Turning Points and Optimisation (AI HL)
A stationary point is where a curve's gradient is momentarily zero - the top of a hill, the bottom of a dip, or a flat spot in between. Finding these points, and then working out whether each is a maximum, a minimum or neither, is the basis of optimisation: using calculus to find the best possible value of some quantity, such as the largest area or the lowest cost. It's part of the broader Differentiation topic.
37 questions on this sub-topic.
Two methods
Covered under IB syllabus reference SL5.6: values of \(x\) where the gradient of a curve is zero, solving \(f'(x)=0\), local maximum and minimum points, and the awareness that a local extremum need not be the greatest or least value on the whole domain.
Finding stationary points
Differentiate, set \(f'(x)=0\), and solve for \(x\).
Substitute each \(x\)-value back into \(f(x)\) to get the full coordinate pair. Not in the formula booklet - it's a method, not a formula.
Optimisation method
Write the quantity as a function of one variable, using a constraint to eliminate the other.
Differentiate, solve \(f'(x)=0\), confirm it's the required max/min, then answer the question actually asked. Not in the formula booklet - applied method.
Need the full syllabus wording and formula-booklet reference table? See Differentiation.
Worked examples
For \(f(x) = x^3 - 12x.\)
(a) Find \(f'(x).\)
(b) Find the interval on which \(f\) is decreasing.
Worked solution
(a) \(f'(x) = 3x^2 - 12.\) M1 A1
(b) \(f'(x) < 0 \Rightarrow x^2 < 4.\) M1 A1
Decreasing on \((-2, 2).\) A1
For \(y = x^3 - 3x^2.\)
(a) Find \(\dfrac{d^2y}{dx^2}.\)
(b) Find the coordinates of the point of inflexion.
Worked solution
(a) \(\dfrac{dy}{dx} = 3x^2 - 6x,\ \dfrac{d^2y}{dx^2}\) M1
\(= 6x - 6.\) A1
(b) \(6x - 6 = 0 \Rightarrow x\) M1
\(= 1.\) A1
\(y = -2\); inflexion \((1, -2).\) A1
Common mistakes
- Assuming every solution of \(f'(x)=0\) is a maximum or minimum. Some are points of inflexion where \(f''(x)=0\) too, and the second derivative test needs a different check (the sign of \(f'(x)\) either side) whenever \(f''(x)=0\).
- Stopping at the \(x\)-value. The question usually asks for coordinates or a maximum value, not just where the stationary point occurs - substitute back into the original function to finish the problem.
- Forgetting the domain in optimisation problems. A calculus maximum can fall outside the values that actually make sense in context (a negative length, for example) - always check the constraint before stating a final answer.
Ready to practise properly?
36 turning-point and optimisation questions, marked instantly like the real exam.
Quick answers
How do you find a stationary point?
Differentiate the function, set \(f'(x) = 0\), and solve for \(x\). Substitute each \(x\)-value back into \(f(x)\) to get the full coordinate pair.
How do you tell if a stationary point is a maximum or a minimum?
Check the sign of the second derivative there: \(f''(x) < 0\) means a local maximum, \(f''(x) > 0\) means a local minimum. If \(f''(x) = 0\), use a sign check of \(f'(x)\) either side of the point instead. See the parent Differentiation page's GDC guidance for graphing a function to check turning points visually.