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Locate a polynomial turning point numerically

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After a coarse table, what should happen next?

Year 11, VCE Units 1 & 2, Chapter 5, 4 difficulty levels, Calculator allowedBuild a worksheet

Estimating a turning point

Use a table to find where nearby outputs reach a local high or low, then refine the input values around that location

\[\text{coarse table}\;\rightarrow\;\text{smaller step}\;\rightarrow\;\text{reported coordinates}\]
  1. Compare nearby outputs A greatest nearby output suggests a local maximum and a least nearby output suggests a local minimum \(\displaystyle P(2)=15<P(2.25)=\frac{259}{16}\)
  2. Refine the neighbourhood After a coarse table, use a smaller input step around the best listed value \(\displaystyle x=2,\;2.25,\;2.5\)
  3. Keep coordinate roles The input is the horizontal coordinate and P(x) is the vertical coordinate \(\displaystyle (x,P(x))=\left(2.25,\frac{259}{16}\right)\)
  4. Round at the end Use the final retained table row and round both coordinates only to the requested precision \(\displaystyle x\approx2.3,\quad P(x)\approx16.2\)

Worked example

Consider the cubic model shown. Use a table to estimate its local maximum; give the coordinate pair to one decimal place

\[P(x)=-8x^3+33x^2-27x+1\]
  1. \(\displaystyle P(1)=-1,\quad P(2)=15,\quad P(3)=1\)The coarse table places the greatest nearby value near x=2.
  2. \(\displaystyle P(1.5)=\frac{31}{4},\quad P(2.5)=\frac{59}{4}\)A half-unit step checks the neighbourhood around the coarse best input
  3. \(\displaystyle P(2.25)=\frac{259}{16}=16.1875\)The refined table has its greatest listed value at x=2.25
  4. \(\displaystyle \left(x,P(x)\right)\approx\left(2.3,16.2\right)\)Rounding the retained input and output to one decimal place gives the reported coordinate

Common mistakes

Using the coarse value as final
✕\(\displaystyle P(2)=15\quad\Rightarrow\quad(x,P(x))=(2,15)\)✓\(\displaystyle P(2.25)=\frac{259}{16}>P(2)=15\quad\Rightarrow\quad\text{refine near }x=2\)
A coarse table identifies a neighbourhood; a smaller step is needed before reporting the final estimate
Swapping the coordinates
✕\(\displaystyle x=\frac{259}{16},\quad P(x)=2.25\)✓\(\displaystyle (x,P(x))=\left(2.25,\frac{259}{16}\right)\)
The input is the horizontal coordinate and the output is the vertical coordinate
Rounding too early
✕\(\displaystyle P(2.25)\approx16\)✓\(\displaystyle P(2.25)=\frac{259}{16}=16.1875\)
Keep exact table values until the requested rounding step

Difficulty levels

introChoose an interval between known features and complete one coarse value table
standardRefine one neighbourhood and report a maximum or minimum coordinate pair
challengeCompare multiple stationary-point neighbourhoods or use equal-output information to narrow the interval
examComplete several refinements and interpret a maximum or minimum using the stated domain

Questions about this topic

Why is a coarse table not enough?+

The largest or smallest listed output may be close to the stationary point but need not occur at the exact input. A smaller input step gives a closer estimate

Which coordinate is horizontal?+

The input x is the horizontal coordinate. The value P(x) is the vertical coordinate

Why must the output be rounded after the table is complete?+

Early rounding can change which nearby value is greatest or least. Retain the calculation values until the requested precision is applied

Can I use a calculator?+

A calculator is allowed unless a question is marked no calculator

Which year level and chapter is this?+

Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 5: Higher-degree polynomials

In this chapter

Interpret and determine a transformed power functionConnect a factorised higher-degree rule and its graphSolve a higher-degree polynomial equation or inequalityAnalyse a parameterised polynomial familyBracket and approximate a polynomial root by bisectionApproximate solutions from intersections of two graphsLocate a polynomial turning point numerically