◀ ▲ ▶Branches / Analysis / Proposition: Continuous Real Functions on Closed Intervals Take Maximum and Minimum Values within these Intervals
The following proposition is an important result about how continuous functions behave on closed real intervals. It turns out that once a function is continuous, there is always a constant such that it cannot be exceeded by the function values, no matter which function we will take. This works only for closed intervals.
Proposition: Continuous Real Functions on Closed Intervals Take Maximum and Minimum Values within these Intervals
Let \([a,b]\) be a closed real interval and let \(f:[a,b]\to\mathbb R\) be an arbitrary continuous real function. Then there are two numbers $p,q\in[a,b]$ with
$$\begin{array}{rcl}f(p)&=&\max\{f(x):~x\in[a,b]\},\\f(q)&=&\min\{f(x):~x\in[a,b]\}.\end{array}$$
Notes
 This proposition does not work if the interval is open.
 For instance, the function $f:(0,1]\to\mathbb R$, $f(x)=\frac 2x$ is continuous and it takes a minimum at $x=1$ (which is $2$), but does not take a maximum on the open interval $(0,1].$
 Another example is the function $g:(0,1)\to\mathbb R$, $g(x)=x.$ It neither takes a maximum nor a minimum for all values of $x\in(0,1)$.
Table of Contents
Proofs: 1 Corollaries: 1
Mentioned in:
Chapters: 1
Explanations: 2
Proofs: 3
Thank you to the contributors under CC BYSA 4.0!
 Github:

References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983