If outputs get closer and closer to a bound, must a maximum or minimum exist?
My prediction and reason
For y = x, predict the maximum and minimum on [0, 2], then on (0, 2). Decide whether removing only the endpoints changes the answer.
Evidence and observations
For the straight segments joining water-level points (0, 1), (2, 5), (4, 2), (6, 4), record the rising and falling intervals.
Record the global maximum and minimum water levels and their times. Compare the local low at time 4 with the global low.
For y = x on (0, 2), choose an output near 2 and then a larger allowed one; do the same near 0 with a smaller allowed one.
Explain or give a counterexample
Transfer
For f(x) = (x − 2)² + 1 on [3, 5], find both extrema. Explain why the unrestricted vertex cannot provide the minimum on this new domain.
Exit task
For f(x) = x on (0, 2], state the maximum and minimum, if they exist, and justify the endpoint decisions.
5 min predict open and closed intervals; 12 min read the water graph; 8 min explain attainment; 12 min paper extrema; 3 min exit check.
If outputs get closer and closer to a bound, must a maximum or minimum exist?
For y = x, predict the maximum and minimum on [0, 2], then on (0, 2). Decide whether removing only the endpoints changes the answer.
Evidence tasks
- For the straight segments joining water-level points (0, 1), (2, 5), (4, 2), (6, 4), record the rising and falling intervals.
- Record the global maximum and minimum water levels and their times. Compare the local low at time 4 with the global low.
- For y = x on (0, 2), choose an output near 2 and then a larger allowed one; do the same near 0 with a smaller allowed one.
Reference reasoning
Read a graph from left to right: increasing means greater inputs give greater outputs within the stated interval. In the piecewise-linear water model, the maximum is 5 at time 2 and the minimum is 1 at time 0; the local low 2 at time 4 is not the global minimum. An extreme value must be attained. For y = x on (0, 2), any proposed output x can be exceeded by (x + 2)/2 and undercut by x/2, both still allowed. Thus neither a maximum nor a minimum exists there, although 0 and 2 are bounds.
Describe the monotonic intervals and the minimum on the given domain.
f(x)=(x−2)2+1,x∈[0,5] - The vertex is at (2, 1); the function decreases on [0, 2] and increases on [2, 5].
- The minimum is 1, attained at x = 2.
Compare the extrema before and after removing the endpoints.
f(x)=x;x∈[0,2] or x∈(0,2) - On [0, 2], the minimum is 0 at x = 0 and the maximum is 2 at x = 2.
- On (0, 2), neither endpoint is allowed, and every allowed output has a larger and a smaller allowed output. Neither extreme exists.
Transfer and exit task
For f(x) = (x − 2)² + 1 on [3, 5], find both extrema. Explain why the unrestricted vertex cannot provide the minimum on this new domain.
For f(x) = x on (0, 2], state the maximum and minimum, if they exist, and justify the endpoint decisions.
The maximum is 2 at x = 2. There is no minimum: 0 is excluded, and for any allowed x, x/2 is allowed and gives a smaller output.
- Distinguishes an attained maximum from an unattained lower bound.
- Gives a reason valid for every proposed minimum rather than only a few decimal samples.
Questions
- Is 2 the value, or the input where the value occurs?
- If a student proposes 0.001 as the minimum on (0, 2), can you always find a smaller allowed value?
Expected responses
- Students may label the local low at time 4 as the minimum. Compare it with the included starting point.
- Students may use 0 and 2 as extrema on the open interval. Check the membership of their input points.
Confusions
- For the water example, the connecting straight segments are a stated model assumption.
- A finite graph window cannot establish the global extrema of a function on an unbounded domain.
Board plan
- Monotonicity is stated on an interval.
- Extreme value + attaining input.
- A bound is an extreme only when attained.
