Turning Points and Optimisation (AA SL)
A curve turns where its gradient is momentarily zero. Finding these stationary points and working out whether each is a maximum, a minimum, or a point of inflexion is one of the most exam-heavy skills in Differentiation - it's also the engine behind every optimisation problem, from maximising the area of a fenced garden to minimising the cost of a can.
18 questions on this sub-topic.
Finding and classifying stationary points
Covered under IB syllabus reference SL5.8. There's no formula-booklet entry to look up here - the whole skill is a fixed method you apply to \(f'(x)\) and \(f''(x)\).
Stationary points
Solve \(f'(x)=0\), then use \(f''(x)\) (or a sign table for \(f'(x)\)) to classify each solution as a maximum, minimum or point of inflexion.
Points of inflexion
A point of inflexion is where \(f''(x)=0\) and the concavity changes sign either side of it. It can have a zero gradient (making it also a stationary point) or a non-zero gradient - so \(f''(x)=0\) alone is never enough to prove one exists.
Need the full syllabus wording and the rest of differentiation? See Differentiation.
Worked examples
Find the \(x\)-coordinate of the stationary point of \(y = x^{2} - 8x + 1\).
Worked solution
\(\dfrac{dy}{dx}=2x-8.\) A1
\(2x-8=0\Rightarrow x=4.\) M1 A1
For \(y = x^3 - 6x^2 + 9x\):
(a)(i) Find the coordinates of the stationary point where \(x>2.\)
(a)(ii) Classify its nature using the second derivative.
(b)(i) Find the coordinates of the other stationary point.
(b)(ii) Classify its nature using the second derivative.
Worked solution
(a)(i) Attempt to solve \(dy/dx=0\) via factorising: \(\dfrac{dy}{dx}=3x^2-12x+9=3(x-1)(x-3)=0\Rightarrow x=1,3.\) M1
Points \((1,4)\) and \((3,0)\); the stationary point with \(x>2\) is \((3,0).\) A1
(a)(ii) \(\dfrac{d^2y}{dx^2}=6x-12.\) At \(x=3\): M1
\(6>0\Rightarrow\) minimum. A1
(b)(i) The other stationary point is \((1,4).\) A1
(b)(ii) At \(x=1\): \(-6<0\Rightarrow\) maximum. A1
For \(f(x) = x^3 - 3x^2 - 9x + 2\):
(a)(i) Find the stationary point with \(x<1\).
(a)(ii) Find the stationary point with \(x>1\).
(b)(i) Classify the stationary point with the smaller \(x\)-value, by examining the sign of \(f'(x)\).
(b)(ii) Classify the stationary point with the larger \(x\)-value.
Worked solution
(a)(i) \(f'(x) = 3x^2 - 6x - 9 = 3(x-3)(x+1).\) M1
\(f'(x)=3(x-3)(x+1).\) A1
(a)(ii) Stationary at \(x = 3\) and \(x = -1.\) A1
(b)(i) For \(x < -1\): \(f' > 0\); for \(-1 < x < 3\): \(f' < 0\); for \(x > 3\): \(f' > 0\). R1
(b)(ii) So \(x = -1\) is a local maximum and \(x = 3\) is a local minimum. A1
Common mistakes
- Stopping at \(f'(x)=0\). Solving for \(x\) only gives half the answer - a stationary "point" needs a \(y\)-coordinate too, found by substituting back into the original function, not \(f'(x)\).
- Reading the sign of \(f''(x)\) backwards. \(f''(x)>0\) means concave up, which is a minimum; \(f''(x)<0\) means concave down, which is a maximum. Mixing these up is the single most common error on this skill.
- Assuming \(f''(x)=0\) proves a point of inflexion. It doesn't - the concavity has to actually change sign either side of that point. \(y=x^4\) at \((0,0)\) has \(f''(0)=0\) but is a minimum, not a point of inflexion, because the concavity stays positive on both sides.
Ready to practise properly?
18 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 back into the original function to find the \(y\)-coordinate.
How do you tell a maximum from a minimum?
Find \(f''(x)\) at the stationary point. If \(f''(x) > 0\) it is a local minimum; if \(f''(x) < 0\) it is a local maximum. If \(f''(x) = 0\) the second derivative test is inconclusive and you need a sign table for \(f'(x)\) instead. A GDC's minimum/maximum tool can also confirm this graphically - see using your GDC.