### Slides: C2 - Chapter 9 - Differentiation

```C2 Chapter 9 Differentiation
Dr J Frost ([email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */)
Last modified: 5th October 2013
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)
Example Exam Question
Edexcel C2 June 2010

1
1
−
−?  2
2
a)
= 2
b)  > 32 ?
Showing a function is increasing/decreasing
Show that f(x) = x3 + 24x + 3 (x ϵ ℝ) is an
increasing function.
′  =  +
?

is always positive for all . Thus ′  >
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) = x3 + 3x2 – 9x is a decreasing function.
3x2 + 6x – 9 < 0
Thus -3 < x < 1?
2
Find the values of x for which the function
f(x) = x + (25/x) is a decreasing function.
1 – (25/x2) < 0
Thus -5 < x < 5?
Questions
C2 pg 130
Exercise 9A
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’.
Finding turning points
Edexcel C2 May 2013 (Retracted)
(2,9)
?
Although it might be interest
to know if this is
a minimum point or a maximum point...
Do we have a minimum or maximum point?
Method 1:
Consider the points immediately
before and after the
? stationary point.
Method 2:
Use the second-order derivative to
see whether the gradient
is
?
increasing or decreasing.
Do we have a minimum or maximum point?
(2, -48)
Find the coordinates of the turning point on the curve with
equation y = x4 – 32x. Determine whether this is a minimum or
maximum point.
Method 1
Value of x
Gradient
Shape
Method 2
x<2
e.g. x = 1.9
x=2
e.g. -4.56
?
?0
?
?
We can see from this
shape that this is a
minimum point.
x>2
e.g. x = 2.1
2
2
= 12x2 ?
When x = 2,
2
2
= 48.
?
?
e.g. 5.04
?
2
>
2
0, so a minimum
? point.
What are the advantages of each method?
Points of inflection
Stationary point
of inflection
(“saddle point”)
Non-stationary
point of inflection
A point of inflection is a point where the curve
changes from concave to convex (or vice versa).

We can see that when = 0, we might not have a maximum

or minimum, but a point of inflection instead.
At A Level, you won’t see non-stationary points of inflection.
Stationary Points of inflection
So how can we tell if a stationary point is a point of inflection?
Non-Stationary Points of inflection
(not in the A Level syllabus)

2
2
At this point:
> 0 (i.e. not stationary)
?
= 0 (i.e. gradient is not
?
changing at this point)
Stationary Point Summary
d2y / dx2
Type of Point
<0
Maximum
>0
Minimum
=0
Could be maximum, minimum
or point of inflection. Use
‘Method 1’ to find gradient just
before and after.
y = x4 has a turning point at x = 0.
Show that this is a minimum
point.
dy/dx = 4x3.
d2y/dx2 = 12x2
When x = 0, d2y/dx2 = 0, so we
can’t classify immediately.
?
When x = -0.1, dy/dx = -0.004.
When x = +0.1, dy/dx = +0.004.
Gradient goes from negative to
positive, so minimum point.
Further Examples
Find the stationary points of y = 2x3 – 15x2 + 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. ?
State the range of outputs of 6x – x2
≤9
?
Exercise 9B
Dr Frost
Differentiation – Practical applications
Objectives: Use differentiation in real-life problems that involve
optimisation of some variable.
Optimisation Problems
We have a sheet of A4 paper,
which we want to fold into a
cuboid. What height should
we choose for the cuboid to
maximise the volume?
These are examples of optimisation
problems: we’re trying to
maximise/minimise some quantity
by choosing an appropriate value of
a variable that we can control.
x
y
We have 50m of
fencing, and want to
make a bear pen of the
following shape, but
that maximises the
area. What should we
choose x and y to be?
Breaking down optimisation problems
Suppose that we have 100cm of rope,
that we want to put in the shape of a
minor segment. We want to choose a
radius for this minor segment that
maximises the area covered by the rope.
What radius do we choose?
M
N
r cm
Strategy
1. Form two equations: one representing
the thing we’re trying to maximise (here
the area) and the other representing the
constraint (here the length of rope)
Typically we’d need to write out two
equations (e.g. perimeter and area, or
volume and area) and combine them
together, using given information, to
form the one equation we’d need.
O
2. Use calculus to find out the value of
the variable we’re interested in when we
have a minimum/maximum.

e.g. Find when  = 0.
Example Exam Question
Edexcel C2 May 2011
Example 1
M
N
Suppose that we have 100cm of rope,
that we want to put in the shape of a
minor segment. We want to choose a
radius for this minor segment that
maximises the area covered by the rope.
What radius do we choose?
a) Show that A = 50r – r2
r cm
O
Given that r varies, find:
b) The value of r for which A is a
maximum and show that A is a
maximum.
c) Find the value of angle MON for this
maximum area.
d) Find the maximum area of the sector
OMN.
Example 2

A large tank in the shape of a cuboid is
to be made from 54m2 of sheet metal.
The tank has a horizontal base and no
top. The height of the tank is  metres.
Two of the opposite vertical faces are
squares.
a) Show that the volume, V m3, of the

tank is given by  =  −   .
b) Given that x can vary, use
differentiation to find the maximum
or minimum value of V.
c) Justify whether your value of V is a
minimum or maximum.
```