A stationary point occurs where the derivative is zero. But stationary does not automatically mean maximum or minimum. The derivative tells you that the tangent is horizontal; classification requires more information about how the curve behaves around the point.
1. Finding stationary points
For y=f(x), solve f′(x)=0. Each solution gives a candidate x-coordinate. Substitute it into the original function to obtain the point’s y-coordinate.
Do not report only the x-value if the question asks for stationary points.
2. Worked quadratic example
For y=x²−6x+11, dy/dx=2x−6. Set this to zero:
2x−6=0, so x=3. Then y=9−18+11=2.
The stationary point is (3,2).
3. First-derivative sign test
Check the sign of f′(x) on either side of the stationary x-value.
- Negative → positive: local minimum.
- Positive → negative: local maximum.
- No sign change: not a turning maximum or minimum.
For the quadratic above, 2x−6 is negative before 3 and positive after 3, so (3,2) is a local minimum.
4. Stationary inflection example
For y=x³, dy/dx=3x². The derivative is zero at x=0, giving stationary point (0,0).
But 3x² is positive on both sides of zero. The function increases before and after the point. Thus (0,0) is a stationary point of inflection, not a maximum or minimum.
5. Second derivative test
Where applicable, the second derivative can classify a stationary point x=a:
- f′(a)=0 and f″(a)>0: local minimum.
- f′(a)=0 and f″(a)<0: local maximum.
- f″(a)=0: test is inconclusive; use another method.
6. Why the second derivative test works intuitively
A positive second derivative indicates the gradient is increasing locally, consistent with a curve bending upward. At a horizontal tangent, this produces a local minimum. A negative second derivative indicates locally decreasing gradient and downward bending, consistent with a maximum.
7. Worked cubic with two stationary points
Let y=x³−3x²−9x+5.
dy/dx=3x²−6x−9=3(x−3)(x+1).
Stationary x-values are −1 and 3.
At x=−1, y=−1−3+9+5=10. At x=3, y=27−27−27+5=−22.
Second derivative: d²y/dx²=6x−6.
At x=−1, f″=−12<0, so (−1,10) is a local maximum. At x=3, f″=12>0, so (3,−22) is a local minimum.
8. Stationary versus turning
A stationary point has horizontal tangent. A turning point changes the direction of monotonicity from increasing to decreasing or vice versa. Many school examples are both, but the concepts are not identical.
y=x³ at the origin is stationary but does not turn.
9. Nonstationary turning behaviour
Some functions outside the usual smooth A-Math examples can turn at a sharp point where the derivative does not exist. This reinforces why “turning point” and “stationary point” should not be treated as synonyms in general mathematics.
10. Graph sketching from derivative signs
If f′ is positive, then negative, then positive across successive intervals, the original graph increases, decreases, then increases. This predicts a local maximum followed by a local minimum.
A sign table can therefore convert algebraic derivative factors directly into graph shape.
11. Repeated factors in the derivative
If f′(x) contains a squared factor (x−a)², the derivative may touch zero without changing sign at x=a. This is a clue that the stationary point may be an inflection rather than an extremum.
Always inspect the full derivative sign, not only the fact that a factor becomes zero.
12. Domain boundaries are different
On a restricted domain, a maximum or minimum may occur at an endpoint even though the derivative there is not zero. Stationary-point analysis finds interior horizontal tangents; optimisation over a closed interval must also inspect boundaries.
13. Common mistakes
- Solving f(x)=0 instead of f′(x)=0.
- Finding x but not the corresponding y-coordinate.
- Calling every stationary point a maximum or minimum.
- Using f″(a)=0 to claim an inflection automatically.
- Ignoring endpoints in a restricted-domain extrema problem.
14. Independent practice
- Find and classify the stationary point of y=x²+4x+7.
- Find the stationary points of y=x³−3x.
- Explain why y=x³ has a stationary point at the origin but no local extremum there.
- If f′ changes from positive to negative at x=a, classify the point.
- If f′ changes from negative to positive at x=b, classify the point.
- Why is f″(a)=0 inconclusive by itself?
15. Answers
- dy/dx=2x+4=0 gives x=−2; y=3. f″=2>0, so (−2,3) is a minimum.
- dy/dx=3x²−3=0 gives x=±1. Points: (−1,2) and (1,−2). f″=6x, so (−1,2) maximum and (1,−2) minimum.
- f′=3x² is positive on both sides of zero, so the function continues increasing.
- Local maximum.
- Local minimum.
- Different curve behaviours can have second derivative zero; inspect derivative signs or higher/local behaviour.
16. What mastery looks like
A student has stationary-point control when they can find coordinates, distinguish stationary from turning behaviour, choose a classification test, interpret derivative signs and remember that restricted-domain extrema may also occur at boundaries.
Continue with Tangents and Normals, How to Check a Derivative, or return to the Additional Mathematics Hub.