This post discusses when a real-valued 1-variable function has global extrema. As usual in rigorous mathematics we start from proper definitions of key concepts.

### Proper definitions of global extrema

If a function \(f(x)\) is defined over a domain \(D\), then a *global maximum* of \(f(x)\) is a point \(a\in D\) such that \(f(a)\geq f(x)\) for all \(x\in D\).

Such a maximum is called global (or absolute), because the inequality \(f(a)\geq f(x)\) should hold over the whole domain \(D\), not only around the point \(x=a\) as for a local maximum.

Similarly, a *global minimum* of \(f(x)\) is a point \(a\in D\) such that \(f(a)\leq f(x)\) for all \(x\in D\).

### Functions without global extrema over an open interval

The simple function \(f(x)=x\) has no global extrema over the whole real line, because the set of all possible values of \(f(x)\) is unbounded.

If we restrict the function \(f(x)=x\) to any open interval \((a,b)\), then the values of \(f(x)=x\) are in the bounded set \((a,b)\), but this set does not include its boundary points \(a,b\). Hence \(f(x)=x\) again has no global extrema over any open interval \((a,b)\).

However, the function \(f(x)=x\) on any closed interval \([a,b]\), has the global minimum at \(x=a\) and the global maximum at \(x=b\), because \(f(a)\leq f(x)\leq f(b)\) for all \(x\in[a,b]\).

### Discontinuous functions without global extrema

The function \(f(x)=\left\{ \begin{array}{l} \tan x \mbox{ for } -\frac{\pi}{2}<x<\frac{\pi}{2},\\ 0 \mbox{ for } x=\pm\frac{\pi}{2} \end{array} \right. \) has no global extrema over the closed interval \([-\frac{\pi}{2},\frac{\pi}{2}]\). Indeed, the set of all possible values of \(f(x)\) is the unbounded real line.

A good exercise is to prove that any continuous function \(f(x)\) over a closed interval \([a,b]\) is bounded, namely there are bounds \(c,d\) such that \(c\leq f(x)\leq d\) for all \(x\in [a,b]\).

A proof requires a proper understanding of continuity via the epsilon-delta argument. If this theorem is examinable in your calculus course, you could let us know in the comments, which would certainly promote your university.

### When does a global extremum exist?

It turns out that a bad domain and discontinuity of a function are the only reasons why global extrema do not exist in the examples above. The powerful theorem below covers the exercise on boundedness of continuous functions over closed intervals, which we stated above.

**The Extreme Value Theorem. **

Any continuous function over a closed interval has a global maximum and a global minimum.

**Riddle 16**: is there a continuous surface with exactly 2 global

maxima, but without any global minima over the whole plane?**How to submit**: to write your full answer, simply submit a comment.**Hint**: definitions of global extrema are similar for 2-variable functions.**Warning**: give an example or justify that there is no such a surface.**Prize**: free 1-year access to one of our interactive web tutorials.**Restriction**: only the first correct public answer will be rewarded.

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This article is about continuous functions in analysis. For the statistical theorem in extreme value theory , see Fisher–Tippett–Gnedenko theorem .

Thank you for the reference. Here is the link to Fisher–Tippett–Gnedenko theorem at Wikipedia.

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