Locate maxima and minima - the basis of optimisation questions.
In short: Stationary points are where a function's gradient is zero, so they include maxima, minima and points of inflection. On a GDC you can find them by graphing the function and using the maximum or minimum tool, or by finding where the derivative equals zero. Give both coordinates when the question asks for the point.
| TI-84 Plus CE | Casio fx-CG50 and fx-CG100 | TI-Nspire CX | |
|---|---|---|---|
| Key sequence | Graph, then 2nd → CALC → 3:minimum / 4:maximum (set left bound, right bound, guess). To solve f′(x)=0, enter nDeriv(Y1,X,X) as Y2 and find its zeros. | Graph then G-Solv → MIN / MAX; or solve d/dx(f(x))=0 with SolveN in Run-Matrix. | menu → Analyze Graph → Minimum / Maximum; or define f′ with the derivative template and solve f′(x)=0. |
Tip: Max/min on the graph gives both coordinates at once - usually faster than solving f′(x)=0 by hand.
Tip: Max/min on the graph gives both coordinates at once - usually faster than solving f′(x)=0 by hand.
Tip: Max/min on the graph gives both coordinates at once - usually faster than solving f′(x)=0 by hand.
Here's a real IB-style question that uses exactly this technique.
Find the coordinates of the local maximum and local minimum of f(x) = x³ − 3x² − 9x + 5.
Locate maxima and minima - the basis of optimisation questions. Finding the local maximum and minimum points of a cubic or higher-order polynomial. Checking stationary points found by setting f′(x) = 0 by hand.
1. Graph the function so you can see how many turning points there are. 2. Graph, then 2nd → CALC → 3:minimum / 4:maximum (set left bound, right bound, guess). To solve f′(x)=0, enter nDeriv(Y1,X,X) as Y2 and find its zeros. 3. Substitute each x back into f(x) for the y-coordinate.
1. Graph the function so you can see how many turning points there are. 2. Graph then G-Solv → MIN / MAX; or solve d/dx(f(x))=0 with SolveN in Run-Matrix. 3. Substitute each x back into f(x) for the y-coordinate.
1. Graph the function so you can see how many turning points there are. 2. menu → Analyze Graph → Minimum / Maximum; or define f′ with the derivative template and solve f′(x)=0. 3. Substitute each x back into f(x) for the y-coordinate.
Max/min on the graph gives both coordinates at once - usually faster than solving f′(x)=0 by hand.
In the worked example on this page (4 marks, Paper 2), the marks are: M1: Graph f(x) and use the maximum/minimum tools; A1: Local max (−1, 10); A1: Local min (3, −22).
Questions that need this technique link back to this guide. Try one in practice mode, or see every GDC guide.