Finding and classifying stationary points. Second derivative test. Optimisation problems.
Answer at least 3 of 3 correctly to complete this section.
Imagine you are designing a cardboard box with no lid. You have a fixed sheet of cardboard, and you want the box to hold as much as possible. How do you find the best dimensions? You need stationary points — the places where a function stops increasing and starts decreasing (or vice versa).
Differentiation tells you the gradient. Stationary points are where the gradient is zero. This is one of the most important applications of calculus, and it appears on almost every A-Level paper.
A stationary point of a function f(x) is a point where the gradient is zero:
f′(x)=0
At a stationary point, the curve is momentarily flat — it is neither going up nor going down. There are three types:
| Type | What happens | f′(x) changes from |
|---|---|---|
| Local maximum | Curve goes up then down | positive → 0 → negative |
| Local minimum | Curve goes down then up | negative → 0 → positive |
| Point of inflection | Curve flattens but doesn’t change direction | same sign → 0 → same sign |
Key distinction: a stationary point is where f′(x)=0, not where f(x)=0. Setting f(x)=0 finds the roots (where the curve crosses the x-axis), which is a completely different thing.
Choose a function below and see where the stationary points are:
f(x)=x2−4x+3
Answer at least 3 of 4 correctly to complete this section.
The process is always the same:
Find the stationary points of f(x)=2x3−9x2+12x−4.
Step 1: Differentiate.
f′(x)=6x2−18x+12
Step 2: Set equal to zero.
6x2−18x+12=0
Step 3: Solve. Factor out 6 first:
x2−3x+2=0
(x−1)(x−2)=0
x=1orx=2
Step 4: Find the y-values.
f(1)=2−9+12−4=1
f(2)=16−36+24−4=0
The stationary points are (1,1) and (2,0).
Always simplify before solving. Factoring out common factors (like 6 above) makes the algebra much easier.
Answer at least 3 of 4 correctly to complete this section.
Once you have found the stationary points, you need to classify them. The second derivative test is the standard method.
Evaluate f′′(x) at each stationary point:
| f′′(x) at stationary point | Classification |
|---|---|
| f′′(x)>0 | Local minimum (concave up — like a cup) |
| f′′(x)<0 | Local maximum (concave down — like a hill) |
| f′′(x)=0 | Inconclusive — check further |
Why does this work? The second derivative measures how the gradient is changing. If f′′(x)>0, the gradient is increasing (going from negative to positive), so we are at the bottom of a dip. If f′′(x)<0, the gradient is decreasing (going from positive to negative), so we are at the top of a hill.
Classify the stationary points of f(x)=2x3−9x2+12x−4.
We found f′(x)=6x2−18x+12, with stationary points at x=1 and x=2.
Differentiate again:
f′′(x)=12x−18
At x=1: f′′(1)=12−18=−6<0 → local maximum at (1,1).
At x=2: f′′(2)=24−18=6>0 → local minimum at (2,0).
If the second derivative test is inconclusive, examine the sign of f′(x) on either side of the stationary point:
Common trap: f′′(x)=0 does not automatically mean it’s a point of inflection. For example, f(x)=x4 has f′′(0)=0 but x=0 is a minimum, not a point of inflection. You must always check.
Answer at least 3 of 4 correctly to complete this section.
Optimisation is the most common exam application of stationary points. You are given a real-world scenario and asked to maximise or minimise something.
A farmer has 100 m of fencing and wants to enclose a rectangular field against a straight wall (so only 3 sides need fencing). Find the maximum area.
Let x = the length perpendicular to the wall. Then the side parallel to the wall is 100−2x.
A(x)=x(100−2x)=100x−2x2
Differentiate: A′(x)=100−4x.
Set to zero: 100−4x=0⇒x=25.
Classify: A′′(x)=−4<0 → maximum.
Maximum area: A(25)=25×50=1250 m2.
Check: x must be positive and less than 50 (otherwise no fencing left for the parallel side). x=25 is in this range, so the answer is valid.
Answer at least 3 of 4 correctly to complete this section.
The curve C has equation y=x3−6x2+9x+1.
(a) Find dxdy.
(b) Find the coordinates of the stationary points of C.
(c) Determine the nature of each stationary point.
(a) dxdy=3x2−12x+9
(b) Set dxdy=0:
3x2−12x+9=0
x2−4x+3=0
(x−1)(x−3)=0
x=1 or x=3
At x=1: y=1−6+9+1=5. Point: (1,5).
At x=3: y=27−54+27+1=1. Point: (3,1).
(c) dx2d2y=6x−12
At x=1: f′′(1)=6−12=−6<0 → local maximum.
At x=3: f′′(3)=18−12=6>0 → local minimum.
Answer at least 3 of 4 correctly to complete this section.
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