IB Maths AI SLDifferentiationPaper 1 & 2stationary points~6 min read
Local Minimum & Maximum Points
A curve’s peaks and troughs — its local maximum and minimum points — are where it momentarily flattens out. At each one the gradient is zero. This note finds those points by solving f′(x) = 0, and decides the nature of each — peak or trough — from how the gradient changes around it.
📘 What you need to know
A stationary point is where the gradient is zero: f′(x) = 0. Local maxima and minima are both stationary points.
A local maximum is a peak — the highest point of the curve in its immediate neighbourhood.
A local minimum is a trough — the lowest point of the curve nearby.
Find a stationary point’s x-coordinate by solving f′(x) = 0; get the y-coordinate by substituting that x into f(x).
The nature of a stationary point is whether it is a maximum or a minimum.
At a maximum the gradient changes + → −; at a minimum it changes − → +.
Stationary points
A stationary point is a point on a curve where the gradient is zero — the curve is, for an instant, flat. Both local maximum points (peaks) and local minimum points (troughs) are stationary points.
Stationary points
at a stationary point, f′(x) = 0
local maxima and minima are stationary points — solve f′(x) = 0 to find their x-coordinates
“Local” matters: a local maximum is the highest point in its immediate neighbourhood, but the curve may climb higher elsewhere. The same is true of a local minimum.
Finding the coordinates
To locate the stationary points, differentiate and solve f′(x) = 0. Each solution is the x-coordinate of a stationary point.
For the matching y-coordinate, substitute that x-value back into the original function f(x). Write each stationary point as a coordinate pair (x, y).
Deciding the nature
The nature of a stationary point is whether it is a maximum or a minimum. The quickest test is the sign of the gradient just before and just after the point.
The nature of a stationary point
local maximum: f′(x) changes + → −
local minimum: f′(x) changes − → +check the sign of the gradient just before and just after the point
At a maximum the curve climbs, flattens, then falls — the gradient runs positive, zero, negative. At a minimum it falls, flattens, then climbs. A sketch or a GDC graph confirms the nature at a glance.
At a local maximum the gradient runs positive, zero, negative; at a local minimum it runs negative, zero, positive. At both, the tangent is flat.
🧭 Recipe — finding and classifying stationary points
Differentiate to find f′(x).
Solve f′(x) = 0 — the solutions are the x-coordinates of the stationary points.
Find each y-coordinate by substituting the x-value into f(x); write each point as (x, y).
Decide the nature: check the sign of f′(x) just before and just after each point.
State each point as a local maximum (+ → −) or local minimum (− → +) — a sketch confirms it.
Worked examples
WE 1
Coordinates of a stationary point
Find the coordinates of the stationary point of f(x) = x2 − 8x + 3.
a stationary point is where f′(x) = 0f′(x) = 2x − 82x − 8 = 0 ⇒ x = 4y-coordinate: f(4) = 16 − 32 + 3 = −13stationary point: (4, −13)solve f′(x) = 0 for the x-coordinate, then substitute into f(x) — not f′(x) — for the y-coordinate.
WE 2
Stationary point and its nature
Find the stationary point of f(x) = 5 + 4x − x2 and state its nature.
f′(x) = 4 − 2x4 − 2x = 0 ⇒ x = 2f(2) = 5 + 8 − 4 = 9 ⇒ point (2, 9)nature: f′ is + for x < 2 and − for x > 2gradient changes + → −, so (2, 9) is a local maximuma downward parabola has a single stationary point — its peak.
WE 3
Two stationary points
Find the coordinates of the stationary points of f(x) = x3 − 3x − 1.
f′(x) = 3x2 − 33x2 − 3 = 0 ⇒ x2 = 1 ⇒ x = 1 or x = −1f(1) = 1 − 3 − 1 = −3f(−1) = −1 + 3 − 1 = 1stationary points: (1, −3) and (−1, 1)a cubic typically has two stationary points — solve f′(x) = 0, then find each y-value separately.
WE 4
Stationary points and their nature
Find the stationary points of f(x) = 2x3 − 6x and state the nature of each.
f′(x) = 6x2 − 6 = 6(x − 1)(x + 1)f′(x) = 0 ⇒ x = 1 or x = −1f(−1) = −2 + 6 = 4 ⇒ (−1, 4)f(1) = 2 − 6 = −4 ⇒ (1, −4)nature: at x = −1, f′ changes + → −; at x = 1, − → +(−1, 4) local maximum · (1, −4) local minimumcheck the gradient’s sign on each side — for a positive cubic the left point is the maximum, the right one the minimum.
WE 5
A stationary point with a negative power
The function f(x) = x + 4/x is defined for x > 0. Find its stationary point and state its nature.
rewrite: f(x) = x + 4x−1f′(x) = 1 − 4x−2 = 1 − 4/x21 − 4/x2 = 0 ⇒ x2 = 4 ⇒ x = 2 (x > 0)f(2) = 2 + 4/2 = 4 ⇒ point (2, 4)nature: f′(1) = −3 < 0, f′(4) = 0.75 > 0gradient changes − → +, so (2, 4) is a local minimumrewrite the fraction as a negative power first; with x > 0, only the positive solution is valid.
WE 6
Full question: maximum height of a ball
A ball is thrown, and its height h metres after t seconds is modelled by h = 1 + 14t − 5t2. (a) Find dh/dt. (b) Find the time at which the ball reaches its maximum height. (c) Find that maximum height. (d) Justify that it is a maximum.
(a) differentiate the height modeldh/dt = 14 − 10t(b) maximum height where dh/dt = 014 − 10t = 0 ⇒ t = 1.4 s(c) substitute t = 1.4 into hh = 1 + 14(1.4) − 5(1.4)2 = 1 + 19.6 − 9.8h = 10.8 m(d) dh/dt is + before t = 1.4 and − after(a) 14−10t · (b) t = 1.4 s · (c) 10.8 m · (d) gradient + → −, so it is a maximum“maximum height” signals a stationary point — set the derivative to 0, then justify with the gradient’s sign change.
💡 Top tips
Stationary points come from f′(x) = 0 — differentiate first, then solve.
The y-coordinate comes from the original f(x), not from the derivative.
To classify a point, check the sign of f′(x) just either side: + → − is a maximum, − → + is a minimum.
A cubic usually has two stationary points; a quadratic has exactly one.
Words like “maximum height”, “minimum cost” or “greatest value” are a cue to set the derivative to zero.
⚠ Common mistakes
Finding the y-coordinate from f′(x) instead of the original f(x).
Stopping at the x-coordinate — a stationary point is a full coordinate pair (x, y).
Guessing the nature instead of checking the gradient’s sign change or the graph.
Confusing local and global — a local maximum need not be the highest point of the whole curve.
Dropping a valid solution of f′(x) = 0, or keeping one outside the function’s domain.
Next up: Modelling with Differentiation — using maxima and minima to solve real optimisation problems. The method here is the backbone of all of it: differentiate, solve f′(x) = 0, find the coordinates, then test the nature with the sign of the gradient.
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