TheoremBase

A Continuous Injective Function on a Closed Interval is Strictly Monotone

Analysis level. Injectivity plus continuity on a closed real interval forces strict monotonicity; the whole argument is repeated use of the intermediate value theorem.

Statement

In the setting of Single-Variable Calculus on an Interval, let a,b∈Ra,b\in\mathbb{R} with a<ba<b, let [a,b][a,b] be the closed interval determined by aa and bb, and let f:[a,b]→Rf:[a,b]\to\mathbb{R} be continuous on [a,b][a,b] and injective, that is,

f(x)≠f(y)whenever x,y∈[a,b] and x≠y.f(x)\ne f(y)\qquad\text{whenever }x,y\in[a,b]\text{ and }x\ne y .

Problem. Show that ff is strictly monotone on [a,b][a,b].

Solutions

Log in to submit a solution.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…