MISSION · Calculus
Concavity and Points of Inflection
The first derivative gives the slope, the second gives the curve
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The idea in one sentence
The second derivative measures how the slope itself is changing, which is the bend of the curve rather than its direction.
Two independent questions
Every point on a graph answers two separate questions, and confusing them is the main source of error here.
- Which way is it going? That’s \(f'\). Positive means rising.
- Which way is it bending? That’s \(f''\). Positive means concave up.
They combine into four cases, and each has a recognisable shape:
| \(f'\) | \(f''\) | Shape |
|---|---|---|
| + | + | rising, getting steeper |
| + | − | rising, flattening off |
| − | + | falling, levelling out |
| − | − | falling, getting steeper |
A car’s position gives the picture: \(f'\) is speed, \(f''\) is acceleration. You can be moving forwards while braking — rising and flattening — and that combination is not a contradiction.
Concave up and concave down
Concave up (\(f'' > 0\)): the curve holds water, like \(y = x^2\). The tangent line lies below the curve, and the slope is increasing — it might be going from −5 to −1 to 3, still increasing the whole way.
Concave down (\(f'' < 0\)): the curve sheds water, like a hill. Tangents lie above.
Note that concavity says nothing about whether the function is positive, or rising. A curve can be far below the axis, falling steeply, and still concave up.
Points of inflection
An inflection point is where concavity changes — the curve stops bending one way and starts bending the other.
\[y = x^3 - 3x^2 \quad \Rightarrow \quad y' = 3x^2 - 6x \quad \Rightarrow \quad y'' = 6x - 6\]Setting \(y'' = 0\) gives \(x = 1\). And \(y''\) genuinely changes sign there — negative to the left, positive to the right — so \(x = 1\) is a point of inflection.
That check is necessary. \(y = x^4\) has \(y'' = 12x^2\), which is zero at the origin but never changes sign: the curve is concave up on both sides, so there’s no inflection there. \(f'' = 0\) is a candidate, not a verdict.
The second derivative test
At a stationary point, where \(f' = 0\):
- \(f'' > 0\) — concave up, so it’s a minimum.
- \(f'' < 0\) — concave down, so it’s a maximum.
- \(f'' = 0\) — no information. Fall back to checking the sign of \(f'\) either side.
It’s usually the quickest way to classify a turning point, as long as you remember that the inconclusive case is real.
Sketching a curve
With both derivatives you can draw a graph without plotting points:
- Intercepts — set \(x = 0\), then \(y = 0\).
- Stationary points — solve \(f' = 0\), classify with \(f''\).
- Inflections — solve \(f'' = 0\), confirm the sign changes.
- End behaviour — what happens as \(x \to \pm\infty\).
For \(y = x^3 - 3x^2\): stationary at \(x = 0\) (max) and \(x = 2\) (min), inflection at \(x = 1\) exactly halfway between, and the ends run to \(\pm\infty\). That’s the whole shape.
Common mix-ups
- Reading \(f'' < 0\) as “decreasing”. That’s \(f' < 0\). \(f''<0\) means the slope is decreasing, which the function itself can do while rising.
- Calling every \(f'' = 0\) an inflection without checking the sign change.
- Using the second derivative test when it returns 0 and concluding there’s no turning point.
- Confusing steepest point with inflection. They coincide for these examples because the steepest slope is where the slope stops increasing — which is genuinely the same condition.
Check yourself
Find the inflection point of \(y = x^3 - 3x^2\), and describe a graph where \(f' > 0\) and \(f'' < 0\).
\(y'' = 6x - 6\), zero at x = 1, and the sign flips there, so that’s the inflection. With f′ positive and f″ negative the curve is rising but flattening — heading up, losing steam.
LEVEL 3 · REFLECT
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