Higher Mathematics
Unit 1.3
Using Differentiation
Increasing & Decreasing Functions,
Stationary Points
A function is
increasing
when
f ′(x) > 0
A function is
decreasing
when
f ′(x) < 0
Stationary points
occur when
f ′(x) = 0
y
x
Stationary Points
The nature of a stationary point
depends on the gradient on either side
of it. Stationary points may be one of
many types but always have f ′ (x) = 0
i.e. the gradient is 0 so the function is
horizontal at that point.
Maximum Turning Point
A maximum turning point (max TP) occurs when the gradient either side
of a stationary point changes from positive to negative.
y
TP
x

TP

f ′ (x)
+ ve
0
− ve
x
Graph
Max TP
Minimum Turning Point
A minimum turning point (min TP) occurs when the gradient either side
of a stationary point changes from negative to positive.
y
x
x

TP

f ′ (x)
− ve
0
+ ve
TP
Min TP
Graph
Rising Point of Inflexion
A rising point of inflexion (P I) occurs when the gradient either side of a
stationary point remains positive.
y
x
x

TP

f ′ (x)
+ ve
0
+ ve
PI
Graph
Rising P I
Falling Point of Inflexion
A falling point of inflexion (P I) occurs when the gradient either side of a
stationary point remains negative.
y
x
PI
x

TP

f ′ (x)
− ve
0
+ ve
Graph
Falling P I
Determine the Nature of Stationary
Points
Solve the equation f ′ (x) = 0 and
consider the gradient f ′ (x) around each stationary point:
If f ′ (x) is positive then the graph is increasing
If f ′ (x) is negative then the graph is decreasing
Curve Sketching
In order to sketch the graph of a function we need to
find:
the y-intercept
the x-intercept
the stationary points and their nature
the behaviour of the curve for large positive and
negative x
Curve Sketching
It is important to justify your working for each stage of the process.
y
The intercepts should be clearly
marked on your final sketch.
The coordinates of the
turning points must be clearly
labelled
(0,0)
(3,0)
x
All relevant working must be
noted and clearly understood.
(2, -1)
Maximum & Minimum Values on a
Closed Interval
In a closed interval the maximum and minimum values of a function
y
are either at
a stationary point or at
Local maximum
an end point of the interval.
x
Local minimum
Closed Interval
A closed interval is often written as { −2 < x < 3} or [−2 , 3]
This interval would refer to the values of the function from −2 to 3
Maximum and minimum values are either at a stationary point or
at an end point, so start by finding these points.
Find f (−2) and f (3)
Establish the stationary values for f ′ (x) in the interval from −2 to 3
and determine their nature. Find the value of the function at this
point
You are now in a position to state the local maximum or minimum
value of the function within the given interval.
Problem Solving
Differentiation can be used to solve problems which require
maximum or minimum values.
Problems typically cover topics such as areas, volumes and
rates of change. They often involve having to establish a
suitable formula in one variable and then differentiating to
find a maximum or minimum value. This is known as
Optimisation.
It is important to check the validity of any solutions as often
an answer is either nonexistent (e.g. a negative length or time)
or outside an acceptable interval.
Optimisation
Optimisation is term we use to describe a simple process of
finding a maximum or minimum value for a given situation.
Normally we would have to graph functions and then use our
graph to establish a maximum or minimum.
As we have learned previously, differentiation allows us to
quickly find the required value and is the expected manner of
solving problems like this in Higher mathematics.
Drawing graphs would not be an acceptable solution.
Optimisation
Problems posed will often involve more than one variable.
The process of differentiation requires that we rewrite or rearrange formulae so that there is only one variable – typically
x.
Questions are therefore multi-part where the first part would
involve establishing a formula in x from a given situation, part
2 would involve the differentiation and validation of an
acceptable answer, with part 3 the solution to the problem set.
You would do well to note that although the first part is
important, you can normally expect to complete the rest of the
question even when you cannot justify a formula in part 1.
Speed and Acceleration
Speed and acceleration are physical quantities which involve a rate of
change. Because of this they can be found using differentiation.
Speed is simply the rate of change of distance over time i.e.
Speed = Dist / Time. If the distance is a known function, typically d(t),
then speed is simply d ′(t). This can be called v(t) where v is the speed in a
straight line.
Acceleration is the rate of change of speed over time i.e.
Acceleration = change in Speed / Time. If the speed is a known function,
typically d ′(t), a function of d(t), then the acceleration is the second
derivative of the distance function d ′ ′(t) or more normally v ′ (t). For
simplicity, travel is once again in a straight line.
Problems involving these quantities often involve evaluating a
derived function at a given time or finding a maximum speed
from the derived function using stationary values.