Higher: find stationary points and classify original graph shape
Use differentiation to locate stationary coordinates, then distinguish maxima, minima and non-turns from the original curve’s general shape.
Exam scope and assessment notes
Higher only: Foundation has no section3.4 Calculus requirement. Higher also includes all shared Foundation content. These six original lessons teach variable/local versus interval rates, termwise integer-power differentiation, gradients/rates, stationary coordinates and graph-shape classification, one-dimensional kinematics and other simple practical models. H:3.4C requires both the gradient/rate and stationary-point lessons; H:3.4E requires both the motion and other-practical lessons with worked examples and all Guided, Independent and Transfer stages. The six-lesson split is StudyForA’s teaching design, not a syllabus-prescribed count. Small negative-integer powers with x≠0 conservatively retain integer-power breadth; fractional/trigonometric differentiation, integration and product/quotient/chain rules are not inferred requirements. H:3.4D classifies maxima and minima by general original-graph shape only; a second derivative is used for acceleration in motion, not as a substitute classification test. A derivative zero supplies an input, not the original stationary ordinate or an automatic turning point. Keep every original domain, physical restriction, admissible integer count and endpoint. Continuous relaxation supports differentiation for discrete counts; it does not make noninteger production feasible. Accepted boundary maxima need not be stationary. The separate H3.3D drawn-tangent construction is not replaced or newly credited by calculus. Fixed interactive examples support exact algebraic self-checks and approximate graph exploration; no general expression/root solver or written-work marker is supplied. Written work is not automatically marked. Attempt acknowledgements and revealed references do not award marks, prove a construction or establish mastery. All five Calculus source rows have structural teaching and staged-practice links; exhaustive subskill approval, specialist approval and cumulative formal assessment remain unestablished. The wider Maths A course and other Pearson qualifications remain incomplete. Publisher modular-map tags and uncorrected proof navigation are separate evidence: Book2 Graphs9 map pages386–415 differs from proof start385 and its later calculus subheadings. Full final-book and Teaching Hub body/pagination alignment remains unverified; chapter links confer no full-book or modular teaching credit.
Before you begin
- Differentiate integer-power sums and factor simple expressions.
- Read and sketch quadratic, cubic and quartic graph shapes.
- Distinguish original values from derivative values and retain allowed domains.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For f(x)=x²+4, what is the original stationary point?
What you will learn
- Solve f′(x)=0 and recover every permitted stationary point on the original curve.
- Classify stationary points from the original graph’s bowl, hill or rising-through shape.
- Distinguish local from global extrema and a stationary non-turn from a maximum or minimum.
f(x)=x² − 6x + 11. Original domain: x in all real inputs. Original values use output units; first rates use output units per input unit.
The panels share the same numbered horizontal scale and have independent vertical scales. Original height, derivative value and derivative zero have different meanings. No input, derivative solution or reference point is preselected.
No input is preselected. Choose a point on your own drawing or a rate input you want to check.
Blue: original; purple: first derivative. Vertical scales differ. Filled boundary markers are included; open markers show excluded finite limits or holes. Traces split at original exclusions and are clipped to each panel; offscreen points are omitted rather than moved to a border. A wider view does not change a domain.
Find the full stationary coordinates
At a stationary point of a differentiable curve, the local gradient is zero. Differentiate the original function, solve f′(x)=0 and retain every solution in the original domain. Then substitute each input into f, not f′, to find its original ordinate. The derivative zero point (r,0) is generally not the original stationary point (r,f(r)).
Factor carefully and keep every potentially zero factor. A quartic can have several stationary inputs, so a first root is not necessarily the whole answer. If the original function excludes an input, a derivative expression cannot make it a valid stationary point. Mark the accepted original coordinates on a sketch of the original curve.
Classify by the original curve’s general shape
Inspect the original curve around each stationary point. A bowl-shaped part falls into a point and rises away from it, giving a minimum. A hill-shaped part rises into a point and falls away, giving a maximum. Draw the relevant curve on both sides and use original function values to check the depicted shape. Classification here uses the general graph shape, not a second-derivative test.
A curve can flatten and continue rising through a stationary point, with no turn at all. A zero derivative therefore finds a candidate stationary point rather than automatically proving a maximum or minimum. A constant function has derivative zero throughout its domain: every point is stationary, but it has no isolated turning-point list. A nonzero constant derivative gives no stationary point.
Local behaviour is not automatically global
A local maximum is higher than nearby points; it need not be higher than every point of the full function. A cubic can have a local hill and a local valley while continuing without bound in its end directions. A W-shaped quartic can have two equal minima and a middle local maximum, while its outer arms rise above that middle height.
Keep the original domain and the visible window distinct. A crop can hide a higher part of a curve, and a restricted model can have a largest boundary value without a stationary derivative there. For this lesson’s all-real polynomial examples, use the overall curve shape to justify any global claim; do not infer it from a small view around one point.
Sources for this lesson
Pause and explain
You know only that f′(2)=0. Which classification is justified immediately?
Worked example
For f(x)=x²−6x+11 on all real x, use differentiation to find the stationary point. Recover its full original coordinates, draw the original graph and classify the point from its upward-opening shape. Show the derivative zero separately and do not use a second-derivative classification test.
Show the worked solution
- Differentiate: f′(x)=2x−6. Set the gradient to zero: 2x−6=0 gives x=3, which belongs to the all-real domain.
- Return to the original function: f(3)=9−18+11=2. Mark the original stationary point (3,2), while the corresponding derivative zero is (3,0). These heights have different meanings.
- The original graph is the upward bowl f(x)=(x−3)²+2. It falls into (3,2) and rises away from it; for example f(2)=f(4)=3. Thus the point is a minimum, and the square form shows it is the global minimum on all real x. The classification comes from that original graph shape.
Answer Stationary point (3,2), a minimum and global lowest point on the all-real domain. The derivative is f′=2x−6, with zero point (3,0).
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find every stationary point of f(x)=x³−3x²−9x+5 on all real x. Recover full original coordinates, draw the original cubic around both points and identify the local maximum and local minimum from its rising/falling shape. Mark derivative zeros separately; do not use a second-derivative test.
Give me a hint
Factor f′ completely, retain both inputs and substitute into f. Use original neighbouring heights and the cubic’s general shape to identify its hill and valley.
Compare my reasoning
- Differentiate and factor: f′(x)=3x²−6x−9=3(x+1)(x−3). Its permitted zero inputs are −1 and 3.
- The original heights are f(−1)=−1−3+9+5=10 and f(3)=27−27−27+5=−22. Thus the original stationary points are (−1,10) and (3,−22), while the derivative zeros are (−1,0) and (3,0).
- Draw the original cubic: its hill near x=−1 has neighbouring heights f(−2)=3 and f(0)=5, both below 10; its valley near x=3 has f(2)=−17 and f(4)=−15, both above −22. The overall curve rises into then falls from the first point, and falls into then rises from the second. They are a local maximum and local minimum, not global extrema of this unbounded cubic.
Local maximum (−1,10); local minimum (3,−22). Derivative f′=3(x+1)(x−3). Classification uses the original cubic’s hill and valley shape.
Look for these in your work
- Retain both derivative roots and recover both original heights.
- Draw the original hill and valley rather than classifying from a derivative height.
- Use general original graph shape and keep local/global meanings distinct.
For f(x)=x⁴−8x²+6 on all real x, find all stationary points and draw the original W-shaped graph with its stationary coordinates. Mark the corresponding derivative zeros separately. Classify from the original shape and explain why the middle local maximum is not a global maximum. Do not use a second-derivative test.
Give me a hint
Factor the derivative without cancelling x. Recover every original height, then inspect both valleys, the middle hill and the outer arms of the original quartic.
Compare my reasoning
- Differentiate: f′(x)=4x³−16x=4x(x−2)(x+2). The three stationary inputs are −2,0,2; none may be discarded from the all-real domain.
- Substitution into f gives f(−2)=16−32+6=−10, f(0)=6 and f(2)=−10. Mark (−2,−10), (0,6), (2,−10) on the original graph and (−2,0), (0,0), (2,0) on the derivative graph.
- The original values at x=−3,−2,−1,0,1,2,3 are 15,−10,−1,6,−1,−10,15, supporting the W shape. Its two valleys are minima; f=(x²−4)²−10 shows they share the global lowest height. The middle point is only a local maximum: the outer arms rise above 6, with f(3)=15 already a counterexample to a global maximum claim.
Stationary points: (−2,−10) and (2,−10), both global minima; (0,6), a local maximum only. The original W shape supplies the classification.
Look for these in your work
- Find all three derivative roots without cancellation.
- Recover and label every original stationary coordinate.
- Classify from the W shape and explain the middle point’s local rather than global maximum.
For f(x)=(x−1)³+2 on all real x, expand if needed, find its stationary point by differentiation and draw the original increasing cubic through that point. Mark its derivative zero separately and explain why zero gradient is not automatically a maximum or minimum. Use the original shape for the conclusion.
Give me a hint
Expand into integer powers before differentiating. Compare original heights on either side of the stationary input and see whether the curve actually changes from rising to falling or vice versa.
Compare my reasoning
- Expand f=x³−3x²+3x+1. Differentiate to get f′=3x²−6x+3=3(x−1)², so the stationary input is x=1.
- The original height is f(1)=2. Mark original point (1,2) and derivative zero (1,0). Useful original points include (0,1), (1,2), (2,3).
- The translated cubic rises through the horizontal direction at (1,2), continuing from lower to higher heights rather than turning into a hill or valley. Thus (1,2) is a stationary non-turning point, neither a maximum nor a minimum. A repeated derivative zero alone did not classify the original shape.
Stationary point (1,2), a non-turning point. The original increasing cubic passes through its horizontal direction without a maximum or minimum.
Look for these in your work
- Differentiate the expanded integer-power form and find the stationary input.
- Return to the original function for height 2.
- Use the original rising-through shape to explain the absence of a maximum or minimum.
Common mistakes
- Stopping at a zero-gradient input or giving the derivative height zero as the original ordinate.
- Losing one zero factor or calling every stationary point a maximum or minimum.
- Using a second-derivative test instead of original graph shape, or calling a local maximum the largest value globally.
What can you explain now?
For f(x)=7 on all real inputs, what does f′(x)=0 say about stationary points and isolated turning points?
Compare with the explanation
Every input is stationary, but the horizontal graph has no isolated turning points.
The rate is zero everywhere, so it is not a finite list of isolated derivative roots. The graph does not form a hill or valley turn at a singled-out input.
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Make it stick
Explain why f′=0 is followed by substitution into f and inspection of the original graph. Contrast a local hill, a global valley, a stationary non-turn and a constant graph.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.