Maxima,minima and context boundaries
Local turns describe nearby behaviour. A global maximum or minimum compares every admitted value in the stated domain or context,including the effects of endpoints,holes and unbounded directions.
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
- Complete mathematical stationary-point lists and derivative sign charts
- Original domain versus open/closed contextual bounds
- Exact rational/surd comparison and one-sided graph behaviour
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Can an included interval endpoint give a global maximum without being stationary?
What you will learn
- Select context-admitted stationary candidates from the complete mathematical list.
- Include permitted endpoints and retain one-sided open-boundary,hole and infinity behaviour.
- Compare exact or certified value bounds and distinguish attained extrema from unattained bounds or unbounded directions.
- Report all attaining inputs,units and the limits of an invented model.
Maxima,minima and context boundaries: Worked
Local turns describe nearby behaviour. A global maximum or minimum compares every admitted value in the stated domain or context,including the effects of endpoints,holes and unbounded directions.
Study the method and its limits
List every relevant candidate
First find the complete mathematical stationary list and retain original-domain restrictions. Then select stationary inputs admitted by the context and include any permitted finite endpoints. Calculate original function heights for all of them.
For each excluded endpoint or original hole,state the appropriate one-sided limiting behaviour. For an unbounded domain,consider the behaviour as the input tends to either infinity. A finite graph window is a drawing choice,not a restriction on the mathematical domain.
Separate a bound from an attained extremum
Compare candidate heights and limiting bounds. A finite least value is a minimum only if some admitted input attains it. A finite unattained lower bound is not a minimum;use the same distinction for an upper bound and maximum. An infinite direction means unbounded behaviour,not an attained infinite extremum.
An included endpoint can give a global extremum without being stationary. A constant function on an open interval attains both its maximum and minimum at every admitted input,so open endpoints alone do not decide attainment.
Support the comparison
Keep exact rational and surd heights when they are available. For other algebraic stationary inputs,use certified value intervals to justify a strict comparison. Close rounded decimals do not establish equality;an unresolved comparison must remain for review.
Give the attaining input or all attaining inputs,units and model context with each extremum. Distinguish mathematical conclusions about an invented formula from claims that a physical or economic model is realistic.
Current stage: 0 of 2 case attempts recorded;0 fixed comparisons agree. These counts do not award marks or mastery.
Original formula: y = x^3-3*x
Include allowed endpoints after filtering the mathematical stationary list.
Stated context: [0,2]. Original mathematical restrictions still apply. Quantity unit: dimensionless.
- Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries.
- Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers.
- State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained 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
My contextual candidate and boundary comparison
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.
List every relevant candidate
First find the complete mathematical stationary list and retain original-domain restrictions. Then select stationary inputs admitted by the context and include any permitted finite endpoints. Calculate original function heights for all of them.
For each excluded endpoint or original hole,state the appropriate one-sided limiting behaviour. For an unbounded domain,consider the behaviour as the input tends to either infinity. A finite graph window is a drawing choice,not a restriction on the mathematical domain.
Separate a bound from an attained extremum
Compare candidate heights and limiting bounds. A finite least value is a minimum only if some admitted input attains it. A finite unattained lower bound is not a minimum;use the same distinction for an upper bound and maximum. An infinite direction means unbounded behaviour,not an attained infinite extremum.
An included endpoint can give a global extremum without being stationary. A constant function on an open interval attains both its maximum and minimum at every admitted input,so open endpoints alone do not decide attainment.
Support the comparison
Keep exact rational and surd heights when they are available. For other algebraic stationary inputs,use certified value intervals to justify a strict comparison. Close rounded decimals do not establish equality;an unresolved comparison must remain for review.
Give the attaining input or all attaining inputs,units and model context with each extremum. Distinguish mathematical conclusions about an invented formula from claims that a physical or economic model is realistic.
Sources for this lesson
Pause and explain
Is a finite lower bound a minimum when no admitted input attains it?
Worked example
Original y = x^3-3*x. Include allowed endpoints after filtering the mathematical stationary list. Stated context [0,2]. Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers. Original A = -w^2+8*w. Positive rectangle widths attain a maximum area but not zero area. Stated context (0,8). Unit: square units. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers.
Show the worked solution
- Case 1: For y=x^3-3x,the mathematical stationary inputs are-1 and1. Only1 lies in[0,2].
- Case 1: The contextual candidates are x=0,1,2 with heights0,-2,2.
- Case 1: The global minimum on[0,2] is-2 at1;the global maximum is2 at the included endpoint2.
- Case 2: For a rectangle with perimeter16,width w gives length8-w and area A=w(8-w)=-w^2+8w. Positive sides require0<w<8.
- Case 2: The derivative-2w+8 is0 at w=4,where A=16 square units. The area increases before4 and decreases afterwards.
- Case 2: Area approaches0 as w approaches either excluded endpoint,but no admitted rectangle has area0. The maximum16 is attained at4;the lower bound0 is unattained,so there is no minimum area.
Answer Case 1: An endpoint can give a global extremum without being a mathematical stationary point. For y=x^3-3x,the mathematical stationary inputs are-1 and1. Only1 lies in[0,2]. The contextual candidates are x=0,1,2 with heights0,-2,2. The global minimum on[0,2] is-2 at1;the global maximum is2 at the included endpoint2. The local maximum at-1 belongs to the full graph but not this model interval. Case 2: Keep degenerate endpoint shapes outside the stated model rather than claiming an attained zero-area rectangle. For a rectangle with perimeter16,width w gives length8-w and area A=w(8-w)=-w^2+8w. Positive sides require0<w<8. The derivative-2w+8 is0 at w=4,where A=16 square units. The area increases before4 and decreases afterwards. Area approaches0 as w approaches either excluded endpoint,but no admitted rectangle has area0. The maximum16 is attained at4;the lower bound0 is unattained,so there is no minimum area. Units are arbitrary consistent length and area units. This geometric model assumes an ideal rectangle.
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Original y = 2*x+1. A finite lower bound need not be an attained minimum. Stated context (0,3]. Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers. Original y = 3. Both global extrema are attained at every admitted input,not just a finite point list. Stated context (-2,2). Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers.
Give me a hint
Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers.
Compare my reasoning
- Case 1: For y=2x+1,the derivative2 is positive everywhere and there are no stationary inputs.
- Case 1: On0<x<=3,the height approaches1 at the excluded lower endpoint and equals7 at the included upper endpoint3.
- Case 2: For y=3 and-2<x<2,the derivative is0 at every admitted input.
- Case 2: Every admitted input has height3. Therefore the minimum and maximum are both3,and both are attained at every input in the open interval.
Case 1: The global maximum is7 at3. The lower bound is1 but is not attained,so there is no global minimum. For y=2x+1,the derivative2 is positive everywhere and there are no stationary inputs. On0<x<=3,the height approaches1 at the excluded lower endpoint and equals7 at the included upper endpoint3. An open endpoint limit is not a point on the contextual graph. Case 2: Open endpoints do not prevent extrema when the same value is already attained throughout the interval. For y=3 and-2<x<2,the derivative is0 at every admitted input. Every admitted input has height3. Therefore the minimum and maximum are both3,and both are attained at every input in the open interval. No finite list of selected coordinates represents all attaining inputs;strict extrema and isolated turns are absent.
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^3-3*x. Two local turns do not imply global extrema on the whole real domain. No extra contextual interval;retain the original mathematical domain. Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers. Original y = 1/x. Keep an attained endpoint minimum separate from an unbounded direction. Stated context (0,2]. Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers. Original y = x^3-6*x. Compare an exact irrational stationary height with both allowed endpoint values. Stated context [0,2]. Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers.
Give me a hint
Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers.
Compare my reasoning
- Case 1: For y=x^3-3x,the local maximum is2 at-1 and local minimum is-2 at1.
- Case 1: On the full real domain,the leading cubic term makes y tend to positive infinity as x increases without bound,and to negative infinity as x decreases without bound.
- Case 2: For y=1/x on0<x<=2,the derivative-1/x^2 is negative and has no zero.
- Case 2: The value at the included endpoint2 is1/2. As x approaches0 from the right,the function increases without bound.
- Case 3: For y=x^3-6x,the full stationary inputs are plus or minus sqrt(2). Only the positive input belongs to[0,2].
- Case 3: The candidate heights are0 at0,-4sqrt(2) at sqrt(2),and-4 at2. Since sqrt(2)>1,the stationary height is less than-4.
Case 1: The function has no global maximum and no global minimum,despite its two local turns. For y=x^3-3x,the local maximum is2 at-1 and local minimum is-2 at1. On the full real domain,the leading cubic term makes y tend to positive infinity as x increases without bound,and to negative infinity as x decreases without bound. A finite drawing window cannot establish global boundedness. Case 2: The global minimum is1/2 at2;there is no global maximum and no finite upper bound. For y=1/x on0<x<=2,the derivative-1/x^2 is negative and has no zero. The value at the included endpoint2 is1/2. As x approaches0 from the right,the function increases without bound. Positive infinity is not an attained value or an admitted input. Case 3: The global minimum is exactly-4sqrt(2) at sqrt(2);the global maximum is0 at the included endpoint0. For y=x^3-6x,the full stationary inputs are plus or minus sqrt(2). Only the positive input belongs to[0,2]. The candidate heights are0 at0,-4sqrt(2) at sqrt(2),and-4 at2. Since sqrt(2)>1,the stationary height is less than-4. Retain exact heights for the comparison rather than comparing rounded surds alone.
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^3/3+5/x-1/(2*x^2). Retain all mathematical roots before selecting model candidates and comparing certified value intervals. Stated context [1/10,2]. Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers. Original y = x^2+2/x. Compare the admitted minimum with both closed boundary heights while retaining the original zero exclusion. Stated context [1/2,3]. Unit: dimensionless. Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers.
Give me a hint
Retain all mathematical stationary candidates before selecting contextual ones. Record the allowed closed endpoints and the excluded/open boundaries. Enter candidate heights,one-sided bounds and infinity behaviour with inclusion decisions. Compare them explicitly and place your own admitted graph markers. State the lower and upper bounds separately from whether they are attained,including all attaining inputs or the all-inputs case. Then submit independent final decisions and explained answers.
Compare my reasoning
- Case 1: Use the same three mathematical roots of x^5-5x+1=0 from the stationary lesson. On[1/10,2],reject the negative root and retain the two positive roots.
- Case 1: The included endpoint heights are1/3000 at1/10 and121/24 at2. The interior stationary heights are approximately12.503 and4.226.
- Case 1: Comparing all four admitted candidates gives the global minimum1/3000 at1/10 and the global maximum approximately12.503 at x approximately0.200.
- Case 2: For y=x^2+2/x,the original zero exclusion remains. The derivative is2x-2/x^2,and its sole real zero is1.
- Case 2: On[1/2,3],the candidate heights are17/4 at1/2,3 at1,and29/3 at3.
- Case 2: The global minimum is3 at1 and the global maximum is29/3 at3.
Case 1: Use certified value intervals to support these distinct comparisons. Rounded graph readings communicate the result but do not certify equal or unequal algebraic values. Use the same three mathematical roots of x^5-5x+1=0 from the stationary lesson. On[1/10,2],reject the negative root and retain the two positive roots. The included endpoint heights are1/3000 at1/10 and121/24 at2. The interior stationary heights are approximately12.503 and4.226. Comparing all four admitted candidates gives the global minimum1/3000 at1/10 and the global maximum approximately12.503 at x approximately0.200. This is an invented contextual interval;no physical interpretation or specialist approval is inferred. Case 2: A reciprocal term does not prevent closed-interval comparison when the full admitted interval avoids its pole. For y=x^2+2/x,the original zero exclusion remains. The derivative is2x-2/x^2,and its sole real zero is1. On[1/2,3],the candidate heights are17/4 at1/2,3 at1,and29/3 at3. The global minimum is3 at1 and the global maximum is29/3 at3. The original domain exclusion still matters when interpreting or extending the graph beyond this context.
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?
Is a finite lower bound a minimum when no admitted input attains it?
Compare with the explanation
No.
An unattained finite bound is separate from an attained minimum;give the attainment decision explicitly.
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Make it stick
Is a finite lower bound a minimum when no admitted input attains it? No. An unattained finite bound is separate from an attained minimum;give the attainment decision explicitly. 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.