Differentiate the gradient itself: The first derivative f'(x) is the gradient — how fast y is changing.
The second derivative f''(x) is the derivative of f'(x) — how fast the gradient itself is changing.
Picture a car: f' is its speed, f'' is its acceleration. A positive f'' means the gradient is getting steeper (curve concave up, like the inside of a valley); a negative f'' means the gradient is dropping (curve concave down, like the top of a hill).
A GDC is allowed throughout — you can get f'' at a point numerically, but you should be able to differentiate twice by hand too.
IB-style question — find the second derivative
A drone's height above the ground is modelled by h(t) = t³ − 6t² + 9t metres, where t is the time in seconds.
Find h''(t) and state the units.
Step by step
- Differentiate once for the velocity h'(t).
- Differentiate again for the acceleration h''(t).
- h'' is the rate of change of velocity, so it is an acceleration.
Final answer
h''(t) = 6t − 12, measured in metres per second per second (m s⁻²).
IB-style question — read off concavity
For the drone above, h''(t) = 6t − 12.
Is the height curve concave up or concave down at t = 1 s?
Step by step
- Substitute t = 1 into h''.
- h''(1) < 0, so the gradient is decreasing here.
Final answer
Concave down at t = 1 s (the height curve is dome-shaped there — the drone is slowing its climb).
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Where the bend switches direction: A point of inflexion is where the curve changes its concavity — from concave up to concave down, or the other way round.
Since concavity is controlled by the sign of f'', an inflexion happens where f'' changes sign. The first step is to solve f''(x) = 0, then check the sign actually switches across that point (it usually does in AI questions).
Real-world meaning: on a population or sales curve, the inflexion is the moment growth stops accelerating and starts slowing — the 'fastest growth' instant.
IB-style question — inflexion of an exponential model
The number of users (in thousands) of a new app t months after launch is modelled by N(t) = 2t(4 − et), for 0 ≤ t ≤ 2.
Find the value of t at the point of inflexion.
Step by step
- Expand first: N(t) = 8t − 2t·eᵗ. Differentiate using the product rule on 2t·eᵗ.
- Differentiate again (product rule on the last term).
- Set N''(t) = 0 and factor out −2eᵗ (which is never 0).
- Solve for t.
Final answer
Algebraically t = −2, which lies outside 0 ≤ t ≤ 2 — so within the modelled window there is NO inflexion; N'' stays negative (always concave down). Always check the answer is in the domain and interpret it.
IB-style question — a cubic with an inflexion
Find the x-coordinate of the point of inflexion of f(x) = x³ − 6x² + 5x + 2, and confirm concavity changes.
Step by step
- Differentiate twice.
- Solve f''(x) = 0.
- Check the sign of f'' on each side: f''(1) = −6 (concave down), f''(3) = 6 (concave up).
Final answer
Point of inflexion at x = 2; concavity changes from down to up, so it is a genuine inflexion.