5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

Syllabus
2020
Topic
5.5
Level

Compare Every Candidate for Global Extrema

For a function on a closed interval [a,b][a,b], an absolute extremum can occur only at an interior critical point or at an endpoint. If the function is continuous on [a,b][a,b], the Extreme Value Theorem also guarantees that an absolute minimum and maximum exist. The Candidates Test locates them by comparing all possible outputs.

  1. Find every cc in (a,b)(a,b) where f(c)=0f'(c)=0 or f(c)f'(c) does not exist, provided f(c)f(c) exists.
  2. Add the endpoints aa and bb to the candidate list.
  3. Evaluate the original function ff at every candidate.
  4. The greatest output is the absolute maximum value; the least is the absolute minimum value. State each value and the input where it occurs.

For f(x)=x2+2x3f(x)=x^2+2x-3 on [3,2][-3,2], f(x)=2x+2=0f'(x)=2x+2=0 gives the interior critical point x=1x=-1. The candidates are therefore 3-3, 1-1, and 22.

Candidate xx f(x)f(x) Conclusion
3-3 00 neither global extreme
1-1 4-4 absolute minimum
22 55 absolute maximum

Compare values of ff, not values of ff'. Endpoints must be checked even though the usual interior derivative test does not classify them as local critical points. If the largest or smallest output occurs at several candidates, report every location.