IB Maths AI HLDifferentiationPaper 1 & 2~7 min read
Local Minimum & Maximum Points
A stationary point is where the curve momentarily levels off — f′(x) = 0. But flat alone isn’t the full story: is the curve sitting at the bottom of a dip (a local minimum) or the top of a hump (a local maximum)? That’s the point’s nature, and you decide it by checking how the gradient behaves on either side.
📘 What you need to know
Stationary point: anywhere f′(x) = 0 — the gradient is zero.
Local minimum: the lowest point in its immediate neighbourhood; gradient goes − → 0 → +.
Local maximum: the highest point in its immediate neighbourhood; gradient goes + → 0 → −.
“Local” means nearby — the curve may go higher or lower elsewhere.
Nature = whether a stationary point is a max or a min.
On the GDC: plot the graph and use “solve for minimum / maximum” to read the coordinates straight off.
What “local” really means
At every local max or min the curve flattens out, so f′(x) = 0 there. The word local is the key qualifier: it’s the highest (or lowest) point just around that spot, not necessarily over the whole graph.
local maximumtop of a humpGreatest value of f(x) in the local vicinity. Curve rises into it, then falls.
local minimumbottom of a dipLowest value of f(x) in the local vicinity. Curve falls into it, then rises.
Condition for any stationary pointf′(x) = 0
✗ not in booklet — it’s a definition you must remember
Many functions shoot off to ±∞ for large x, so a local maximum or minimum is rarely the “everywhere” maximum or minimum. Always read it as “the turning point near here”, not “the biggest value the function ever reaches”.
Deciding the nature
Once you’ve found a stationary point, you classify it by watching the sign of the gradient as you pass through it from left to right.
How the gradient flips through each turning point
Local max: gradient + → −. Local min: gradient − → +. Both sit exactly where f′=0.
Stationary point
Gradient before
At point
Gradient after
Curve does
Local maximum
positive (+)
0
negative (−)
increasing → decreasing
Local minimum
negative (−)
0
positive (+)
decreasing → increasing
🧠 “Max from above, min from below”
Picture walking left to right along the curve. At a max you climb up then drop down (+ then −). At a min you drop down then climb up (− then +). The sign of f′ tells you which.
Finding them on the GDC
For exam questions you usually locate stationary points graphically — quick and reliable on Paper 2.
🧭 Recipe — coordinates & nature of stationary points
Ploty = f(x) on your GDC and sketch it as part of your solution.
Solve for minimum using the graphing menu — read off the x and y coordinates of each minimum point.
Solve for maximum the same way for each maximum point.
State each point as coordinates and label its nature (local min or local max) — the shape on the graph makes this obvious.
🤔 Why doesn’t f′=0 alone tell you the nature?
Zero gradient just says “momentarily flat”. A hilltop and a valley floor are both flat at the very top/bottom — identical in that instant. The difference is entirely in what happens on either side: whether the curve was climbing or falling on the way in. That’s why nature comes from the sign change of f′, not from f′=0 by itself.
Worked examples
All parts use the SME function f(x) = x(x2 − 27).
WE 1
Expand and find f′(x)
Multiply out first, then differentiate term by term.
f(x) = x(x² − 27) = x³ − 27xf′(x) = 3x² − 27
WE 2
Find the x-coordinates of the stationary points
Stationary means f′(x) = 0 — solve the equation.
3x² − 27 = 0x² = 9 → x = ±3x = −3 and x = 3
WE 3
Find the y-coordinates
Substitute each x back into the original f(x) = x3 − 27x.
Sketch y = x(x2 − 27) on the GDC; the cubic rises, turns, falls, turns, rises. Read the shape (or check the sign change of f′).
At x = −3: gradient + → − (top of hump)At x = 3: gradient − → + (bottom of dip)(−3, 54) local maximum(3, −54) local minimum
WE 5
Confirm the nature using a sign check of f′
Test a value just either side of each stationary point in f′(x) = 3x2 − 27.
Near x = −3: f′(−4) = 48 − 27 = 21 (+), f′(−2) = 12 − 27 = −15 (−)+ → − ⟹ local maximum ✓Near x = 3: f′(2) = 12 − 27 = −15 (−), f′(4) = 48 − 27 = 21 (+)− → + ⟹ local minimum ✓The sign-change check agrees with the GDC sketch.
💡 Top tips
Always give coordinates, not just x — substitute back into f(x) for the y-value.
Sketch the graph on your GDC and include it — the nature is usually obvious from the shape.
Nature = sign change of f′: + → − is a maximum, − → + is a minimum.
Use “solve for min/max” on Paper 2 for fast, accurate coordinates.
Remember “local” — these are turning points nearby, not the function’s overall extremes.
⚠ Common mistakes
Stopping at the x-value — a stationary point is a coordinate, so find y too.
Swapping max and min — + → − is a maximum (hump), − → + is a minimum (dip).
Assuming local = global — the curve may climb higher or drop lower elsewhere.
Subbing into f′ for the y-value — that gives the gradient (0), not the coordinate; use f(x).
Sign slips with negatives — double-check (−3)³ = −27, not 27.
Next up — Modelling with Differentiation. You can now locate turning points and classify them as maxima or minima. The final topic of the unit puts that to work on real optimisation problems: writing a quantity in terms of one variable, setting its derivative to zero to find the best value, and justifying it as a true minimum or maximum — often with the second derivative instead of a sign check.
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