Turning Points and Optimisation (AA HL)
Once you can differentiate, the real payoff is using that skill to answer "what's the biggest/smallest possible value?" questions - maximum area, minimum cost, the peak of a trajectory. This page covers finding and classifying stationary points, spotting points of inflexion, and setting up optimisation problems from a constraint. It's part of the broader Differentiation topic.
21 questions on this sub-topic.
The method
Covered under IB syllabus reference SL5.8. Neither technique below is a formula-booklet entry - both are applications of the derivative rules you already know.
Classifying stationary points
Solve \(f'(x) = 0\) to find candidate points. At each one, check the sign of \(f''(x)\): positive means a local minimum, negative means a local maximum. If \(f''(x) = 0\), fall back to checking the sign of \(f'(x)\) either side of the point instead.
Not in the formula booklet - techniqueOptimisation
Write the quantity to optimise as a function of one variable using a constraint, differentiate, set equal to zero, and check the result is a genuine maximum or minimum - or that an endpoint isn't actually better.
Not in the formula booklet - techniqueNeed the full syllabus wording and formula-booklet reference table? See Differentiation.
Worked examples
For \(f(x)=x^3-6x^2+5\):
(a) Find \(f''(x).\)
(b) Find the point of inflexion.
Worked solution
(a) \(f'(x) = 3x^2 - 12x,\) M1
\(f''(x) = 6x - 12.\) A1
(b) \(f''(x) = 0 \Rightarrow x = 2.\) M1
\(f(2) = -11,\) point \((2, -11).\) A1
For \(f(x) = x^3 - 6x^2 + 9x\), find and classify the stationary points.
(a)(i) State the coordinates of one stationary point.
(a)(ii) State the coordinates of the other stationary point.
Worked solution
Stationary points occur where the gradient is zero, so we need \(f'(x)\). \(f'(x)=3x^2-12x+9.\) A1
\(3(x^2-4x+3)=3(x-1)(x-3)=0\;\Rightarrow\;x=1,\,3.\) M1 A1
The sign of the second derivative tells us the concavity, hence max vs min. \(f''(x)=6x-12.\) M1
At \(x=1\): \(f''(1)=-6<0\Rightarrow\) local maximum at \((1,4)\). A1
At \(x=3\): \(f''(3)=6>0\Rightarrow\) local minimum at \((3,0)\). A1
Common mistakes
- Assuming \(f'(x)=0\) is enough. Solving \(f'(x)=0\) only finds candidate stationary points - you still need the second derivative (or a sign change) to say whether each one is a maximum, minimum, or neither.
- Forgetting \(f''(x)=0\) doesn't automatically mean a point of inflexion. You must also check that \(f''(x)\) actually changes sign either side of that point - if it doesn't, it isn't a genuine inflexion.
- Not checking endpoints in optimisation problems. On a restricted domain, the true maximum or minimum can sit at an endpoint rather than at a stationary point - always compare the stationary value against the boundary values.
Ready to practise properly?
21 turning-points-and-optimisation questions, marked instantly like the real exam.
Quick answers
How do you classify a stationary point?
Find where \(f'(x) = 0\), then check the sign of \(f''(x)\) at that point: \(f''(x) > 0\) means a local minimum, \(f''(x) < 0\) means a local maximum. If \(f''(x) = 0\) the second derivative test is inconclusive and you need a sign change of \(f'(x)\) instead.
What is a point of inflexion?
A point where the concavity of the curve changes, found by solving \(f''(x) = 0\) and checking \(f''(x)\) actually changes sign either side of it.
Need a calculator refresher? See Using your GDC on the full Differentiation page.