Academic Skills Advice
More Differentiation: Stationary Points
You need to be able to find a stationary point on a curve and decide whether it is a turning
point (maximum or minimum) or a point of inflexion.
Maximum turning point
Point of inflexion
Minimum turning point
Finding the Stationary Point:
Looking at the 3 diagrams above you should be able to see that at each of the points
shown the gradient is 0 (i.e. the curve goes flat). In lesson 6 we said that differentiation
gives us the gradient function. So to find any stationary points we need to differentiate first
then put the resulting gradient function equal to zero.
When
𝒅𝒚
𝒅𝒙
= 𝟎 (i.e. the gradient is zero) you have a stationary point.
In short: find the 1st differential, put it equal to zero and solve.
Examples:

Find the stationary point on the curve 𝒚 = 𝟑𝒙𝟐 − 𝟏𝟖𝒙 − 𝟕:
Step 1: Differentiate to give:
𝑑𝑦
𝑑𝑥
= 6𝑥 − 18
Step 2: Put this equal to zero:
6𝑥 − 18 = 0
Step 3: Solve:
6𝑥 = 18
𝑥=3
We now know that the stationary point is when 𝑥 = 3.
If we require the full co-ordinate we need to find the value of 𝑦 when 𝑥 = 3.
(We are looking for 𝑦 so we need to use the original equation, 𝑦 = 3𝑥 2 − 18𝑥 − 7)
Original equation:
𝑦 = 3𝑥 2 − 18𝑥 − 7
Substitute 𝑥 = 3:
𝑦 = 3(3)2 − 18(3) − 7
Solve:
𝑦 = −34
So the stationary point is at
(𝟑, −𝟑𝟒)
© H Jackson 2011 / ACADEMIC SKILLS
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
Find the stationary point on the curve 𝒚 =
𝑑𝑦
Step 1: Differentiate to give:
𝑑𝑥
𝟏
𝟑
𝒙𝟑 + 𝟐𝒙𝟐 − 𝟏𝟐𝒙 + 𝟓:
= 𝑥 2 + 4𝑥 − 12
Step 2: Put this equal to zero:
𝑥 2 + 4𝑥 − 12 = 0
Step 3: Solve:
(𝑥 + 6)(𝑥 − 2) = 0
𝑥 = −6 𝑜𝑟 𝑥 = 2
We now know that there are 2 stationary points: one at 𝑥 = −6 and one at 𝑥 = 2.
Next find the corresponding 𝑦 values, using the equation, 𝑦 =
1
1
3
𝑥 3 + 2𝑥 2 − 12𝑥 + 5
When 𝑥 = −6:
𝑦 = 3 (−6)3 + 2(−6)2 − 12(−6) + 5
𝑦 = 77
When 𝑥 = 2:
𝑦 = 3 (2)3 + 2(2)2 − 12(2) + 5
𝑦= −
1
So the 2 stationary points are at (−𝟔, 𝟕𝟕) and (𝟐, −
𝟐𝟓
𝟑
25
3
)
In the above examples we have found out where the stationary points are but we don’t
know what type they are (i.e. maximum, minimum or point of inflexion). We now need to
go a step further (and differentiate again) to find the type.
Type of Stationary point:
To find the type of stationary point we need to differentiate again (i.e. find the 2nd
𝑑2 𝑦
differential, denoted by 𝑑𝑥 2 ). This measures the change in gradient and can help to decide
whether a stationary point is a maximum, minimum or point of inflexion.
After finding where the stationary point is, find the 2nd differential and if:
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
>0
then you have a minimum (positive is minimum).
<0
then you have a maximum (negative is maximum).
=0
then it could be maximum, minimum or point of inflexion.
In the last case – differentiate again:
If
𝑑3 𝑦
𝑑𝑥 3
≠0
then it is a point of inflexion.
Remember: 1st differential tells us where the stationary point is.
2nd differential tells us what type the stationary point is.
© H Jackson 2011 / ACADEMIC SKILLS
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Examples:

Find the stationary point on the curve 𝒚 = 𝒙𝟐 − 𝟒𝒙 + 𝟐:
Use 1st differential to find where the Stationary point is:
Step 1: Differentiate to give:
𝑑𝑦
𝑑𝑥
= 2𝑥 − 4
Step 2: Put this equal to zero:
2𝑥 − 4 = 0
Step 3: Solve:
2𝑥 = 4
𝑥=2
Find corresponding 𝑦 value:
𝑦 = 𝑥 2 − 4𝑥 + 2
Substitute 𝑥 = 2:
𝑦 = 22 − 4(2) + 2
Solve:
𝑦 = −2
So the stationary point is at
(𝟐, −𝟐)
Use 2nd differential to find what type the Stationary point is:
𝑑2 𝑦
𝑑𝑥 2

=2
This is positive (>0) so the turning point must be a minimum.
Find the stationary points on the curve 𝒇(𝒙) = 𝒙𝟑 − 𝟑𝒙𝟐 + 𝟒:
Use 1st differential to find where the Stationary point is:
Step 1: Differentiate to give:
𝑓 ′ (𝑥) = 3𝑥 2 − 6𝑥
Step 2: Put this equal to zero:
3𝑥 2 − 6𝑥 = 0
Step 3: Solve:
3𝑥(𝑥 − 2) = 0
𝑥 = 0 𝑜𝑟 𝑥 = 2
Find corresponding 𝑦 values:
𝑓(𝑥) = 𝑦 = 𝑥 3 − 3𝑥 2 + 4
Substitute 𝑥 = 0:
Substitute 𝑥 = 2:
𝑦 = (0)3 − 3(0)2 + 4
𝑦 = (2)3 − 3(2)2 + 4
So the stationary points are at
(𝟎, 𝟒) and (2, 0)
© H Jackson 2011 / ACADEMIC SKILLS
When using function
notation, 𝑓 ′ (𝑥) means
the 1st differential.
𝑦=4
𝑦=0
3
Use 2nd differential to find what type the Stationary points are:
𝑓 ′′ (𝑥) = 6𝑥 − 6
When using function
notation, 𝑓 ′′ (𝑥) means
the 2nd differential.
This looks a bit different from the 1st example as there is an 𝑥 in the 2nd differential. We
now need to substitute each value of 𝑥 in, separately, to find out what type of stationary
point each one is.
When 𝑥 = 0:
When 𝑥 = 2:
𝑓 ′′ (𝑥) = 6(0) − 6
𝑓 ′′ (𝑥) = −6
(∴ maximum)
𝑓 ′′ (𝑥) = 6(2) − 6
𝑓 ′′ (𝑥) = +6
(∴ minimum)
So (0, 4) is a maximum and (2, 0) is a minimum.

Find the stationary point on the curve 𝒚 = 𝒙𝟑 :
Step 1: Differentiate to give:
𝑑𝑦
𝑑𝑥
= 3𝑥 2
Step 2: Put this equal to zero:
3𝑥 2 = 0
Step 3: Solve:
𝑥=0
Find corresponding 𝑦 value:
𝑦 = 𝑥3
Substitute 𝑥 = 0:
𝑦 = (0)3
Solve:
𝑦= 0
So the stationary point is at
(𝟎, 𝟎)
Use 2nd differential to find what type the Stationary point is:
𝑑2 𝑦
𝑑𝑥 2
When 𝑥 = 0:
𝑑2 𝑦
𝑑𝑥 2
= 6𝑥
=0
so could be a max, min or point of inflexion.
Use 3rd differential to decide:
𝑑3 𝑦
𝑑𝑥 3
=6
The 3rd differential is not 0 so it is a point of inflexion.
© H Jackson 2011 / ACADEMIC SKILLS
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