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Year 12 Maths - Methods (Unit 3 & Unit 4) Applications of differentiation

Types of stationary points

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Theory

A stationary point of a curve is where \(f'(x)=0\). The first-derivative test classifies it as a local maximum (\(f'\) changes \(+\to-\)), a local minimum (\(f'\) changes \(-\to+\)), or a stationary point of inflection (\(f'\) keeps the same sign either side). Build a sign table of \(f'\) around each point and read off the type.

A stationary point of \(y=f(x)\) is a point where the gradient is zero: \(f'(x)=0\), so the tangent is horizontal. There are exactly three kinds. A local maximum is a peak — the curve rises up to it and then falls. A local minimum is a trough — the curve falls to it and then rises. A stationary point of inflection is a flat spot where the curve levels off (\(f'=0\)) but does not turn around: it keeps going in the same direction while its concavity changes.

The first-derivative test decides which one you have by looking at the sign of \(f'(x)\) just to the left and just to the right of the stationary point. A change from positive to negative means the curve stops rising and starts falling — a maximum; negative to positive means a minimum; the same sign on both sides means the curve never turns, so it is a stationary point of inflection.

This test works for every differentiable curve in the course — polynomials and power functions, and curves built from \(e^{x}\), \(\ln x\) and simple trigonometric functions. It is the “which type” companion to actually locating the stationary points by solving \(f'(x)=0\).

Key idea. Solve \(f'(x)=0\) to find the stationary points, then test the sign of \(f'\) either side of each: \(+\to-\) is a maximum, \(-\to+\) is a minimum, no sign change is a stationary point of inflection.
Local maximum and local minimumA cubic with a local maximum on the left (the gradient changes from positive to negative) and a local minimum on the right (the gradient changes from negative to positive). x y max min
Local maximum (\(f'\) goes \(+\to-\)) and local minimum (\(f'\) goes \(-\to+\))
Stationary point of inflectionA cubic y=x^3 with a horizontal tangent at the origin where the gradient is positive on both sides, so the curve levels off but keeps rising: a stationary point of inflection. x y flat
A stationary point of inflection: \(f'=0\) but \(f'\) is positive either side, so the curve levels off yet keeps rising

A stationary point is where the first derivative is zero:

\[f'(x)=0 \quad\Longrightarrow\quad \text{stationary point (horizontal tangent)}\]
f(x)=0

The first-derivative test reads the sign change of \(f'\) across the point:

\[+\to-\ :\ \text{local maximum}\qquad -\to+\ :\ \text{local minimum}\qquad \text{no change}\ :\ \text{stationary inflection}\]
+-

A sign table lays this out. For example, with \(f'(x)=3(x-1)(x+1)\):

\[\begin{array}{c|ccccc} x & <-1 & -1 & (-1,1) & 1 & >1\\ \hline f'(x) & + & 0 & - & 0 & +\\ \text{shape} & \nearrow & \text{max} & \searrow & \text{min} & \nearrow \end{array}\]
Repeated factor. A squared factor in \(f'\), such as the \(x^{2}\) in \(f'(x)=3x^{2}(x-2)\), does not change sign, so that stationary point is a stationary point of inflection, not a turning point.

How to classify a stationary point with the first-derivative test

  1. Differentiate. Find \(f'(x)\) and, where possible, write it in factorised form — the factors show exactly where the sign can change.
  2. Solve \(f'(x)=0\). The solutions are the \(x\)-coordinates of the stationary points.
  3. Build a sign table. Choose a test \(x\) in each interval between and beyond those solutions, and record whether \(f'\) is positive or negative there.
  4. Classify. Across each stationary point: \(+\to-\) is a local maximum, \(-\to+\) is a local minimum, and no sign change is a stationary point of inflection. Substitute back to get the \(y\)-coordinate.
Reading a given graph or \(f'\). If you are shown the graph, a peak is a maximum and a trough is a minimum; a flat spot that keeps going the same way is a stationary inflection. If you are given \(f'(x)\) already factorised, you can classify each stationary point straight from the signs of its factors — no need to find \(f\).
Example 1 — Classify a cubic
Find and classify the stationary points of \(y=x^{3}-3x\).
Solution
Differentiate and factorise — find \(f'(x)\):
\(f'(x)\)\(=\)\(3x^{2}-3\)
\(=\)\(3(x-1)(x+1)\)
Stationary points — solve \(f'(x)=0\):
\(3(x-1)(x+1)\)\(=\)\(0\)
\(x\)\(=\)\(-1\) or \(1\)
At \(x=-1\)
\(y\)\(=\)\((-1)^{3}-3(-1)\)
\(=\)\(-1+3=2\)
\(f'(-2)\)\(=\)\(3(-3)(-1)=9\ (>0)\)
\(f'(0)\)\(=\)\(3(-1)(1)=-3\ (<0)\)
\(\therefore\) \(+\to-\): local maximum at \((-1,2)\)
At \(x=1\)
\(y\)\(=\)\((1)^{3}-3(1)\)
\(=\)\(1-3=-2\)
\(f'(0)\)\(=\)\(3(-1)(1)=-3\ (<0)\)
\(f'(2)\)\(=\)\(3(1)(3)=9\ (>0)\)
\(\therefore\) \(-\to+\): local minimum at \((1,-2)\)
f(x)=3(x-1)(x+1)
Example 2 — Classify from a given \(f'\)
A function has \(f'(x)=(x-1)(x-3)\). Classify its stationary points.
Solution
Stationary points — \(f'(x)=0\) where each factor is zero:
\((x-1)(x-3)\)\(=\)\(0\)
\(x\)\(=\)\(1\) or \(3\)
Sign table — test the sign of each factor in every interval:
\(x<1\)\((-)(-)\)\(f'>0\)
\(1\((+)(-)\)\(f'<0\)
\(x>3\)\((+)(+)\)\(f'>0\)
\(\therefore\) \(+\to-\) at \(x=1\) (local maximum); \(-\to+\) at \(x=3\) (local minimum)
f(x)=(x-1)(x-3)
Example 3 — A repeated factor: stationary inflection
A function has \(f'(x)=3x^{2}(x-2)\). Find and classify its stationary points.
Solution
Stationary points — solve \(f'(x)=0\) for \(3x^{2}(x-2)=0\):
\(3x^{2}(x-2)\)\(=\)\(0\)
\(x\)\(=\)\(0\ (\text{double})\) or \(2\)
Sign table — the squared factor \(x^{2}\) never changes sign:
\(x<0\)\((+)(-)\)\(f'<0\)
\(0\((+)(-)\)\(f'<0\)
\(x>2\)\((+)(+)\)\(f'>0\)
\(\therefore\) no sign change at \(x=0\) (stationary point of inflection); \(-\to+\) at \(x=2\) (local minimum)
Repeated factor gives a stationary inflectionThe curve flattens to a horizontal tangent at the origin (a stationary point of inflection, no turn) and has a local minimum lower down at (2,-4). x y
f(x)=3x2(x-2)
Example 4 — Two types at once
A function has \(f'(x)=(x+1)(x-2)^{2}\). Classify its stationary points.
Solution
Stationary points — solve \(f'(x)=0\) for \((x+1)(x-2)^{2}=0\):
\((x+1)(x-2)^{2}\)\(=\)\(0\)
\(x\)\(=\)\(-1\) or \(2\ (\text{double})\)
Sign table — the squared factor \((x-2)^{2}\) is never negative:
\(x<-1\)\((-)(+)\)\(f'<0\)
\(-1\((+)(+)\)\(f'>0\)
\(x>2\)\((+)(+)\)\(f'>0\)
\(\therefore\) \(-\to+\) at \(x=-1\) (local minimum); no sign change at \(x=2\) (stationary point of inflection)
f(x)=(x+1)(x-2)2

Common pitfalls

Do not swap maximum and minimum. The gradient goes \(+\to-\) at a maximum (rising then falling) and \(-\to+\) at a minimum. A quick check: a maximum is a peak, a minimum is a trough.
A stationary inflection is not a turning point. If \(f'\) keeps the same sign either side of the stationary point, the curve does not turn — it is a stationary point of inflection, not a max or a min. A repeated (squared) factor in \(f'\) is the usual giveaway.
Stationary points need \(f'=0\). A curve can have a point of inflection where \(f'\neq0\) (an ordinary inflection with a sloping tangent). Only inflections with a horizontal tangent are stationary points of inflection.

Frequently asked questions

What is a stationary point?

A point where the gradient is zero, \(f'(x)=0\), so the tangent is horizontal. Solving \(f'(x)=0\) locates them.

What are the three types of stationary point?

A local maximum (peak), a local minimum (trough), and a stationary point of inflection (a flat spot where the curve does not turn).

How does the first-derivative test classify them?

Check the sign of \(f'\) either side: \(+\to-\) is a maximum, \(-\to+\) is a minimum, and the same sign both sides is a stationary point of inflection.

What is a stationary point of inflection?

A point where \(f'=0\) but \(f'\) does not change sign, so the curve levels off yet keeps going the same way while its concavity changes. It comes from a repeated factor in \(f'\).

How do you build a sign table?

Solve \(f'(x)=0\), then pick a test value in each interval between and beyond the solutions and record whether \(f'\) is positive or negative there.

Does the test work for \(e^x\), \(\ln x\) and trig curves?

Yes. Differentiate, solve \(f'(x)=0\), then test the sign of \(f'\) either side of each stationary point — the method is the same for every differentiable curve.