IB Maths AA SLTopic 5 — CalculusPaper 1 & 2~9 min read
Derivatives & Graphs
Given the graph of f(x), can you sketch f′(x) and f″(x)? You don’t need the equation — just read off the features. Stationary points become roots. Inflection points become turning points. Concavity becomes sign. Once you know the translations, this is a skill, not a calculation.
📘 What you need to know
Stationary points of f(x) become roots of f′(x).
Points of inflection of f(x) become turning points of f′(x) AND roots of f″(x).
f(x) concave up ↔ f″(x) positive · f(x) concave down ↔ f″(x) negative.
Going backwards (from f′(x) to f(x)) loses some info — the y-intercept and roots of f(x) can’t be recovered.
The big picture
Each level shows what the next derivative does at the same x-value. Read down, then read across.
f(x), f′(x), and f″(x) stacked together
From f(x) to f′(x)
If f(x) has…
Then f′(x) has…
a stationary point (max, min, or inflection)
a root (touches/crosses x-axis)
a point of inflection
a turning point (max or min)
an increasing interval
positive y-values
a decreasing interval
negative y-values
a concave-up interval
an increasing section
a concave-down interval
a decreasing section
From f(x) to f″(x)
If f(x) has…
Then f″(x) has…
a point of inflection
a root (crosses x-axis)
a concave-up interval
positive y-values
a concave-down interval
negative y-values
If you’re going from f(x) to f″(x), it’s easiest to find f′(x) first, then differentiate again. The full chain f(x) → f′(x) → f″(x) is just two applications of the same rules.
Going backwards: from f′(x) to f(x)
If f′(x) has…
Then f(x) has…
a root
a stationary point
positive values
an increasing section (not necessarily positive)
negative values
a decreasing section (not necessarily negative)
⚠️
You can’t recover everything from f′(x)
Without knowing one specific point on f(x), you can’t determine: the y-intercept, the actual roots of f(x), or where it’s positive/negative. You can sketch the shape, but not the exact position.
🧠
“Sign of f′ = direction of f”
Wherever f′(x) is positive, f(x) is going up. Wherever f′(x) is negative, f(x) is going down. Where f′(x) crosses zero, f(x) is momentarily flat.
Worked examples
WE 1
Describe features of f′(x) from f(x)
The graph of y = f(x) has a local max at x = 1, a local min at x = 4, and a non-stationary point of inflection at x = 2.5. Describe the corresponding features of f′(x).
Translate each feature using the rules.Local max at x = 1:f′(1) = 0 → root of f′(x)Local min at x = 4:f′(4) = 0 → root of f′(x)Inflection at x = 2.5:f′ has a turning point at x = 2.5f′(x) has roots at x = 1 and x = 4, with a turning point at x = 2.5f′(x) looks like a parabola with roots at the stationary points!
WE 2
Match algebraic to graphical features
For f(x) = x³ − 3x², find the x-coordinates where f′(x) = 0 and where f″(x) = 0. Describe what these mean for the graph of y = f(x).
f′(x) = 0f′(x) = 3x² − 6x = 3x(x − 2) = 0x = 0 or x = 2 → stationary points of f(x)f″(x) = 0f″(x) = 6x − 6 = 0 → x = 1 → point of inflectionstationary at x = 0, 2 · inflection at x = 1f(x) goes: increasing → max at x=0 → decreasing → inflection at x=1 → decreasing → min at x=2 → increasing
WE 3
Find concavity intervals from f″(x)
Given f″(x) = 6x − 12, find the intervals where the graph of y = f(x) is concave up and concave down.
concave upf″(x) > 0 → 6x − 12 > 0 → x > 2concave downf″(x) < 0 → 6x − 12 < 0 → x < 2concave up: x > 2 · concave down: x < 2x = 2 is the point of inflection — concavity switches there!
WE 4
Identify f(x) features from f′(x)
The graph of y = f′(x) is a parabola with roots at x = −1 and x = 3, opening upward. Describe what f(x) looks like.
stationary pointsRoots of f′(x):x = −1 and x = 3 → stationary points of f(x)classify themf′(x) is negative between the roots → f(x) decreasing on (−1, 3)f′(x) positive outside → f(x) increasing for x < −1 and x > 3So at x = −1: increasing → decreasing → MAXAt x = 3: decreasing → increasing → MINf(x): max at x = −1, min at x = 3cubic shape with two turning points — but we can’t tell where it crosses the y-axis!
WE 5
Full feature analysis from algebra
For y = x⁴ − 4x³, find the stationary points and points of inflection. State what f′(x) looks like at each x-value.
stationary points (f′ = 0)f′(x) = 4x³ − 12x² = 4x²(x − 3)f′ = 0 at x = 0 (double root) and x = 3inflection candidates (f″ = 0)f″(x) = 12x² − 24x = 12x(x − 2)f″ = 0 at x = 0 and x = 2classify x = 0Both f′(0) = 0 AND f″(0) = 0 → could be horizontal inflection or stationary point — test sign change of f″:f″(−1) = 36 > 0, f″(1) = −12 < 0 → sign changes → horizontal inflection at x = 0classify x = 3f″(3) = 12(3)(1) = 36 > 0 → local minimumother inflection at x = 2f″(2) = 0 with sign change → non-stationary inflection at x = 2horizontal inflection (0, 0), local min (3, −27), inflection (2, −16)x = 0 is the rare case where stationary and inflection happen together!
💡 Top tips
Stationary points → roots, inflection → turning point on f′, concavity → sign of f″. Memorise these three.
Sketch from left to right, marking key x-values with vertical guide lines.
Check by differentiating if the function is given algebraically — it confirms your sketch.
Polynomial degree drops by 1 each time you differentiate. Cubic f(x) → quadratic f′(x) → linear f″(x).
Going backwards loses info — sketch the shape, but don’t claim exact y-values.
Use your GDC on Paper 2 to plot all three for comparison.
⚠ Common mistakes
Confusing roots of f′(x) with roots of f(x). They’re at completely different x-values usually.
Putting turning points of f′(x) at stationary points of f(x). Turning points of f′ are at inflection points of f.
Treating “f(x) positive” as “f(x) increasing”. They’re unrelated — direction is set by f′, not by f’s height.
Trying to sketch f(x) precisely from f′(x) without an extra known point. The y-intercept is unknown.
Forgetting that polynomial degree drops. If f(x) is degree n, f′(x) is degree n−1.
🎉 You’ve finished the Further Differentiation series! You can differentiate anything (chain, product, quotient), test stationary points, find concavity and inflection, and sketch the full f, f′, f″ family. Calculus is now a working toolkit.
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