Absolute maximum and minimum values: attainment and domains
Distinguish attained extrema from bounds, compare all candidates on closed intervals, and handle open or unbounded domains.
Content updated: October 22, 2026
An absolute maximum or minimum must be attained at a point in the specified domain. A limit at an excluded endpoint may only be a bound, while a nondifferentiable point can still give an absolute minimum.
Objectives: check existence and attainment, compare candidates on closed intervals, and distinguish open-interval questions and local extrema.
Prerequisites: local extrema (English lesson); continuity on a closed interval (English lesson) and limits.
On this page: Idea · Method · Examples · Practice · Further learning
1. Bounds and attained values
Let $f$ be defined on a nonempty set $K$. A number $M$ is its absolute maximum value if $f(x)\le M$ for every $x\in K$ and there is an $x_1\in K$ with $f(x_1)=M$. An absolute minimum value $m$ satisfies the corresponding conditions $f(x)\ge m$ and equality at a point in $K$. Write $\max_K f$ and $\min_K f$.
An upper or lower bound alone is insufficient: identify a point attaining it. A local extremum compares nearby points, while an absolute maximum or minimum compares all of $K$. Changing the domain may create or remove an absolute maximum or minimum.
2. Methods on closed and open intervals
By the extreme value theorem (Weierstrass's theorem), a function continuous on a finite closed interval $[a;b]$ attains both an absolute maximum and an absolute minimum. Being merely defined there is insufficient: $g(x)=x$ for $0\le x<1$, with $g(1)=0$, is defined on $[0;1]$ but does not attain the upper bound $1$, so has no absolute maximum.
For a continuous function on a closed interval, list the two endpoints, interior points where the derivative is $0$, and interior points where it does not exist. When the candidate sets are finite, calculate every value and compare. If an absolute extremum occurs in the interior and the derivative exists, Fermat's theorem forces the derivative to be $0$. Do not omit a corner such as that of $|x|$. Infinite candidate sets require further reasoning; a finite list is not automatic.
On an open interval or unbounded domain, also examine boundary limits and limits at infinity to determine bounds for the whole domain, then check whether they are attained inside it. Do not automatically treat a limit at an excluded point as a function value. Conversely, do not reject a value merely because it equals a boundary limit: for example, $x^2(x-1)^2$ on $(0;2)$ has minimum $0$ at $1$, although its limit at $0^+$ is also $0$.
3. Three worked examples
Example 1. A minimum but no maximum on R
For $f(x)=x^2-x+1=(x-1/2)^2+3/4$, we have $f(x)\ge3/4$, with equality at $x=1/2$. Thus $\min_{\mathbb R}f=3/4$. As $x\to+\infty$, $f(x)\to+\infty$, so the function is unbounded above and has no absolute maximum.
Example 2. A polynomial on a closed interval
Let $f(x)=x^3-3x^2+1$ on $[0;4]$. Continuity ensures attainment of both extrema. The derivative $3x(x-2)$ has roots $0,2$; the distinct candidates are $0,2,4$. We have
Hence the minimum is $-3$ at $2$ and the maximum is $17$ at $4$. The absolute minimum here is also the local minimum value; the maximum occurs at an endpoint and is not an interior local maximum. On the whole of $\mathbb R$, the function has local maximum $1$ at $0$ and local minimum $-3$ at $2$, but no absolute maximum or minimum, since it is unbounded below and above in the two directions at infinity.
Example 3. Endpoints excluded from the domain
Find the absolute maximum and minimum of $f(x)=x^3-3x+2$ on $(0;2)$. Its derivative $3(x-1)(x+1)$ is negative on $(0;1)$ and positive on $(1;2)$; the only stationary input in the interval is $1$. The value and boundary limits are
We have $f(x)=(x-1)^2(x+2)\ge0$, with equality at $1$. On the other hand, $f(x)-4=(x-2)(x+1)^2<0$ for $0<x<2$. Thus the minimum is $0$ at $1$, while the supremum is $4$ but is not attained, so there is no maximum. One cannot choose the “largest remaining value” after discarding the limit $4$.
4. Practice exercises
- Find the absolute maximum and minimum of $x^2$ on $[-1;2]$.
- Find the absolute maximum and minimum of $|x-1|$ on $[0;3]$. Why must $x=1$ be considered?
- Find the absolute maximum and minimum of $f(x)=x$ on $(0;1]$.
Show answers and hints
- The minimum is $0$ at $0$; the maximum is $4$ at $2$.
- The minimum is $0$ at $1$; the maximum is $2$ at $3$. The function is not differentiable at $1$, but that candidate cannot be omitted.
- The maximum is $1$ at $1$. There is no minimum, since each $x>0$ in the domain has a smaller input $x/2$; the lower bound $0$ is not attained.
5. Common mistakes
- Assuming that being defined on a closed interval ensures extrema without checking continuity.
- Omitting endpoints or nondifferentiable points.
- Confusing an unattained bound with an absolute maximum or minimum.
- Treating a point outside the domain as a place of attainment.
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