#math/calculus #flashcards/math The `extreme value theorem` states: %%::%%If $f$ is continuous on a closed, bounded interval $[a,b]$, then $f$ attains both a maximum and a minimum value on $[a,b]$. Both hypotheses matter. The function $f(x)=x$ on the open interval $(0,1)$ attains neither endpoint bound. On the closed interval $[0,1]$, the discontinuous function defined by $f(0)=0$ and $f(x)=1/x$ for $x>0$ is unbounded above. **Proof sketch.** Continuity on $[a,b]$ first forces $f$ to be bounded. Otherwise, choose $x_{n}\in[a,b]$ with $|f(x_{n})|>n$; a convergent subsequence has a limit $x_{*}\in[a,b]$, contradicting continuity at $x_{*}$. Now let $M=\sup f([a,b])$ and choose $x_{n}$ with $f(x_{n})>M-1/n$. A convergent subsequence $x_{n_{k}}\to x_{*}$ satisfies $f(x_{*})=M$ by continuity. Applying the same argument to $-f$ gives a minimum. $\blacksquare$ The theorem guarantees existence; [[3B4 - Closed Interval Test]] gives a practical way to find the extrema.