Learn stationary points the easy way, with plain English intuition, worked examples and an auto marked practice test. Find local maxima, minima and stationary points of inflection for VCE Maths Methods Units 3 and 4.
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Flat. That is the one word that gives away a stationary point. As a graph climbs and dips, there are special moments where, just for an instant, it stops going up or down and runs perfectly level. Those flat spots are where the most interesting things happen: the highest profit, the lowest cost, the turning of the tide. Stationary points are how calculus finds them, and the whole trick is asking one simple question of the gradient.
What a stationary point really is
A stationary point is any place on a curve where the gradient is zero. The tangent line there is horizontal, so the curve is momentarily flat. That is the entire definition. Because the gradient of f is given by its derivative f′(x), finding stationary points means finding where the derivative equals zero.
f′(x)=0
Every stationary point comes from solving this one equation. The most common mistake in the whole topic is skipping it: students differentiate, then forget to actually set the derivative equal to zero before solving.
The three kinds, and how to tell them apart
Finding where f′(x)=0 tells you that the curve is flat, but not which way it is flat. There are three possibilities, and knowing the difference is called finding the nature of the stationary point.
A local maximum is a peak: the curve rises up to it and then falls away. A local minimum is a valley: the curve drops down to it and then climbs back. A stationary point of inflection is the odd one out: the curve flattens for an instant but then carries on in the same direction it was already heading, like a brief pause on a staircase.
The word “local” matters. A local maximum is the highest point in its own neighbourhood, not necessarily the highest point on the whole graph.
The cubic f(x) = x³ − 3x² is flat at both marked points: a local maximum at (0, 0) where it rises then falls, and a local minimum at (2, −4) where it falls then rises.
The sign diagram of the derivative
The cleanest way to find the nature is a sign diagram of f′(x). The plan in plain English: the sign of the gradient on each side of the flat spot tells you whether the curve is rising or falling there, and that reveals the shape.
Mark each stationary point on a number line, then test a value of f′(x) in each interval between and beyond them. You only care about the sign, plus or minus, not the size.
Gradient goes positive then negative (+0−): the curve rises then falls, so you have a local maximum.
Gradient goes negative then positive (−0+): the curve falls then rises, so you have a local minimum.
Gradient keeps the same sign on both sides (+0+ or −0−): the curve never turns around, so you have a stationary point of inflection.
A handy warning sign appears in the algebra. When f′(x) has a repeated (squared) factor, such as (x−2)2, that factor cannot change sign, so the stationary point it produces is a point of inflection rather than a turning point. Spotting this early saves you from miscounting turning points.
How to actually do it
The same short recipe works every time.
Differentiate to get f′(x), then set f′(x)=0.
Solve for x. These are the x coordinates of the stationary points.
Substitute each x back into the original f(x) to get the matching y value, so you have full coordinates.
Build a sign diagram of f′(x) and read off the nature of each point.
Two traps catch students again and again. The first is stopping at step 2 and handing in only the x value when the question asks for coordinates or for the maximum value. If a question asks for the maximum value, the answer is the y coordinate, not the point and not the x. The second is forgetting step 1’s equals zero, which leaves you solving f(x)=0 by accident and finding intercepts instead of stationary points.
Lock it in with active recall
Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
What equation locates every stationary point of f?
f′(x)=0 — the gradient is zero where the curve is flat. Differentiate first, then set the derivative to zero.
A sign diagram of f′(x) reads +0− across a stationary point. What is its nature?
A local maximum — the curve rises then falls, so it reaches a peak.
What does −0+ in the sign diagram of f′(x) tell you?
A local minimum — the curve falls then rises, so it bottoms out in a valley.
The gradient has the same sign on both sides of a flat spot (+0+ or −0−). What kind of point is it?
A stationary point of inflection — the curve flattens but keeps heading the same way, so it never turns around.
What algebraic feature in f′(x) warns you a stationary point is an inflection, not a turning point?
A repeated (squared) factor, such as (x−2)2. A squared factor cannot change sign, so the gradient does not flip across that point.
A question asks for the maximum value of f. What number do you give?
The y coordinate of the local maximum — not the x value and not the full point. Substitute the stationary x back into f(x).
Recall · Rates of Change and Optimisation
The second-derivative test: where f′(x)=0, what do f′′(x)>0 and f′′(x)<0 tell you?
f′′(x)>0 is concave up like a cup ∪ — a local minimum; f′′(x)<0 is concave down like a cap ∩ — a local maximum. If f′′(x)=0, fall back to a sign diagram of f′.
Recall · Chain Rule
To find stationary points of a composite like sin(sin2x), what rule gives you f′(x)?
The chain rule: (f∘g)′(x)=f′(g(x))⋅g′(x). Differentiate the composite, then set the result to zero as usual.
See this recipe in action in the Worked Examples tab, then test yourself in Try It.
Worked examples
Worked Example 1Stationary points of a composite, from a real exam
With f(x)=sin(x) and g(x)=sin(2x), find the x-values of the stationary points of f∘g on x∈[0,2π], and the range of f∘g on that interval.
1
Form the composite (f∘g)(x)=sin(sin2x) and differentiate with the chain rule.
(f∘g)′(x)=cos(sin2x)⋅2cos(2x)
2
Since sin(2x)∈[−1,1], which never reaches ±2π, the factor cos(sin2x) is always positive, so it is never zero. Stationary points therefore need cos(2x)=0.
cos(2x)=0⟹2x=2π+kπ
3
Solve for x within [0,2π].
x=4π,43π,45π,47π
4
For the range, sin(2x)∈[−1,1], so sin(sin2x) ranges over [−sin1,sin1].
range=[−sin1,sin1]
Answer
x=4π,43π,45π,47π;range=[−sin1,sin1]
VCAA 2024 Mathematical Methods Exam 2, Section B Q5b
Worked Example 2Verifying a shared turning point, from a real exam
For f(x)=12−x(x−2)2 and g(x)=12x−3x2, verify that f(x) and g(x) both have a turning point at P, and give the co-ordinates of P.
1
Differentiate g(x)=12x−3x2 and set the derivative to zero to find its turning point.
g′(x)=12−6x=0⟹x=2,g(2)=24−12=12
2
Expand and differentiate f(x)=12−(x3−4x2+4x), then check x=2.
f′(x)=−(3x2−8x+4),f′(2)=−(12−16+4)=0
3
Evaluate f(2) to confirm both curves turn at the same point.
f(2)=12−2(0)2=12⟹P=(2,12)
Answer
P=(2,12)
VCAA 2023 Mathematical Methods Exam 1, Q9b
Worked Example 3A cubic with two stationary points
Find the coordinates and nature of all stationary points of f(x)=x3−3x2.
1
Differentiate, then set the derivative equal to zero. This is the step students most often forget.
f′(x)=3x2−6x=3x(x−2)=0
2
Solve for x. These are the x values where the curve is momentarily flat.
x=0orx=2
3
Find the matching y values so you can give full coordinates, not just x.
f(0)=0,f(2)=8−12=−4
4
Build a sign diagram of f′(x). Test a point in each interval: at x=−1, f′=9>0; at x=1, f′=−3<0; at x=3, f′=9>0.
+0−0+
5
Read the nature off the sign change. Positive to negative is a local maximum; negative to positive is a local minimum.
(0,0) is a local maximum,(2,−4) is a local minimum
Answer
(0,0) local maximum,(2,−4) local minimum
Worked Example 4A stationary point of inflection
Find the stationary point of f(x)=x3+1 and state its nature.
1
Differentiate and set equal to zero.
f′(x)=3x2=0⟹x=0
2
Find the y value to give the full coordinate.
f(0)=1
3
Sign diagram of f′(x)=3x2. At x=−1, f′=3>0; at x=1, f′=3>0. The sign does NOT change.
+0+
4
Same sign on both sides means the curve flattens but keeps heading the same way. That is a stationary point of inflection.
(0,1) is a stationary point of inflection
Answer
(0,1) stationary point of inflection
Worked Example 5A repeated factor warns you
The derivative of a function is f′(x)=(x+1)(x−2)2. Find the nature of each stationary point.
1
Set f′(x)=0 to locate the stationary points.
(x+1)(x−2)2=0⟹x=−1orx=2
2
Sign diagram. Across x=−1 the simple factor flips the sign; at x=−2, f′<0 and at x=0, f′>0.
x=−1:−0+
3
The squared factor (x−2)2 is never negative, so the sign of f′(x) does not change across x=2. At x=1.5, f′>0 and at x=3, f′>0.
x=2:+0+
4
Negative to positive gives a local minimum; no sign change gives a stationary point of inflection.
x=−1 local minimum,x=2 stationary point of inflection
Answer
x=−1 local minimum,x=2 stationary point of inflection
Practice questions
Practice test
Try it yourself
9 questions, 14 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.Let g:R→R be defined by g(x)=4x3−3x4. Find the coordinates of both stationary points of g.
2marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
Solving g′(x)=12x2−12x3=12x2(1−x)=0 gives x=0 and x=1,
with coordinates (0,0) and (1,1). Both coordinates are required; giving only
the x-values lost marks.
VCAA 2025 Mathematical Methods Exam 2, Section B Q1a
Q2.Let f:R→R, f(x)=x(x−2)(x+1). Find the coordinates of the stationary points of f.
2marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
Differentiating f(x)=x3−x2−2x gives f′(x)=3x2−2x−2, with stationary
points at x=31±7 and exact y-values −2720±147.
Exact answers were required; some students gave only the x-values.
VCAA 2023 Mathematical Methods Exam 2, Section B Q1b
Q3.Consider the case where w>0. Find, in terms of w, the coordinates of the minimum point of the graph of y=(x−1)(x−w).
2marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
By symmetry the minimum lies on the axis of symmetry, midway between the roots
x=1 and x=w, so x=21+w. Substituting gives
y=(2w−1)(21−w)=−4(w−1)2,
so the minimum point is (21+w,−4(w−1)2). The report
notes students using symmetry generally found the x-coordinate correctly, but many
made errors combining fractions for the y-coordinate.
VCAA 2025 Mathematical Methods Exam 1, Q9b.i
Q4.The x coordinates of the stationary points of f(x)=x3−12x are:
1mark
Need a hint?
Stationary points come from solving f′(x)=0, not f(x)=0. Differentiate first, then solve.
Show worked solution
f′(x)=3x2−12=0 gives x2=4, so x=±2. Option C comes from solving f(x)=0 instead of f′(x)=0, the most common slip on this type of question.
Q5.A function has f′(x)=3(x−1)2. The point where x=1 is a:
1mark
Need a hint?
A squared factor can never go negative, so ask whether the gradient actually changes sign across x=1.
Show worked solution
f′(x)=3(x−1)2≥0 for all x, so the derivative is zero at x=1 but does not change sign. No sign change means a stationary point of inflection. A repeated (squared) factor is the warning sign that it is not a turning point.
Q6.The sign diagram of f′(x) reads +0− across x=3. At x=3 the graph of f has a:
1mark
Need a hint?
Positive gradient means rising, negative means falling. Sketch what rising-then-falling looks like.
Show worked solution
The gradient is positive (curve rising) then zero then negative (curve falling). Rising then falling means the curve reaches a peak, so x=3 is a local maximum. Reading the sign change the wrong way round is what produces the local minimum distractor.
Q7.The local minimum of f(x)=x2−6x+5 has coordinates:
1mark
Need a hint?
The question asks for coordinates, so once you find x, substitute it back into f(x) for the y value.
Show worked solution
f′(x)=2x−6=0 gives x=3, then f(3)=9−18+5=−4, so the point is (3,−4). Stopping at x=3 and forgetting to substitute back for the y value is a frequent error; the question asks for coordinates, so both numbers are needed.
Q8.The maximum value of f(x)=−x2+4x+1 is:
1mark
Need a hint?
The maximum *value* is a single number, the y coordinate, not the x value and not the full point.
Show worked solution
f′(x)=−2x+4=0 gives x=2, and f(2)=−4+8+1=5. The maximum value is the y coordinate, 5. Option B gives the coordinates of the turning point rather than the value, and option A mistakes the x coordinate for the answer.
Q9.For f(x)=x3−3x2−9x, find the coordinates of all stationary points and state the nature of each.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
Set the derivative equal to zero:
f′(x)=3x2−6x−9=3(x2−2x−3)=3(x−3)(x+1)=0
so x=3 or x=−1.
The y values are f(3)=27−27−27=−27 and f(−1)=−1−3+9=5.
Sign diagram of f′(x): at x=−2, f′(x)>0; at x=0, f′(x)=−9<0; at x=4, f′(x)>0. So the pattern is +0−0+.
Across x=−1 the gradient goes positive to negative, so (−1,5) is a local maximum. Across x=3 the gradient goes negative to positive, so (3,−27) is a local minimum.
Frequently asked questions
How do I find stationary points of a function?
Differentiate the function to get the gradient, set that derivative equal to zero, and solve for x. Each solution is the x coordinate of a stationary point. Substitute each x back into the original function to get the full coordinates.
How do I tell whether a stationary point is a maximum, minimum, or inflection?
Use a sign diagram of the derivative. Test the sign of the gradient just before and just after the point. Positive then negative is a local maximum, negative then positive is a local minimum, and the same sign on both sides is a stationary point of inflection.
What is the difference between a stationary point of inflection and a turning point?
At a turning point the curve changes direction, rising to falling or falling to rising, so the gradient changes sign. At a stationary point of inflection the curve flattens but keeps heading the same way, so the gradient does not change sign. A repeated squared factor in the derivative is the usual giveaway for an inflection.
