01 · A quadratic minimum
Find and classify the stationary point of f(x) = x² − 6x + 5.
Hint
Solve 2x − 6 = 0.
Worked solution
(3,−4), a minimum since f″ = 2 > 0.
Understand · explore · practise
Stationary points: maxima, minima and inflection
Find stationary-point coordinates and classify them using sign changes or second derivatives. Handle inconclusive tests, domain restrictions, local extrema and ranges.
Before you startDifferentiation, factorisation, equations and graph sketching
01 / Find the coordinates
Solve f′(x) = 0, then return to f(x).
A stationary point is a point on the curve with derivative zero. Find its x-coordinate from the derivative equation; find its y-coordinate from the original function.
f(x) = x³ − 3x + 2
f′(x) = 3(x − 1)(x + 1)
x = −1 or 1
Stationary points: (−1,4), (1,0)
Explore the examples in the model. Each shows the function, its derivative signs and its second derivative at the stationary point. A flat tangent by itself does not settle the classification.
Compare different flat pointsMove at your pace
For x², the stationary point is (0,0). The first derivative is negative just to the left and positive just to the right. The point is a minimum. The second derivative there is 2. The second-derivative test confirms this classification.
Compare flat points that are minima, maxima and inflections
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / The first-derivative test
Look immediately on both sides of the candidate.
f′ changes + to −: local maximum
f′ changes − to +: local minimum
No sign change: no local turn
For f′(x) = 3(x − 1)(x + 1), the signs are +, −, + across the intervals split by −1 and 1. Thus (−1,4) is a local maximum and (1,0) a local minimum.
Use a sufficiently small neighbourhood with no other zero or domain break. Factor signs give a justification across the interval; isolated rounded calculator values are weaker evidence.
03 / The second-derivative test
Use it only after establishing a stationary point.
If f′(a) = 0:
f″(a) > 0 ⇒ local minimum
f″(a) < 0 ⇒ local maximum
For the cubic above, f″(x) = 6x. At x = −1 it is −6, confirming the maximum; at x = 1 it is 6, confirming the minimum.
A positive second derivative at a point with non-zero first derivative says the slope is increasing. It does not make that point a minimum.
04 / When f″ = 0
Zero means the test gives no decision.
At x = 0, each of x⁴, −x⁴ and x³ has f′ = 0 and f″ = 0. Their behaviour is different:
x⁴: derivative 4x³ changes − to + → minimum
−x⁴: derivative −4x³ changes + to − → maximum
x³: derivative 3x² stays positive → no turn
The cubic has a stationary inflection because its concavity changes across zero. Do not label every f″ = 0 input as an inflection; x⁴ is a counterexample.
05 / Stationary inflection
The curve flattens without reversing its direction.
For f(x) = (x − 2)³ + 1, f′(x) = 3(x − 2)². It is zero at x = 2 and positive on both sides. The curve rises through (2,1).
Also f″(x) = 6(x − 2) changes from negative to positive, confirming a change in concavity. This is a stationary inflection.
By contrast, x³ + x has an inflection at zero with gradient 1. “Inflection” describes changing concavity; “stationary” adds the separate condition f′ = 0.
06 / A cubic derivative
Factorisation can reveal three stationary points.
Let f(x) = x⁴ − 4x³ − 2x² + 12x + 1. Then:
f′(x) = 4(x³ − 3x² − x + 3)
= 4(x − 3)(x − 1)(x + 1)
You can find a root of the cubic derivative with the factor theorem, divide out its factor and solve the remaining quadratic. The stationary x-values are −1, 1 and 3.
f(−1) = −8; f(1) = 8; f(3) = −8
f″(x) = 12x² − 24x − 4
f″(−1) = 32; f″(1) = −16; f″(3) = 32
There are minima at (−1,−8),(3,−8) and a local maximum at (1,8). Since f(x) = [(x − 1)² − 4]² − 8, its range is [−8,∞). The local maximum is not a global maximum.
07 / Restricted domains and reciprocal branches
Classify only points that belong to the original curve.
For f(x) = x + 4/x, x ≠ 0, f′(x) = 1 − 4/x². It vanishes at x = ±2. Since f″(x) = 8/x³, (−2,−4) is a local maximum and (2,4) a local minimum.
The asymptote at zero separates the branches. These are not a greatest and least value over the whole domain.
A root function with a boundary
For g(x) = x − 4√x, x ≥ 0, g′(x) = 1 − 2/√x for x > 0. The stationary point is (4,−4) and g″(x) = x−3/2 > 0 there. The boundary x = 0 must be considered separately. In fact g(x) = (√x − 2)² − 4, so its least value is −4 and its range is [−4,∞).
08 / Local versus global
Check the whole allowed domain for a greatest or least value.
For f(x) = 4x − x² on 0 ≤ x ≤ 5, the stationary candidate is x = 2. Compare it with both endpoints:
f(0) = 0; f(2) = 4; f(5) = −5
The maximum is 4 and the minimum is −5, so the range is [−5,4]. The minimum occurs at a boundary, not a stationary point.
A positive-domain maximum
For g(x) = 18 − 16/x − x², x > 0, g′(x) = 16/x² − 2x. The stationary condition gives x³ = 8, so x = 2 and g(2) = 6. The derivative is positive before 2 and negative after it, proving a global maximum on the positive domain. The function tends to −∞ at both domain ends, giving range (−∞,6].
09 / Your turn
Coordinates and a justified classification are both needed.
State the relevant domain.
02 · A quadratic maximum
Find the greatest value of g(x) = 7 + 4x − x² for real x.
Hint
The stationary input is x = 2.
Worked solution
g(2) = 11; g″ = −2 < 0.
The downward quadratic has global maximum 11 and range (−∞,11].
03 · Two stationary points
Find and classify the stationary points of x³ − 12x + 1.
Hint
The derivative is 3(x − 2)(x + 2).
Worked solution
(−2,17): local maximum.
(2,−15): local minimum.
The second derivative is 6x.
04 · Zero second derivative
Classify the stationary point of y = x⁴ at the origin.
Hint
Use f′ = 4x³ on both sides.
Worked solution
It is a minimum: the derivative changes from negative to positive. f″(0) = 0 is inconclusive by itself.
05 · Shifted inflection
Find and classify the stationary point of f(x) = (x + 1)³ − 2.
Hint
f′ = 3(x + 1)².
Worked solution
(−1,−2), a stationary inflection.
The first derivative stays positive on both sides, and f″ = 6(x + 1) changes sign.
06 · A reciprocal restriction
Find stationary points of f(x) = x + 9/x, x > 0.
Hint
Solve 1 − 9/x² = 0, then apply the domain.
Worked solution
(3,6), a minimum since f″(3) = 18/27 > 0.
The algebraic root −3 is outside the domain.
07 · A quartic range
Find the least value and range of x⁴ − 4x².
Hint
Factor f′ = 4x(x² − 2) and compare the stationary values.
Worked solution
Minima at x = ±√2, with value −4.
Local maximum at x = 0, with value 0.
Range: [−4,∞).
Also x⁴ − 4x² = (x² − 2)² − 4.
08 · Boundary comparison
Find the range of f(x) = x² − 2x on 0 ≤ x ≤ 3.
Hint
Compare x = 1 with the endpoints.
Worked solution
f(0) = 0, f(1) = −1, f(3) = 3.
Range: [−1,3].
09 · Is it stationary?
f(x) = x³ + x has f″(0) = 0. Is the origin stationary?
Hint
Check f′(0).
Worked solution
No: f′(0) = 1. It is a non-stationary inflection because concavity changes, but its tangent is not horizontal.
10 · A false inference
A student finds one local maximum and claims it is the greatest value of a cubic with positive leading coefficient. Explain the problem.
Hint
Consider large positive x.
Worked solution
Such a cubic is unbounded above as x increases. A local maximum only compares nearby values, not all values on the real line.
10 / Recap
A flat tangent starts the investigation.
- Find x from f′ = 0 and y from f.
- Derivative sign changes classify local maxima and minima.
- Apply the second-derivative test only at stationary points.
- f″ = 0 is inconclusive.
- Check concavity before calling a point an inflection.
- Respect domain restrictions and compare boundaries for global extrema.
Section 1 of 10 · Find the coordinates