01 Curvature, inflections and derivative graphs Verify a concavity change at a point on the curve, distinguishing inflections from stationary points. ↗ Question directory1. Foundation · Find the inflection point of x³. 2. Standard · Find the inflection point of x³−6x²+3x. 3. Transfer · Show that x³+x has an inflection at zero but no horizontal tangen 4. Foundation · Find f″(−1) for f=x³. 5. Foundation · Does x² have an inflection? 6. Foundation · Explain why zero is not an inflection of x⁴. 7. Foundation · Where is x³−3x² concave up? 8. Standard · Find the inflection abscissae of x⁴−6x². 9. Standard · If f′ has a strict local maximum at x=2 and is differentiable wit 10. Standard · Can zero be an inflection point of 1/x? 11. Standard · At the inflection of x³−3x²+4, find the tangent slope. 12. Transfer · For f=x⁴+kx², determine when it has two distinct inflection point 13. Transfer · For f=x⁴, the tangent at zero is y=0. Compare with x³ and explain ← →
What you will be able to explain Verify a concavity change at a point on the curve, distinguishing inflections from stationary points. Read second-derivative signs and parameter degeneracies. Curvature is change in slope A curve may rise while bending downward; height and curvature answer different questions.
A rising derivative means slopes increase; an inflection requires a reversal of curvature at a curve point.
f ′ ′ > 0 : concave up f''>0:\ \text{concave up} f ′′ > 0 : concave up A rising derivative means slopes increase; an inflection requires a reversal of curvature at a curve point.
( x 4 + k x 2 ) ′ ′ = 12 x 2 + 2 k (x^4+kx^2)''=12x^2+2k ( x 4 + k x 2 ) ′′ = 12 x 2 + 2 k A rising derivative means slopes increase; an inflection requires a reversal of curvature at a curve point.
k < 0 : x = ± − k / 6 k<0:\ x=\pm\sqrt{-k/6} k < 0 : x = ± − k /6 When it applies The point must belong to the curve; a zero second derivative is only a candidate. Counterexample: x⁴ has f″(0)=0 without a sign change; 1/x has a domain hole at zero.
Curve, first/second derivative signs, tangent crossing and coordinate table. Pause at these common mistakes Confusing an inflection with a horizontal tangent. Ignoring a hole in the domain. PREDICT → EXPLORE → EXPLAIN → TRANSFER
Make a prediction, then explore the relationship. Lesson question: Predict the inflections of x⁴+kx² as k crosses zero.
Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.
Numerical data for the diagram; printable Input Second derivative Concavity -1.5 15 Up / 向上 0 -12 Down / 向下 1.5 15 Up / 向上
Explain: Use the table and f″=12x²+2k to distinguish two simple roots from the double root at zero.
Transfer: Explain why the disappearance of two inflections does not eliminate the minimum at zero.
My prediction, exploration and explanation Try it. Leave your reasoning visible. Use one hint at a time. A correction explains what changed, not just the final answer.
Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.
Find the inflection point of x³. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Require a change of concavity at a point on the curve.
Hint 2+ Start from this intermediate relation.
Worked solution+ Require a change of concavity at a point on the curve.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
x < 0 : f ′ ′ < 0 ; x > 0 : f ′ ′ > 0 x<0:f\prime\prime<0;\ x>0:f\prime\prime>0 x < 0 : f ′′ < 0 ; x > 0 : f ′′ > 0 A zero second derivative alone does not prove inflection.
(0,0), since f″=6x changes sign.
Checks and common pitfalls: A zero second derivative alone does not prove inflection.
Reasoning checklist · self / teacher assessment Require a change of concavity at a point on the curve. Give the intermediate relations and explain the conclusion with its restrictions. A zero second derivative alone does not prove inflection. Think first. Reveal a hint when the class is ready.
Find the inflection point of x³−6x²+3x. f = x 3 − 6 x 2 + 3 x f=x^3-6x^2+3x f = x 3 − 6 x 2 + 3 x First step provided — complete the reasoning: After locating the change of f″, evaluate f rather than f′.
Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ After locating the change of f″, evaluate f rather than f′.
Hint 2+ Start from this intermediate relation.
f ′ ′ = 6 x − 12 = 6 ( x − 2 ) f''=6x-12=6(x-2) f ′′ = 6 x − 12 = 6 ( x − 2 ) Worked solution+ After locating the change of f″, evaluate f rather than f′.
Establish and simplify this mathematical relation.
f ′ ′ = 6 x − 12 = 6 ( x − 2 ) f''=6x-12=6(x-2) f ′′ = 6 x − 12 = 6 ( x − 2 ) Establish and simplify this mathematical relation.
f ( 2 ) = 8 − 24 + 6 = − 10 f(2)=8-24+6=-10 f ( 2 ) = 8 − 24 + 6 = − 10 The inflection point includes both coordinates.
(2,−10).
Checks and common pitfalls: The inflection point includes both coordinates.
Reasoning checklist · self / teacher assessment After locating the change of f″, evaluate f rather than f′. Give the intermediate relations and explain the conclusion with its restrictions. The inflection point includes both coordinates. Think first. Reveal a hint when the class is ready.
Show that x³+x has an inflection at zero but no horizontal tangent there. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Test curvature separately from tangent slope.
Hint 2+ Start from this intermediate relation.
f ′ ′ = 6 x changes sign f''=6x\text{ changes sign} f ′′ = 6 x changes sign Worked solution+ Test curvature separately from tangent slope.
Establish and simplify this mathematical relation.
f ′ ′ = 6 x changes sign f''=6x\text{ changes sign} f ′′ = 6 x changes sign Establish and simplify this mathematical relation.
f ′ ( 0 ) = 1 ; y = x f'(0)=1;\ y=x f ′ ( 0 ) = 1 ; y = x An inflection need not be stationary.
Inflection (0,0), tangent y=x.
Checks and common pitfalls: An inflection need not be stationary.
Reasoning checklist · self / teacher assessment Test curvature separately from tangent slope. Give the intermediate relations and explain the conclusion with its restrictions. An inflection need not be stationary. Think first. Reveal a hint when the class is ready.
Find f″(−1) for f=x³. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Differentiate twice, keeping the sign of the input.
Hint 2+ Start from this intermediate relation.
f ′ = 3 x 2 ; f ′ ′ = 6 x f'=3x^2;\ f''=6x f ′ = 3 x 2 ; f ′′ = 6 x Worked solution+ Differentiate twice, keeping the sign of the input.
Establish and simplify this mathematical relation.
f ′ = 3 x 2 ; f ′ ′ = 6 x f'=3x^2;\ f''=6x f ′ = 3 x 2 ; f ′′ = 6 x Establish and simplify this mathematical relation.
f ′ ′ ( − 1 ) = − 6 f''(-1)=-6 f ′′ ( − 1 ) = − 6 The first derivative value 3 is not the curvature value.
The requested value is -6.
Checks and common pitfalls: The first derivative value 3 is not the curvature value.
Think first. Reveal a hint when the class is ready.
Does x² have an inflection? Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Look for a change of concavity rather than a minimum.
Hint 2+ Start from this intermediate relation.
Worked solution+ Look for a change of concavity rather than a minimum.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
no concavity change \text{no concavity change} no concavity change A minimum can occur with no inflection.
No; f″=2 is positive everywhere.
Checks and common pitfalls: A minimum can occur with no inflection.
Reasoning checklist · self / teacher assessment Look for a change of concavity rather than a minimum. Give the intermediate relations and explain the conclusion with its restrictions. A minimum can occur with no inflection. Think first. Reveal a hint when the class is ready.
Explain why zero is not an inflection of x⁴. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Check the sign on both sides of the zero.
Hint 2+ Start from this intermediate relation.
Worked solution+ Check the sign on both sides of the zero.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
f ′ ′ ( x ) = 12 x 2 > 0 ( x ≠ 0 ) f''(x)=12x^2>0\ (x\ne0) f ′′ ( x ) = 12 x 2 > 0 ( x = 0 ) The necessary candidate equation is not a sufficient criterion.
f″=12x² has no sign change.
Checks and common pitfalls: The necessary candidate equation is not a sufficient criterion.
Reasoning checklist · self / teacher assessment Check the sign on both sides of the zero. Give the intermediate relations and explain the conclusion with its restrictions. The necessary candidate equation is not a sufficient criterion. Think first. Reveal a hint when the class is ready.
Where is x³−3x² concave up? f = x 3 − 3 x 2 f=x^3-3x^2 f = x 3 − 3 x 2 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Solve the strict positive-second-derivative inequality.
Hint 2+ Start from this intermediate relation.
Worked solution+ Solve the strict positive-second-derivative inequality.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
6 x − 6 > 0 ⟺ x > 1 6x-6>0\iff x>1 6 x − 6 > 0 ⟺ x > 1 Increasing and concave up are separate properties.
For x>1.
Checks and common pitfalls: Increasing and concave up are separate properties.
Reasoning checklist · self / teacher assessment Solve the strict positive-second-derivative inequality. Give the intermediate relations and explain the conclusion with its restrictions. Increasing and concave up are separate properties. Think first. Reveal a hint when the class is ready.
Find the inflection abscissae of x⁴−6x². f = x 4 − 6 x 2 f=x^4-6x^2 f = x 4 − 6 x 2 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Factor the second derivative and inspect each sign change.
Hint 2+ Start from this intermediate relation.
f ′ ′ = 12 ( x 2 − 1 ) f''=12(x^2-1) f ′′ = 12 ( x 2 − 1 ) Worked solution+ Factor the second derivative and inspect each sign change.
Establish and simplify this mathematical relation.
f ′ ′ = 12 ( x 2 − 1 ) f''=12(x^2-1) f ′′ = 12 ( x 2 − 1 ) Establish and simplify this mathematical relation.
signs: + , − , + ; x = ± 1 \text{signs: }+,-,+;\ x=\pm1 signs: + , − , + ; x = ± 1 The two roots of f″ both cause concavity changes here.
x=−1 and x=1.
Checks and common pitfalls: The two roots of f″ both cause concavity changes here.
Reasoning checklist · self / teacher assessment Factor the second derivative and inspect each sign change. Give the intermediate relations and explain the conclusion with its restrictions. The two roots of f″ both cause concavity changes here. Think first. Reveal a hint when the class is ready.
If f′ has a strict local maximum at x=2 and is differentiable with f″ positive left and negative right, what happens to f? f ′ ′ > 0 ( x < 2 ) ; f ′ ′ < 0 ( x > 2 ) f''>0\ (x<2);\ f''<0\ (x>2) f ′′ > 0 ( x < 2 ) ; f ′′ < 0 ( x > 2 ) Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Translate the rate graph’s increase/decrease into f’s curvature.
Hint 2+ Start from this intermediate relation.
f ′ rises then falls f'\text{ rises then falls} f ′ rises then falls Worked solution+ Translate the rate graph’s increase/decrease into f’s curvature.
Establish and simplify this mathematical relation.
f ′ rises then falls f'\text{ rises then falls} f ′ rises then falls Establish and simplify this mathematical relation.
f changes from concave up to concave down f\text{ changes from concave up to concave down} f changes from concave up to concave down A maximum of f′ is not automatically a maximum of f.
f has an inflection at x=2, provided f is defined and continuous there.
Checks and common pitfalls: A maximum of f′ is not automatically a maximum of f.
Reasoning checklist · self / teacher assessment Translate the rate graph’s increase/decrease into f’s curvature. Give the intermediate relations and explain the conclusion with its restrictions. A maximum of f′ is not automatically a maximum of f. Think first. Reveal a hint when the class is ready.
Can zero be an inflection point of 1/x? f = 1 / x ; x ≠ 0 f=1/x;\ x\ne0 f = 1/ x ; x = 0 The rational denominator must be nonzero; excluded inputs are not points on its curve. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.
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Hint 1+ Check the domain before interpreting a sign change.
Hint 2+ Start from this intermediate relation.
f ′ ′ = 2 / x 3 f''=2/x^3 f ′′ = 2/ x 3 Worked solution+ Check the domain before interpreting a sign change.
Establish and simplify this mathematical relation.
f ′ ′ = 2 / x 3 f''=2/x^3 f ′′ = 2/ x 3 Establish and simplify this mathematical relation.
0 ∉ dom ( f ) 0\notin\operatorname{dom}(f) 0 ∈ / dom ( f ) A concavity change across a hole does not create an inflection point there.
No; zero is not a point on the curve.
Checks and common pitfalls: A concavity change across a hole does not create an inflection point there.
Reasoning checklist · self / teacher assessment Check the domain before interpreting a sign change. Give the intermediate relations and explain the conclusion with its restrictions. A concavity change across a hole does not create an inflection point there. Think first. Reveal a hint when the class is ready.
At the inflection of x³−3x²+4, find the tangent slope. f = x 3 − 3 x 2 + 4 f=x^3-3x^2+4 f = x 3 − 3 x 2 + 4 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Find the curvature change first, then evaluate the first derivative.
Hint 2+ Start from this intermediate relation.
f ′ ′ = 6 ( x − 1 ) ⟹ x = 1 f''=6(x-1)\implies x=1 f ′′ = 6 ( x − 1 ) ⟹ x = 1 Worked solution+ Find the curvature change first, then evaluate the first derivative.
Establish and simplify this mathematical relation.
f ′ ′ = 6 ( x − 1 ) ⟹ x = 1 f''=6(x-1)\implies x=1 f ′′ = 6 ( x − 1 ) ⟹ x = 1 Establish and simplify this mathematical relation.
f ′ ( 1 ) = 3 − 6 = − 3 f'(1)=3-6=-3 f ′ ( 1 ) = 3 − 6 = − 3 The second derivative is zero there, but the tangent slope need not be.
The requested value is -3.
Checks and common pitfalls: The second derivative is zero there, but the tangent slope need not be.
Think first. Reveal a hint when the class is ready.
For f=x⁴+kx², determine when it has two distinct inflection points. f = x 4 + k x 2 f=x^4+kx^2 f = x 4 + k x 2 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Classify the second-derivative roots and the boundary k=0.
Hint 2+ Start from this intermediate relation.
f ′ ′ = 12 x 2 + 2 k f''=12x^2+2k f ′′ = 12 x 2 + 2 k Worked solution+ Classify the second-derivative roots and the boundary k=0.
Establish and simplify this mathematical relation.
f ′ ′ = 12 x 2 + 2 k f''=12x^2+2k f ′′ = 12 x 2 + 2 k Establish and simplify this mathematical relation.
k < 0 : x = ± − k / 6 , + , − , + ; k ≥ 0 : no sign change k<0:x=\pm\sqrt{-k/6},\ +,-,+;\ k\ge0:\ \text{no sign change} k < 0 : x = ± − k /6 , + , − , + ; k ≥ 0 : no sign change At k=0, the double root gives no inflection.
Exactly k<0, at x=±√(−k/6).
Checks and common pitfalls: At k=0, the double root gives no inflection.
Reasoning checklist · self / teacher assessment Classify the second-derivative roots and the boundary k=0. Give the intermediate relations and explain the conclusion with its restrictions. At k=0, the double root gives no inflection. Think first. Reveal a hint when the class is ready.
For f=x⁴, the tangent at zero is y=0. Compare with x³ and explain why crossing behavior supports only the latter’s inflection. f = x 4 ; g = x 3 f=x^4;\ g=x^3 f = x 4 ; g = x 3 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Applications of derivatives Curvature, inflections and derivative graphs Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Use both a tangent comparison and a curvature check for these two polynomials.
Hint 2+ Start from this intermediate relation.
x 4 ≥ 0 ; x 3 changes sign at 0 x^4\ge0;\ x^3\text{ changes sign at }0 x 4 ≥ 0 ; x 3 changes sign at 0 Worked solution+ Use both a tangent comparison and a curvature check for these two polynomials.
Establish and simplify this mathematical relation.
x 4 ≥ 0 ; x 3 changes sign at 0 x^4\ge0;\ x^3\text{ changes sign at }0 x 4 ≥ 0 ; x 3 changes sign at 0 Establish and simplify this mathematical relation.
( x 4 ) ′ ′ = 12 x 2 ; ( x 3 ) ′ ′ = 6 x (x^4)''=12x^2;\ (x^3)''=6x ( x 4 ) ′′ = 12 x 2 ; ( x 3 ) ′′ = 6 x Do not promote a visual crossing rule to a universal test without assumptions.
x⁴ stays above its tangent and has no concavity change; x³ crosses its tangent and f″ changes sign.
Checks and common pitfalls: Do not promote a visual crossing rule to a universal test without assumptions.
Reasoning checklist · self / teacher assessment Use both a tangent comparison and a curvature check for these two polynomials. Give the intermediate relations and explain the conclusion with its restrictions. Do not promote a visual crossing rule to a universal test without assumptions. Think first. Reveal a hint when the class is ready.
