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Find the local maximum and minimum values of f.

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

TOPIC 01

2021 JM02

These local values are not the global bounds of the cubic on the real line.

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

01 / Standard#Your turn

Find the local maximum and minimum values of f.

f(x)=2x3−9x2+12x−5f(x)=2x^3-9x^2+12x-5

Official paper · jm02-2021 · 2(a)(ii) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 8

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Factor the first derivative to find stationary points.
Hint 2
Read the derivative signs on the three intervals.
Worked solution
  1. The stationary abscissae are 1 and 2.

    f′(x)=6(x−1)(x−2)=0  ⟺  x=1,2f'(x)=6(x-1)(x-2)=0\iff x=1,2
  2. The signs are positive, negative, positive.

    x<1:f′>0;1<x<2:f′<0;x>2:f′>0x<1:f'>0;\quad1<x<2:f'<0;\quad x>2:f'>0
  3. Evaluate f at the two points and classify the changes.

    f(1)=0 (local maximum),f(2)=−1 (local minimum)f(1)=0\text{ (local maximum)},\quad f(2)=-1\text{ (local minimum)}

Local maximum 0 at x=1; local minimum −1 at x=2.

Checks and common pitfalls: These local values are not the global bounds of the cubic on the real line.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The stationary abscissae are 1 and 2.
    f′(x)=6(x−1)(x−2)=0  ⟺  x=1,2f'(x)=6(x-1)(x-2)=0\iff x=1,2
  • The signs are positive, negative, positive.
    x<1:f′>0;1<x<2:f′<0;x>2:f′>0x<1:f'>0;\quad1<x<2:f'<0;\quad x>2:f'>0
  • Evaluate f at the two points and classify the changes.
    f(1)=0 (local maximum),f(2)=−1 (local minimum)f(1)=0\text{ (local maximum)},\quad f(2)=-1\text{ (local minimum)}

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Curriculum and source notes ↗