑑𝑦
= 4𝑥 3 − 32
𝑑𝑥
Another example
Set 3 Paper 1 Q16
𝑑𝑦
= 12𝑥 2 + 12𝑥 + 3 = 0
𝑑𝑥
2 + 4𝑥 + 1 = 0
? 4𝑥
Find
𝑥2 at stationary point(s)
2𝑥 + 1 = 0
1
𝑥=−
2
3
2
1
1
1
= 4 − find
+ 6 stationary
−
+ 3 − point
+ 5 = 4.5
?𝑦Hence
2
2
2
𝑑𝑦
When 𝑥 = −0.51, 𝑑𝑥 = 12 −0.51
𝑑𝑦
−0.49, 𝑑𝑥
2
+ 12 −0.51 + 3 = 0.0012
When 𝑥 = ? Determine
= 0.0012 type
Both positive, so a point of inflection.
of point
Grad
before
At
stationary
point
Grad
after
Test Your Understanding
Q
Find the stationary points of 𝑦 = 2𝑥 3 − 15𝑥 2 + 24𝑥 + 6 and
determine which of the points are maximum/minimum/points of
inflection.
(1, 17) is a maximum point.
(4, -10) is a maximum point. ?
N
State the range of outputs of 𝟔𝒙 − 𝒙𝟐
𝑓 𝑥 ≤9
?
Exercises
(From textbook Page 197)
1 For each of the curves give below
𝑑𝑦
𝑑𝑦
(a) Find 𝑑𝑥 and the value(s) of 𝑥 for which 𝑑𝑥 = 0
(b) Classify the point(s) that correspond to these 𝑥-values.
(c) Find the corresponding 𝑦 value.
(d) Sketch the curve.
(i)
(ii)
(iii)
(iv)
(v)
(vi)
𝑦
𝑦
𝑦
𝑦
𝑦
𝑦
= 𝑥 3 + 6𝑥 2 + 12𝑥 + 8
= 3𝑥 4 + 4𝑥 3
= 3 + 4𝑥 3 − 𝑥 4
= 3𝑥 5 − 5𝑥 3
= 4𝑥 3 2 − 𝑥
= 𝑥 3 − 3𝑥 2 + 3𝑥 + 1
−𝟐, 𝟎 Point of inflection.
?
−𝟏, −𝟏 Min, (𝟎, 𝟎) Point
? of inflection.
(𝟎, 𝟑) Point of inflection.?(𝟑, 𝟑𝟎) Max
(−𝟏, 𝟐) Max (𝟎, 𝟎) Point?of inflec (𝟏, −𝟐) Min
(𝟎, 𝟎) Point of inflection.? 𝟏. 𝟓, 𝟔. 𝟕𝟓 Max
(𝟏, 𝟐) Point of inflection.?
2 (i) Find the position and nature of any stationary points of the curve
𝑦 = 𝑥 3 − 3𝑥 2 + 3𝑥 + 2
Point of inflection at (1,3)
?
(ii) Sketch the curve.