AP Calculus AB — calculusStandard1 min read

Absolute and Relative Extrema

Find maximum and minimum values using the first and second derivative tests for AP Calculus AB.

Tutor AI Team
Tutor AI Team

Finding extrema is one of the most important applications of derivatives in AP Calculus.

Critical Points

Where f′(x)=0f'(x) = 0 or f′(x)f'(x) is undefined.

First Derivative Test

At a critical point cc:

  • f′f' changes from + to −: local max.
  • f′f' changes from − to +: local min.
  • No sign change: no extremum.

Second Derivative Test

At a critical point where f′(c)=0f'(c) = 0:

  • f′′(c)>0f''(c) > 0: local min.
  • f′′(c)<0f''(c) < 0: local max.
  • f′′(c)=0f''(c) = 0: inconclusive.

Absolute Extrema on $[a,b]$

Closed Interval Method:

  1. Find all critical points in (a,b)(a,b).
  2. Evaluate ff at critical points and endpoints.
  3. Largest = absolute max. Smallest = absolute min.

Extreme Value Theorem

A continuous function on a closed interval has both an absolute max and min.

Practice Problems

    1. Find all extrema: f(x)=x3−3x+2f(x) = x^3 - 3x + 2 on [−2,3][-2, 3].
    1. Use the second derivative test on f(x)=x4−4x3f(x) = x^4 - 4x^3.

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Key Takeaways

  • ✓

    Critical points: f′=0f' = 0 or undefined.

  • ✓

    First derivative test: sign change of f′f'.

  • ✓

    Second derivative test: sign of f′′f'' at critical point.

  • ✓

    Closed interval: check endpoints too.

Tags:
#ap#calculus#applications#extrema#optimization

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