Turning Points and Optimisation (AI SL)
At a turning point the gradient of a curve is momentarily zero, which is exactly what \(f'(x)=0\) lets you locate. This page shows how to find and classify those points, then how the same idea is used to maximise or minimise a quantity in a real-world context - part of the wider Differentiation topic.
19 questions on this sub-topic.
Locating and classifying turning points
Covered under IB syllabus reference SL5.6. This isn't a formula-booklet result - it's a direct application of the derivative you already know how to find.
Stationary points
\(f'(x)=0\)
Solve this equation to find the \(x\)-values where the gradient is momentarily zero. Substitute back into \(f(x)\) to get the full coordinates.
Maximum, minimum, or neither?
Check the sign of \(f'(x)\) either side, or use your GDC's graphing tools.
A local maximum isn't automatically the greatest value of a function over its whole domain - always check the domain before stating a final answer.
For calculator steps to locate a maximum or minimum on the graph screen, see the parent topic's GDC guidance.
Worked examples
Find the coordinates of the stationary point of \(y = x^2 - 6x + 5.\)
Worked solution
\(\dfrac{dy}{dx} = 2x - 6 = 0 \Rightarrow x\) M1 \(= 3.\) A1
\(y = 9 - 18 + 5\) M1 \(= -4\), so \((3, -4).\) A1
A shop's daily revenue is \(R(p) = -4p^2 + 240p\) dollars, where \(p\) is the price of an item in dollars.
(a) Find \(R'(p).\)
(b)(i) Find the price that maximises revenue.
(b)(ii) Find the maximum revenue.
Worked solution
(a) \(R'(p) = -8p + 240.\) M1
\(R'(p) = -8p+240.\) A1
(b)(i) \(-8p + 240 = 0 \Rightarrow p = 30.\) M1
\(p = 30.\) A1
(b)(ii) \(R(30) = -4(900) + 240(30) = 3600\), so $3600. A1
Common mistakes
- Assuming the first solution of \(f'(x)=0\) is the one wanted. A cubic (or a piecewise context function) can have more than one stationary point - check which one the question is actually asking for, a maximum, a minimum, or a specific \(x\)-value.
- Stopping at the \(x\)-value instead of answering the question. In optimisation problems the derivative usually gives you the input (a price, a length, a time), but the question often wants the resulting maximum or minimum value itself - read the last line of the question again before writing a final answer.
- Ignoring a sensible real-world domain. A price, a length, or a time cannot be negative. If solving \(f'(x)=0\) produces a value outside the domain the context allows, it should be rejected.
Ready to practise properly?
20 turning-point and optimisation questions, marked instantly like the real exam.
Quick answers
How do you find the coordinates of a turning point?
Differentiate the function, set \(f'(x)=0\), and solve for \(x\). Substitute that \(x\)-value back into the original function to find the corresponding \(y\)-coordinate.
Is a local maximum always the greatest value of a function?
No. A local maximum or minimum only describes the function's behaviour near that point - it is not necessarily the greatest or least value over the whole domain, so always check the domain before giving a final answer.