IB Maths AA HLTopic 5 — CalculusPaper 1 & 2~7 min read
Derivatives & Graphs
Knowing the graph of f, you can sketch f′ and f″ — and vice versa. Two key relationships: roots of f′ correspond to stationary points of f; turning points of f′ correspond to points of inflection of f. The shapes of f, f′, f″ are linked through increasing/decreasing and concave-up/concave-down behaviour.
📘 What you need to know
Roots of f′ = stationary points of f (turning points or horizontal POIs of f).
Roots of f″ = points of inflection of f (where concavity changes).
Turning points of f′ = points of inflection of f (since at POI, f′ has a max or min).
f increasing ⇔ f′ > 0; f decreasing ⇔ f′ < 0.
f concave up ⇔ f″ > 0 ⇔ f′ increasing; f concave down ⇔ f″ < 0 ⇔ f′ decreasing.
Going from f to f′ to f″ is straightforward: each step you “differentiate the picture” once.
Going from f′ back to f works too, but: y-intercept and exact vertical position of f cannot be recovered (that’s a constant of integration).
Cubic POI shortcut: for any cubic, the POI is at the midpoint of its two stationary points (if it has them).
From f → f′
Sketching f′ from f is about reading off WHERE the slopes are zero, positive, and negative.
Where f…
… f′ has…
has a stationary point (max, min, or horizontal POI)
a root (crosses or touches the x-axis)
has a (non-stationary) point of inflection
a turning point (max or min of f′ itself)
is increasing on an interval
positive on the same interval
is decreasing on an interval
negative on the same interval
is concave up on an interval
increasing on the same interval
is concave down on an interval
decreasing on the same interval
From f → f″
The second derivative is “the derivative of the derivative” — apply the same rules a second time, going from f′ to f″.
Where f…
… f″ has/is…
has a point of inflection
a root (crosses the x-axis)
is concave up on an interval
positive on the same interval
is concave down on an interval
negative on the same interval
The visual chain f → f′ → f″
The three graphs aligned: f’s stationary points (x=±1) become f′’s roots; f’s POI (x=0) becomes f′’s minimum AND f″’s root. The vertical alignment makes the relationships obvious.
Going backwards: f′ → f
You can also reverse the chain — given the graph of f′, sketch f. The same rules apply in reverse, but with one caveat.
Where f′…
… f has/is…
has a root (touches or crosses)
a stationary point (location only — type depends on sign change)
is positive on an interval
increasing on the same interval
is negative on an interval
decreasing on the same interval
has a turning point (max or min)
a point of inflection (concavity changes)
Caveat — what cannot be recovered: the y-intercept of f, the exact vertical position of the curve, and the x-intercepts of f are NOT determined by f′ alone. The shape is fixed, but the curve can slide vertically (the constant of integration). Same goes for going from f″ back to f′.
🧭 Recipe — sketch f′ given the graph of f
Mark every stationary point of f — these become the x-intercepts of f′.
Mark every (non-stationary) POI of f — these become the turning points (max/min) of f′.
Identify intervals where f is increasing → f′ is positive there; intervals where f decreases → f′ negative.
Identify intervals where f is concave up → f′ is increasing; concave down → f′ decreasing.
Sketch a smooth curve for f′ obeying all the constraints above.
Worked examples
WE 1
Connect features of f, f′, f″ for a known cubic
For f(x) = x³ − 3x: (a) Find the stationary points and POI of f. (b) Hence state the x-intercepts and turning point(s) of f′(x).
Differentiatef′(x) = 3x² − 3 = 3(x − 1)(x + 1)f″(x) = 6x(a) Stationary points (f′ = 0)x = ±1 → f(−1) = 2, f(1) = −2f″(−1) = −6 < 0 → (−1, 2) local maxf″(1) = 6 > 0 → (1, −2) local minPOI (f″ = 0): x = 0; f(0) = 0 → (0, 0)(b) Features of f′x-intercepts of f′: x = −1 and x = 1 (= stationary points of f)turning point of f′: x = 0 (= POI of f); f′(0) = −3 → minimum of f′ at (0, −3)Stationary: (−1, 2) max, (1, −2) min; POI: (0, 0). f′ has roots at x = ±1 and a minimum at (0, −3)f’ is a parabola opening upward — exactly because f is a cubic with positive leading coefficient
WE 2
Quartic — full feature mapping
For f(x) = x⁴ − 4x²: find all stationary points, classify them, and find all points of inflection. Hence state how many x-intercepts f′(x) and f″(x) have.
Differentiate twicef′(x) = 4x³ − 8x = 4x(x² − 2)f″(x) = 12x² − 8 = 4(3x² − 2)Stationary points: f′ = 0→ x = 0, x = ±√2f(0) = 0; f(±√2) = 4 − 8 = −4f″(0) = −8 < 0 → (0, 0) local maxf″(±√2) = 24 − 8 = 16 > 0 → (±√2, −4) local minPOI: f″ = 0 → x = ±√(2/3)Count interceptsf′(x) has 3 x-intercepts (0, ±√2)f″(x) has 2 x-intercepts (±√(2/3))3 stationary points, 2 POIs. f′ has 3 roots; f″ has 2 rootsa quartic typically gives a “W shape” — two minima, one local max in between, two POIs
WE 3
Reverse: deduce features of f from a description
A cubic curve y = f(x) has a local maximum at x = −2 and a local minimum at x = 4. (a) State the x-coordinates where f′(x) = 0. (b) Find the x-coordinate of the point of inflection of f.
(a) Stationary points are roots of f′f′(x) = 0 at x = −2 and x = 4(b) For any cubic, POI is at the midpoint of its two stationary pointsx_POI = (−2 + 4) / 2 = 1Why this worksf′ is a quadratic with roots −2 and 4 → f′(x) = a(x + 2)(x − 4)f″(x) = a(2x − 2) = 0 at x = 1 ✓(a) f′ = 0 at x = −2, 4; (b) POI at x = 1midpoint shortcut works for ALL cubics — saves time on Paper 1
WE 4
Cubic with negative leading coefficient — full chain
For f(x) = −x³ + 6x² − 9x + 2: (a) Find all stationary points and classify them. (b) Find the point of inflection. (c) State the shape (parabolic) of f′(x) — does it open up or down? Where are its roots and turning point?
Differentiatef′(x) = −3x² + 12x − 9 = −3(x − 1)(x − 3)f″(x) = −6x + 12(a) Stationary pointsx = 1: f(1) = −1 + 6 − 9 + 2 = −2; f″(1) = 6 > 0 → local min (1, −2)x = 3: f(3) = −27 + 54 − 27 + 2 = 2; f″(3) = −6 < 0 → local max (3, 2)(b) POI: f″ = 0 → x = 2; f(2) = −8 + 24 − 18 + 2 = 0→ POI at (2, 0)(c) f′(x) = −3x² + 12x − 9 — parabolaleading coeff −3 < 0 → opens DOWNWARDroots at x = 1, 3 (= stationary points of f)vertex (turning point of f′) at x = 2; f′(2) = −12 + 24 − 9 = 3 → max at (2, 3)(a) min (1, −2), max (3, 2); (b) POI (2, 0); (c) f′ is a downward parabola, roots at 1, 3, max at (2, 3)since f’s leading coeff is negative, the cubic falls overall — local MIN comes before local MAX
WE 5
From f′ shape, deduce features of f
The graph of f′(x) is a parabola opening upward, with x-intercepts at x = 0 and x = 4. (a) State the x-coordinates of all stationary points of f and classify each as max or min. (b) Find the x-coordinate of the POI of f.
(a) Roots of f′ → stationary points of f at x = 0 and x = 4f′ is a parabola opening upward → f′ > 0 outside [0, 4], f′ < 0 inside (0, 4)f′(−1) > 0 (f increasing left of 0)f′(2) < 0 (f decreasing between 0 and 4)f′(5) > 0 (f increasing right of 4)Sign change of f′ at each rootat x = 0: f′ goes + → − → local MAX of fat x = 4: f′ goes − → + → local MIN of f(b) POI of f = where f″ changes sign = turning point of f′vertex of parabola f′ is at midpoint of roots: x = 2(a) local MAX at x = 0; local MIN at x = 4. (b) POI at x = 2“upward parabola for f′” means f’ starts +, dips to −, comes back to + — exactly the picture for “max-then-min” on f
WE 6
Constructive: build a function to fit a description
A cubic f(x) = ax³ + bx² + cx + d satisfies: f(0) = 2, f′(−1) = 0, f′(2) = 0, and f″ changes sign through x = 1/2 (negative to positive). Find one such function.
Set up the conditionsf(0) = d = 2f′(x) = 3ax² + 2bx + cf′(−1) = 3a − 2b + c = 0f′(2) = 12a + 4b + c = 0Subtract: (12a + 4b + c) − (3a − 2b + c) = 09a + 6b = 0 → b = −3a/2f″ sign change at x = 1/2 (concave down → up means a > 0)choose a = 2 → b = −33(2) − 2(−3) + c = 0 → 6 + 6 + c = 0 → c = −12Resulting cubicf(x) = 2x³ − 3x² − 12x + 2Verify the featuresf(−1) = −2 − 3 + 12 + 2 = 9 → local max (−1, 9)f(2) = 16 − 12 − 24 + 2 = −18 → local min (2, −18)f(1/2) = 1/4 − 3/4 − 6 + 2 = −9/2 → POI (1/2, −9/2)f(x) = 2x³ − 3x² − 12x + 2 (one valid answer)scaling all coefficients by any positive k gives infinitely many functions with the same shape; “find a” not “find the”
💡 Top tips
Vertical alignment — when sketching f, f′, f″ on three sets of axes, line them up vertically. Stationary points of f sit directly above the roots of f′, which sit directly above where f″ has its sign feature.
Cubic POI midpoint — for any cubic with two stationary points, the POI is at the midpoint of their x-coordinates. Saves time.
Read f′ as “slope graph”: where f’ is high, f is climbing fast; where f’ = 0, f is flat; where f’ is low (negative), f is falling.
Read f″ as “curvature graph”: where f″ > 0, the curve smiles; where f″ < 0, it frowns.
GDC verification — graph all three on the same screen; the qualitative relationships should match your sketch.
⚠ Common mistakes
Confusing “stationary point of f” with “root of f” — root of f is where f = 0 (curve crosses x-axis); stationary point of f is where f′ = 0 (horizontal tangent). Different things.
Thinking f′ has stationary points at f’s roots — no, f′ has roots at f’s stationary points. The relationship is shifted by one derivative.
Recovering exact f from f′ — you cannot. Vertical position is lost (constant of integration). The shape is determined; the height isn’t.
Forgetting the sign-change requirement for POI — turning point of f′ alone is not enough; the f′ must change from increasing to decreasing (or vice versa) for f to have a genuine inflection.
Sketching f′ pointing in the wrong direction — when f is decreasing, f′ is negative (BELOW the x-axis), not pointing downward.
That closes the Techniques & Applications of Differentiation chapter — you now have the full toolkit for differentiating any HL function (chain, product, quotient rules), classifying stationary points (f″ test), finding inflections, and sketching the f-f′-f″ chain. Up next: Integration — the reverse of differentiation. The ∫ symbol replaces d/dx, and instead of “find the slope at every point,” the question becomes “find the area under the curve.”
Need help with Calculus?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
