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Calculus · eight-question check

Read the idea, work independently, then explain what changed.

TOPIC 01

Calculus

Try two tasks per skill independently, then revisit the linked skill.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

For y=x² at x=3, what is the tangent slope?

y=x2; a=3y=x^2;\ a=3
  • Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.
Skills and prerequisite lessons

Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

  1. 9
  2. 6
  3. 3
  4. 0
Working and explanation
BUILD THE REASONING
Hint 1
Differentiate the function, then evaluate its derivative.
Hint 2
Start from this intermediate relation.
f′(x)=2xf'(x)=2x
Worked solution
  1. Differentiate the function, then evaluate its derivative.
  2. Establish and simplify this mathematical relation.
    f′(x)=2xf'(x)=2x
  3. Establish and simplify this mathematical relation.
    f′(3)=6f'(3)=6
  4. The height is 9 but the slope is 6.
6
Checks and common pitfalls: The height is 9 but the slope is 6.
Think first. Reveal a hint when the class is ready.
02 / Foundation#Your turn
Which statement about |x| at zero is correct?
f(x)=∣x∣f(x)=|x|
  • A derivative requires equal finite one-sided limits; an extremum itself does not require differentiability.
Skills and prerequisite lessons
Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
  1. Derivative is 0.
  2. Derivative is 1.
  3. Continuous, but not differentiable.
  4. Not continuous.
Working and explanation
BUILD THE REASONING
Hint 1
Compare left and right difference quotients.
Hint 2
Start from this intermediate relation.
lim⁡h→0−∣h∣/h=−1\lim_{h\to0^-}|h|/h=-1
Worked solution
  1. Compare left and right difference quotients.
  2. Establish and simplify this mathematical relation.
    lim⁡h→0−∣h∣/h=−1\lim_{h\to0^-}|h|/h=-1
  3. Establish and simplify this mathematical relation.
    lim⁡h→0+∣h∣/h=1\lim_{h\to0^+}|h|/h=1
  4. A corner differs from a jump.
Continuous, but not differentiable.
Checks and common pitfalls: A corner differs from a jump.
Think first. Reveal a hint when the class is ready.
03 / Foundation#Your turn
Find f′(0) for f=(3x+1)².
f=(3x+1)2f=(3x+1)^2
  • Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.
Skills and prerequisite lessons
Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
  1. 2
  2. 3
  3. 9
  4. 6
Working and explanation
BUILD THE REASONING
Hint 1
Keep both the outer derivative and inner derivative.
Hint 2
Start from this intermediate relation.
f′=6(3x+1)f'=6(3x+1)
Worked solution
  1. Keep both the outer derivative and inner derivative.
  2. Establish and simplify this mathematical relation.
    f′=6(3x+1)f'=6(3x+1)
  3. Establish and simplify this mathematical relation.
    f′(0)=6f'(0)=6
  4. The inner factor is 3.
6
Checks and common pitfalls: The inner factor is 3.
Think first. Reveal a hint when the class is ready.
04 / Foundation#Your turn
Find the derivative of x²·x³.
f=x2x3f=x^2x^3
  • Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.
Skills and prerequisite lessons
Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
  1. 6x³
  2. 5x⁴
  3. x⁵
  4. 5x⁵
Working and explanation
BUILD THE REASONING
Hint 1
First combine the powers, then differentiate.
Hint 2
Start from this intermediate relation.
f=x5f=x^5
Worked solution
  1. First combine the powers, then differentiate.
  2. Establish and simplify this mathematical relation.
    f=x5f=x^5
  3. Establish and simplify this mathematical relation.
    f′=5x4f'=5x^4
  4. Multiplying 2x and 3x² gives the wrong rule.
5x⁴
Checks and common pitfalls: Multiplying 2x and 3x² gives the wrong rule.
Think first. Reveal a hint when the class is ready.
05 / Foundation#Your turn
At zero, x³ has which behavior?
f=x3f=x^3
  • Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.
Skills and prerequisite lessons
Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
  1. Local minimum
  2. Local maximum
  3. Stationary but no extremum
  4. Derivative does not exist
Working and explanation
BUILD THE REASONING
Hint 1
Compare derivative signs on both sides.
Hint 2
Start from this intermediate relation.
f′=3x2f'=3x^2
Worked solution
  1. Compare derivative signs on both sides.
  2. Establish and simplify this mathematical relation.
    f′=3x2f'=3x^2
  3. Establish and simplify this mathematical relation.
    signs: +,+\text{signs: }+,+
  4. Zero derivative is not a sufficient extremum test.
Stationary but no extremum
Checks and common pitfalls: Zero derivative is not a sufficient extremum test.
Think first. Reveal a hint when the class is ready.
06 / Foundation#Your turn
Find the maximum of x² on [−1,2].
f=x2; −1≤x≤2f=x^2;\ -1\le x\le2
  • Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.
Skills and prerequisite lessons
Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
  1. 0
  2. 1
  3. 2
  4. 4
Working and explanation
BUILD THE REASONING
Hint 1
Compare both endpoints and the stationary point.
Hint 2
Start from this intermediate relation.
f(−1)=1; f(0)=0; f(2)=4f(-1)=1;\ f(0)=0;\ f(2)=4
Worked solution
  1. Compare both endpoints and the stationary point.
  2. Establish and simplify this mathematical relation.
    f(−1)=1; f(0)=0; f(2)=4f(-1)=1;\ f(0)=0;\ f(2)=4
  3. Establish and simplify this mathematical relation.
    max⁡f=4\max f=4
  4. An extremizing coordinate is not the requested value.
4
Checks and common pitfalls: An extremizing coordinate is not the requested value.
Think first. Reveal a hint when the class is ready.
07 / Foundation#Your turn
Which polynomial has an inflection at zero?
  • Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.
Skills and prerequisite lessons
Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
  1. x²
  2. x⁴
  3. x³
  4. x⁶
Working and explanation
BUILD THE REASONING
Hint 1
Check second-derivative signs.
Hint 2
Start from this intermediate relation.
(x3)′′=6x(x^3)''=6x
Worked solution
  1. Check second-derivative signs.
  2. Establish and simplify this mathematical relation.
    (x3)′′=6x(x^3)''=6x
  3. Establish and simplify this mathematical relation.
    (x4)′′=12x2(x^4)''=12x^2
  4. A zero second derivative is only a candidate.
x³
Checks and common pitfalls: A zero second derivative is only a candidate.
Think first. Reveal a hint when the class is ready.
08 / Foundation#Your turn
For x³+x, what is the tangent slope at its inflection?
f=x3+xf=x^3+x
  • Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes.
Skills and prerequisite lessons
Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
  1. 1
  2. 0
  3. 3
  4. Undefined
Working and explanation
BUILD THE REASONING
Hint 1
Use f″ to locate, f′ to find slope.
Hint 2
Start from this intermediate relation.
f′′=6x  ⟹  x=0f''=6x\implies x=0
Worked solution
  1. Use f″ to locate, f′ to find slope.
  2. Establish and simplify this mathematical relation.
    f′′=6x  ⟹  x=0f''=6x\implies x=0
  3. Establish and simplify this mathematical relation.
    f′(0)=1f'(0)=1
  4. An inflection need not have zero slope.
1
Checks and common pitfalls: An inflection need not have zero slope.
Think first. Reveal a hint when the class is ready.
Focus on one question
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