Extreme and intermediate values

Try

For f(x)=(x−1)² on [−1,3], the minimum 0 occurs at x=1 and the maximum 4 occurs at both endpoints. The extreme value theorem guarantees attainment for any continuous function on a nonempty compact real set, even without a formula for the extremizers. It does not locate them or say they are unique. Existence, computation and uniqueness are separate claims.

Predict

Predict what changes on the open interval (−1,3). The minimum still occurs at 1, but the supremum 4 is no longer attained. A continuous function on a bounded domain can miss its largest value because the limiting input is absent. For intermediate values, the domain’s interval structure matters too: a continuous function on two separated pieces may skip all values between their images.

Experiment

Track distinct hypotheses.

Conclusion	Important domain property	What can fail
Attained maximum and minimum	Nonempty compact set	A boundary extremum may be missing
Every intermediate value	Interval plus continuity	A gap or jump can skip values
Unique extremum or root	Additional structure	Existence does not imply uniqueness

Construct an example for each failure column. The table is a dependency map, not a replacement for proof. In particular, a compact two-point domain satisfies the extreme value theorem but does not supply an interval’s intermediate-value conclusion.

Understand

First prove boundedness of the image of compact K under continuous f. Otherwise choose x_n∈K with |f(x_n)|>n. A subsequence converges to a∈K, and continuity would make its outputs converge to finite f(a), contradicting their increasing magnitudes. Let M=sup f(K). Choose x_n with f(x_n)>M−1/n, extract another convergent subsequence, and use continuity to obtain f(a)=M. Apply the same argument to −f for a minimum.

For the intermediate value theorem, suppose f(a)<c<f(b) on [a,b]. Let S consist of points where f≤c and let s=sup S. If f(s)<c, continuity gives points slightly to the right still in S, contradicting the supremum; s<b because f(b)>c. If f(s)>c, continuity excludes S near s, contradicting supremum approximation; s>a because f(a)<c and continuity provides nearby points in S. Hence f(s)=c. Completeness creates s and continuity rules out both alternatives.

Apply

Repair “the intermediate value theorem requires a positive derivative.” No derivative is needed. A continuous oscillating function still cannot jump over a value between two of its values. Conversely, endpoint values of opposite signs alone do not guarantee a zero when continuity is absent: a step from −1 to 1 skips zero.

Bisection can turn the existence idea into nested brackets when a continuous function has opposite endpoint signs. Each retained closed half still brackets a zero and lengths shrink. The common point is a zero by continuity. A finite bisection run yields a bracket, not generally an exact root or proof of uniqueness.

Continue

Optional depth examines continuous images of compact sets and connected sets as general organizing theorems. The arguments here establish their real-line consequences without requiring general topology first.

Write an existence proof with every domain and continuity condition stated before invoking a theorem. Then explain what extra evidence would support uniqueness, such as strict monotonicity for a root. Continue to differentiation: Rolle’s theorem will combine attained extrema with a local derivative argument, and the mean value theorem will connect local slopes to global change.