Types of stationary points
In Year 12 Mathematical Methods (Queensland, QCAA), 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 \(f'(x)=0\).
A stationary point is where the first derivative is zero:
The first-derivative test reads the sign change of \(f'\) across the point:
A sign table lays this out. For example, with \(f'(x)=3(x-1)(x+1)\):
How to classify a stationary point with the first-derivative test
- Differentiate. Find \(f'(x)\) and, where possible, write it in factorised form — the factors show exactly where the sign can change.
- Solve \(f'(x)=0\). The solutions are the \(x\)-coordinates of the stationary points.
- 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.
- 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.
Differentiate, solve \(f'=0\), then test the signs.
| \(f'(x)\) | \(=\) | \(3x^{2}-3=3(x-1)(x+1)\) |
| \(f'(x)=0\) | \(\Rightarrow\) | \(x=-1,\ 1\) |
| \(x=-1:\ f'\) | \(+\to-\) | local max \((-1,2)\) |
| \(x=1:\ f'\) | \(-\to+\) | local min \((1,-2)\) |
Maximum at \((-1,2)\), minimum at \((1,-2)\).
Read the sign of each factor in a table.
| \(x<1\) | \((-)(-)\) | \(f'>0\) |
\(1| \((+)(-)\) | \(f'<0\) | |
| \(x>3\) | \((+)(+)\) | \(f'>0\) |
\(+\to-\) at \(x=1\): maximum. \(-\to+\) at \(x=3\): minimum.
The squared factor \(x^{2}\) never changes sign.
| \(f'(x)=0\) | \(\Rightarrow\) | \(x=0\ (\text{double}),\ x=2\) |
\(x<0,\ 0| \(\) | \(f'<0\) (no change at \(0\)) | |
| \(x>2\) | \(\) | \(f'>0\) |
\(x=0\): no sign change — stationary point of inflection. \(x=2\): \(-\to+\) — local minimum.
The squared factor \((x-2)^{2}\) is never negative.
| \(x<-1\) | \((-)(+)\) | \(f'<0\) |
\(-1| \((+)(+)\) | \(f'>0\) | |
| \(x>2\) | \((+)(+)\) | \(f'>0\) |
\(x=-1\): \(-\to+\) — local minimum. \(x=2\): no sign change — stationary point of inflection.
Common pitfalls
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.