Stationary points,turns and domain exclusions
A stationary point is an admitted point of the original graph where its derivative is zero. Finding derivative zeros is a beginning;you must retain every root,apply the original domain and explain the signs around each admitted point.
Section 4 · Functions (all14rows linked;self-check practice)
Exam scope and assessment notes
Original joint4Mteaching/practice bundle across three lessons and25cases,each with Worked,Guided,Independent and Application stages. Students enter their own graph observations,term-by-term differentiation,proposed derivative,original graph markers,complete cleared root lists,domain/context decisions,sign-chart witnesses and classifications. The global comparison retains stationary values,closed endpoints,open bounds,poles,infinities,attainment and all attaining inputs. Exact rational/surd and certified interval references support bounded self-checks;the finite sparse-curve and degree-five limits are tool bounds,not syllabus exclusions. Original punctures survive simplification and both graphs;constant and stationary-without-turn cases are explicit. Correct separate answers cannot replace incorrect actual working. Complete wrong attempts may study explained references without passing. The complete source and learner-route checks support a structural4Mlink only when all three original teaching/worked/practice records remain intact. All14Functions source rows have reviewed structural teaching/practice links;other Mathematics B source gaps remain explicit in the current course map. The wider Pearson curriculum remains incomplete. Supplied graph stimuli do not award separate4Jcurve-drawing,4Kdrawn-tangent or4Nmotion coverage. Written reasoning,graph judgment,model validity and specialist review remain pending. No official marks,mastery or formal assessment readiness. Single Higher-tier linear4MB1,entry context2026/27only. BookISBN9780435044107 is official metadata only;no verified full book body,chapter/Hub map or MathematicsA/Foundation/modular inheritance.
Before you begin
- Integer-power differentiation and original domains
- Original graph heights and derivative graph signs
- Solving linear/quadratic equations and interpreting approximate graphical roots
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Does a zero derivative alone prove that the original graph turns?
What you will learn
- Retain every real cleared derivative candidate before original-domain and context filtering.
- Calculate exact rational or surd original coordinates and appropriate approximate graphical readings.
- Use a complete derivative sign chart to distinguish maxima,minima and stationary points without a turn.
- Handle a derivative-zero identity and original punctures without fabricating finite stationary points.
Stationary points,turns and domain exclusions: Worked
A stationary point is an admitted point of the original graph where its derivative is zero. Finding derivative zeros is a beginning;you must retain every root,apply the original domain and explain the signs around each admitted point.
Study the method and its limits
Solve completely before filtering
Write the derivative-zero equation. Clearing a reciprocal denominator is valid only away from its original zero exclusion. Solve the cleared equation,retain all its real candidates,and reject candidates excluded by the original expression.
Use exact algebra where appropriate,including exact surds. When a stationary equation is not conveniently factorised,read the intersections with zero graphically and give approximations. A completeness argument must account for all real roots;neither a clipped graph nor a few trial inputs is sufficient.
Link a sign chart to a turn
Place all stationary inputs and original domain holes in increasing order. In each intervening root-free interval,show a permitted witness and the sign of the original derivative. The reference checks that no additional derivative roots occur in the interval;one arbitrary sample without that justification does not establish a sign throughout it.
A change from negative to positive gives a local minimum;positive to negative gives a local maximum. The same sign on both sides gives a stationary point without a turn. Substitute each admitted input into the original function to obtain its graph coordinate. A sign change across a hole does not create an admitted stationary point.
Handle an identity and a context explicitly
If the derivative is identically zero,the original function is constant on the supported domain and every admitted input is stationary. Report all inputs instead of manufacturing a finite list. There is no isolated turn or strict extremum;constant values give non-strict extrema.
Build the complete mathematical stationary list first. For an application,mark each point as admitted or rejected by the stated context without erasing it from the mathematical list.
Current stage: 0 of 2 case attempts recorded;0 fixed comparisons agree. These counts do not award marks or mastery.
Original formula: y = x^2-2*x+1
A negative-to-positive derivative change gives a local minimum.
No extra contextual interval is imposed. Retain the original mathematical domain. Quantity unit: dimensionless.
- Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly.
- For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation.
- Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Read the blue original graph first and enter your visual predictions separately from your calculations. Then reconcile them. Use explicit * and ^ in expressions. For example,3*x^2-2/x^3. Use exact surds where taught;for graphical fifth-degree roots,enter decimal readings. Enter undefined for an excluded point. Lower/upper attaining inputs use all,none,or a semicolon-separated list. The blue graph supplies the original function;it does not count as your own curve-drawing work.
My actual working
My term-by-term differentiation
My complete cleared candidate list
My original-domain derivative sign chart
For sign boundaries,use r1,r2,… from your candidate rows;zero-hole for the original excluded zero;and -infinity/infinity at unbounded ends. Give every cell its own rational witness and derivative value. Constant/all-inputs cases use one original-domain cell unless an original hole splits it.
Original graph and my proposed derivative
Blue is the supplied original graph. Orange shows your entered derivative as a proposal;it is not a correctness signal. Mark the original function points yourself. The drawing window does not restrict the mathematical domain.
Add a candidate row before placing its point. The all-inputs and no-admitted-point cases do not need fabricated stationary markers.
Choose a row,then click the graph or use arrow keys to move its point by0.1. Hold Shift for0.01 steps. You can also enter both marker coordinates in the table. Points outside the window stay in your working and are listed below.
Enter a complete derivative expression to view your proposed rate graph.
Working: Not recorded or changed
My separate final answers
Enter these separately from your working. A correct answer cannot replace a missing or incorrect calculation,marker,root list,sign chart or boundary comparison.
My separate answers: My complete cleared candidate list
Separate answers: Not recorded or changed
Record your current calculation,graph and interpretation working for this case.
What these comparisons cover
- The original function domain survives simplification; context bounds and the drawing window do not change it.
- The finite exact reference supports sparse powers -2 through 3 and cleared derivative degree at most five. These are self-check bounds,not exclusions from the syllabus.
- Sturm root accounting certifies every distinct real cleared root; original-domain exclusions are applied afterwards. A few plotted samples cannot certify completeness.
- First-derivative signs certify turns; a zero derivative alone does not prove a maximum or minimum. Constants have all original inputs stationary and no isolated turn or strict extremum.
- Global attained extrema are separate from unattained finite bounds and unbounded directions. Open endpoints,poles,infinities and the original model boundaries are retained.
- Quadratic-field values use exact arithmetic; other real roots use certified rational intervals. Unresolved finite comparisons require review rather than a rounded guess.
- Mathematical self-checks do not establish specialist approval,model validity or formal assessment readiness; supplied graph stimuli do not award separate curve-drawing or tangent-drawing credit.
- Exact rational and surd entries are required where taught;graphical fifth-degree roots use a0.003 input reading tolerance or less near other roots or the pole. Their height readings use0.01 tolerance.
- Own original-graph markers are checked separately with input tolerance at most0.035 and height tolerance0.15. These finite reading tolerances do not establish mathematical equality.
- Every table is learner-entered. Correct independent answers cannot replace missing or incorrect own graph markers,root lists,sign witnesses or contextual comparison working.
- Cleared roots excluded from the original domain remain in the candidate list with undefined height and no plotted point. The all-inputs case has no invented finite root list.
- Sign-chart boundaries use r1,r2,… for your candidate rows,zero-hole for the original excluded zero,and -infinity/infinity for unbounded ends. Enter a permitted rational witness and its actual derivative value in each cell.
- Global candidate rows retain stationary inputs,original-context endpoints,one-sided hole limits and infinity behaviour. Use all or none for the appropriate attaining-input statements,and separate finite input lists with semicolons.
- Showing an explained reference after a complete attempt does not imply that the attempt passed. Written reasoning,graph judgment,model validity,specialist review and formal assessment remain outside automatic approval.
Specialist review and formal assessment readiness remain pending. The supplied graphs do not award separate curve-drawing,tangent-drawing or motion coverage.
Solve completely before filtering
Write the derivative-zero equation. Clearing a reciprocal denominator is valid only away from its original zero exclusion. Solve the cleared equation,retain all its real candidates,and reject candidates excluded by the original expression.
Use exact algebra where appropriate,including exact surds. When a stationary equation is not conveniently factorised,read the intersections with zero graphically and give approximations. A completeness argument must account for all real roots;neither a clipped graph nor a few trial inputs is sufficient.
Link a sign chart to a turn
Place all stationary inputs and original domain holes in increasing order. In each intervening root-free interval,show a permitted witness and the sign of the original derivative. The reference checks that no additional derivative roots occur in the interval;one arbitrary sample without that justification does not establish a sign throughout it.
A change from negative to positive gives a local minimum;positive to negative gives a local maximum. The same sign on both sides gives a stationary point without a turn. Substitute each admitted input into the original function to obtain its graph coordinate. A sign change across a hole does not create an admitted stationary point.
Handle an identity and a context explicitly
If the derivative is identically zero,the original function is constant on the supported domain and every admitted input is stationary. Report all inputs instead of manufacturing a finite list. There is no isolated turn or strict extremum;constant values give non-strict extrema.
Build the complete mathematical stationary list first. For an application,mark each point as admitted or rejected by the stated context without erasing it from the mathematical list.
Sources for this lesson
Pause and explain
Does simplifying x^3/x to x^2 admit a stationary point at zero?
Worked example
Original y = x^2-2*x+1. A negative-to-positive derivative change gives a local minimum. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers. Original y = -x^2+2*x+3. A positive-to-negative derivative change gives a local maximum. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Show the worked solution
- Case 1: Differentiate y=x^2-2x+1 to get2x-2. Solving2x-2=0 gives x=1.
- Case 1: The original point is(1,0). The derivative is negative for x<1 and positive for x>1.
- Case 2: Differentiate y=-x^2+2x+3 to get-2x+2. The stationary input is1.
- Case 2: At x=1,the original height is4. The derivative is positive to the left and negative to the right.
Answer Case 1: The graph changes from decreasing to increasing,so(1,0) is a local minimum and a turn. Differentiate y=x^2-2x+1 to get2x-2. Solving2x-2=0 gives x=1. The original point is(1,0). The derivative is negative for x<1 and positive for x>1. Show both derivative signs and the original point;the equation dy/dx=0 alone does not classify the point. Case 2: The graph changes from increasing to decreasing,so(1,4) is a local maximum and a turn. Differentiate y=-x^2+2x+3 to get-2x+2. The stationary input is1. At x=1,the original height is4. The derivative is positive to the left and negative to the right. The derivative graph crosses its own horizontal axis at1;the original stationary point has height4.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Original y = x^3. A repeated derivative root with positive sign on both sides is not a turn. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers. Original y = 3. Every admitted input is stationary;there is no isolated turn or strict extremum. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Give me a hint
Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Compare my reasoning
- Case 1: For y=x^3,the stationary equation3x^2=0 has the single distinct real root0.
- Case 1: The original point is(0,0). For every nonzero x,the derivative3x^2 is positive on both sides.
- Case 2: For y=3,the derivative is identically0. The stationary equation0=0 is true for every original input.
- Case 2: There is no isolated turning point:the graph is a horizontal line. Every point is a non-strict local minimum and maximum,but no point is a strict one.
Case 1: The point is stationary without a turn. The graph has a horizontal tangent but does not change from increasing to decreasing or vice versa. For y=x^3,the stationary equation3x^2=0 has the single distinct real root0. The original point is(0,0). For every nonzero x,the derivative3x^2 is positive on both sides. Repeated roots count once as distinct inputs. Do not manufacture two stationary points from x^2=0. Case 2: Use an all-inputs statement instead of an empty or fabricated finite root list. For y=3,the derivative is identically0. The stationary equation0=0 is true for every original input. There is no isolated turning point:the graph is a horizontal line. Every point is a non-strict local minimum and maximum,but no point is a strict one. This unit uses first-derivative turns for isolated local classifications;constant and strict/non-strict statements need explicit language.
Look for these in your work
- My visual graph prediction and calculated interpretation are separate.
- My actual differentiation and original-domain working are explicit.
- Every original candidate and relevant sign or boundary comparison is retained.
- My graph markers are my own entries,not inferred from a correct final answer.
- My independent answers state units,context and attainment where applicable;reasoning and model validity need human review.
Original y = x^3-6*x. Keep both exact irrational stationary inputs and their original function heights. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers. Original y = x^3/3+5/x-1/(2*x^2). Graphically solve a degree-five cleared derivative and account for all three real roots before exclusions. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Give me a hint
Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Compare my reasoning
- Case 1: For y=x^3-6x,the derivative is3x^2-6. Solving gives x=-sqrt(2) and x=sqrt(2).
- Case 1: Use x^2=2 to simplify y=x(x^2-6)=-4x at either stationary input. Thus the points are(-sqrt(2),4sqrt(2)) and(sqrt(2),-4sqrt(2)).
- Case 1: The derivative signs across the three intervals are positive,negative,positive. The negative input gives a local maximum;the positive input gives a local minimum.
- Case 2: For y=x^3/3+5/x-1/(2x^2),the original domain excludes0 and the derivative is x^2-5/x^2+1/x^3.
- Case 2: Multiplying the derivative-zero equation by x^3 is valid only for x!=0. It gives x^5-5x+1=0. Graphical solutions are approximately-1.542,0.200 and1.441.
- Case 2: Their original heights are approximately-4.675,12.503 and4.226. Retain the pole0 between the negative and positive branches.
- Case 2: The derivative signs from left to right across the roots and the pole are positive,negative,positive,negative,positive. The first two stationary points are local maxima;the third is a local minimum.
Case 1: Keep exact surds for calculation and separate approximate graph locations around-1.414 and1.414. For y=x^3-6x,the derivative is3x^2-6. Solving gives x=-sqrt(2) and x=sqrt(2). Use x^2=2 to simplify y=x(x^2-6)=-4x at either stationary input. Thus the points are(-sqrt(2),4sqrt(2)) and(sqrt(2),-4sqrt(2)). The derivative signs across the three intervals are positive,negative,positive. The negative input gives a local maximum;the positive input gives a local minimum. Two decimal markers do not prove that the exact stationary equation has no further real roots. Case 2: Find and record every mathematical stationary point before applying any context restriction. The separate root-accounting certificate checks completeness;three plotted samples alone cannot do that. For y=x^3/3+5/x-1/(2x^2),the original domain excludes0 and the derivative is x^2-5/x^2+1/x^3. Multiplying the derivative-zero equation by x^3 is valid only for x!=0. It gives x^5-5x+1=0. Graphical solutions are approximately-1.542,0.200 and1.441. Their original heights are approximately-4.675,12.503 and4.226. Retain the pole0 between the negative and positive branches. The derivative signs from left to right across the roots and the pole are positive,negative,positive,negative,positive. The first two stationary points are local maxima;the third is a local minimum. The fifth-degree roots are graphical approximations,not claimed exact radicals. Written graph judgment and the completeness argument require review.
Look for these in your work
- My visual graph prediction and calculated interpretation are separate.
- My actual differentiation and original-domain working are explicit.
- Every original candidate and relevant sign or boundary comparison is retained.
- My graph markers are my own entries,not inferred from a correct final answer.
- My independent answers state units,context and attainment where applicable;reasoning and model validity need human review.
Original y = x^3/x. The apparent derivative root zero is excluded by the original expression. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers. Original y = x^3+1/x^2. The stationary input belongs to the mathematical graph but falls outside the supplied context. Stated context [1,3]. Unit: dimensionless. Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Give me a hint
Enter your derivative,clearing multiplier and stationary equation while keeping the original domain. Choose finite,none or all-inputs stationary behaviour explicitly. For a finite case,record every candidate input,its original-domain decision,original height and graph marker. Retain excluded candidates with an explanation. Construct a derivative sign chart with witnesses,separate a reciprocal hole from an actual root,and enter the point classifications and context decisions before your independent final answers.
Compare my reasoning
- Case 1: The original expression y=x^3/x is defined only for x!=0. On that original domain it simplifies to x^2 and has derivative2x.
- Case 1: The apparent derivative-zero candidate is0,but it is excluded by the original expression. There are no stationary points.
- Case 1: Retain an open hole at(0,0). The derivative is negative on the negative branch and positive on the positive branch,but there is no admitted point between those branches.
- Case 2: For y=x^3+1/x^2,the derivative is3x^2-2/x^3. On x!=0,the stationary equation becomes3x^5-2=0.
- Case 2: The sole real solution is x=(2/3)^(1/5),approximately0.922,with original height approximately1.960. The derivative changes from negative to positive on its positive branch,so it is a mathematical local minimum.
- Case 2: The stated model context is[1,3]. The stationary input is outside it and is not a contextual candidate.
Case 1: A sign change across an excluded input is not an original stationary turning point. The original expression y=x^3/x is defined only for x!=0. On that original domain it simplifies to x^2 and has derivative2x. The apparent derivative-zero candidate is0,but it is excluded by the original expression. There are no stationary points. Retain an open hole at(0,0). The derivative is negative on the negative branch and positive on the positive branch,but there is no admitted point between those branches. The simplified expression must not silently replace the original domain. Case 2: Preserve the mathematical stationary point in the full list,then explicitly reject it for this context. For y=x^3+1/x^2,the derivative is3x^2-2/x^3. On x!=0,the stationary equation becomes3x^5-2=0. The sole real solution is x=(2/3)^(1/5),approximately0.922,with original height approximately1.960. The derivative changes from negative to positive on its positive branch,so it is a mathematical local minimum. The stated model context is[1,3]. The stationary input is outside it and is not a contextual candidate. A context restriction does not erase a valid stationary point from the mathematical graph.
Look for these in your work
- My visual graph prediction and calculated interpretation are separate.
- My actual differentiation and original-domain working are explicit.
- Every original candidate and relevant sign or boundary comparison is retained.
- My graph markers are my own entries,not inferred from a correct final answer.
- My independent answers state units,context and attainment where applicable;reasoning and model validity need human review.
Common mistakes
- Confusing the original height with its local gradient.
- Deleting a real mathematical derivative candidate merely because a model later rejects it.
- Treating zero derivative as an automatic turn or a reciprocal hole as an admitted point.
- Using a few plotted samples as a complete stationary-root proof.
- Ignoring included endpoints,open boundaries,infinity or whether a bound is attained.
- Allowing correct final answers to replace missing or wrong own graph and sign-chart working.
What can you explain now?
Does simplifying x^3/x to x^2 admit a stationary point at zero?
Compare with the explanation
No.
The original expression excludes zero,so its apparent derivative candidate is not an original stationary point.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Does simplifying x^3/x to x^2 admit a stationary point at zero? No. The original expression excludes zero,so its apparent derivative candidate is not an original stationary point. Does a fixed self-check establish specialist approval or formal assessment readiness? No. Written reasoning,graph judgment and model validity still need human review.
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.