Academic Skills Advice
Differentiation: Stationary Points Summary
There are 3 types of stationary points: maximum, minimum and point of inflexion.
Maximum turning point
Point of inflexion
Minimum turning point
Differentiate once (1st derivative):
This tells you where there are any stationary points.
The first derivative is the gradient function and tells you the gradient at any point. At a
stationary point the gradient = 0 so:
Put
𝑑𝑦
𝑑𝑥
= 0 and find the value of 𝑥.
Use the value(s) of 𝑥 to find the corresponding 𝑦 value(s);
You now have the co-ordinates of the stationary point(s).
Differentiate again (2nd derivative):
This tells you the type of stationary point(s) you have found. Substitute your value(s) of 𝑥
into the 2nd derivative and:
If
If
If
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 0 and
= 0 and
= 0 and
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.
Otherwise (or as an alternative to the above) you need to find the gradient at either side of
the stationary point to decide whether it is a maximum or minimum.
© H Jackson 2011 - 2015 / Academic Skills
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e.g. Find, and classify, the stationary points for the following curve:
𝒙𝟑
𝒚=
− 𝒙𝟐 − 𝟏𝟓𝒙 + 𝟐
𝟑
𝑦=
𝒅𝒚
Find 𝒅𝒙
∴
Put
𝒅𝒚
𝒅𝒙
=𝟎
and solve to find the value of 𝒙
𝑥3
− 𝑥 2 − 15𝑥 + 2
3
𝑑𝑦
= 𝑥 2 − 2𝑥 − 15
𝑑𝑥
𝑥 2 − 2𝑥 − 15 = 0
(𝑥 + 3)(𝑥 − 5) = 0
𝒙 = −𝟑 𝒐𝒓 𝒙 = 𝟓
Find the corresponding 𝒚 values
𝑦=
𝑥3
− 𝑥 2 − 15𝑥 + 2
3
When 𝒙 = −𝟑
𝑦=
(−3)3
3
− (−3)2 − 15(−3) + 2
𝒚 = 𝟐𝟗
When 𝒙 = 𝟓
𝑦=
(5)3
3
− (5)2 − 15(5) + 2
We have found that there are 2 stationary points:
𝒚 = −𝟓𝟔. 𝟑
(−𝟑, 𝟐𝟗) and (𝟓, −𝟓𝟔. 𝟑)
Next we need to classify them (find the type):
𝒅𝟐 𝒚
Find 𝒅𝒙𝟐
𝑑𝑦
= 𝑥 2 − 2𝑥 − 15
𝑑𝑥
∴
Substitute your value(s) of 𝒙 in
𝑑2𝑦
= 2𝑥 − 2
𝑑𝑥 2
When 𝒙 = −𝟑
𝑑2 𝑦
𝑑𝑥 2
= 2(−3) − 2 = −8
∴ Maximum
When 𝒙 = 𝟓
𝑑2 𝑦
𝑑𝑥 2
© H Jackson 2011 - 2015 / Academic Skills
= 2(5) − 2 = 8
∴ Minimum
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