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.
Think of a car: Picture a car: f' is its speed, f'' is its acceleration.
Positive f''
- The gradient is getting steeper
- Curve concave up, like the inside of a valley
Negative f''
- The gradient is dropping
- Curve concave down, like the top of a hill
GDC and by hand: 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.
Finding an inflexion
Solve f''(x) = 0
Concavity is controlled by the sign of f'', so an inflexion happens where f'' changes sign. The first step is to solve f''(x) = 0.
Check the sign switches
Check the sign of f'' 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.