Why an Existence Guarantee Matters
A function may have values that get closer and closer to a highest level without ever reaching it. The Extreme Value Theorem rules out that possibility when the function is continuous on a closed interval: the function must actually attain both a greatest value and a least value there.
This guarantee is useful even before we know how to find the extrema. It tells us that the search has a definite answer: under the theorem’s conditions, there is at least one input where the highest value occurs and at least one where the lowest value occurs. The theorem does not, by itself, tell us which inputs they are.
The word attains is important. An absolute maximum is a function value \(f(c)\) at an input \(c\) in the domain, not merely a number that the function’s values approach. More than one input can attain the same absolute maximum or minimum.
The Extreme Value Theorem
For a nondegenerate interval, the hypotheses have two parts: \(f\) must be continuous at every point of the interval, and the domain under consideration must be a closed, bounded interval \([a,b]\). The endpoints are included. If \(a=b\), the interval contains just one point, and both the minimum and maximum are simply \(f(a)\).
The theorem makes an existence claim, not a location claim. It does not say that the extrema occur at the endpoints, that each occurs at only one input, or that a particular method has already identified where they occur.
Worked Example: A Maximum and Minimum from a Square
Worked Example: A Maximum and Minimum from a Square
Consider \(f(x)=5-(x-2)^2\) on \([0,4]\). Since \(0\le x\le4\), subtracting \(2\) gives \(-2\le x-2\le2\). Therefore, \(0\le(x-2)^2\le4\), and so
This proves that no function value on the interval is below \(1\) or above \(5\). At \(x=2\),
At the endpoints,
Thus the absolute maximum value is \(5\), attained at \(x=2\), and the absolute minimum value is \(1\), attained at both endpoints. The bounds and substitutions verify both claims.
Why Continuity and the Closed Interval Guarantee Attainment
The key idea in the proof is that any sequence of inputs in a closed, bounded interval has a subsequence that converges to a point still in that interval. Continuity then carries the limiting behavior of the inputs over to the function values. Here is the argument in detail.
Bisect \([a,b]\). At least one of the two closed halves contains infinitely many terms of a given sequence in \([a,b]\). Keep such a half and bisect it again. Repeating this produces nested closed intervals, each containing infinitely many terms, whose lengths are \((b-a)/2^k\) after \(k\) bisections. Choose terms with increasing indices, taking the \(k\)th chosen term from the \(k\)th interval. The intervals have a single common point \(c\in[a,b]\), and their lengths approach \(0\). The chosen subsequence therefore converges to \(c\).
Suppose, to the contrary, that the function values were unbounded in absolute value. We could choose \(x_n\in[a,b]\) with \(|f(x_n)|>n\) for each positive integer \(n\). By Step 1, some subsequence \(x_{n_k}\) converges to \(c\in[a,b]\). Since \(f\) is continuous at \(c\), \(f(x_{n_k})\) converges to the finite value \(f(c)\), so those function values must be bounded. This contradicts \(|f(x_{n_k})|>n_k\). Thus the function values are bounded.
Because the function values are bounded and the interval is nonempty, their set has a supremum \(M\), meaning the least upper bound. For each positive integer \(n\), the definition of supremum gives an input \(x_n\in[a,b]\) with \(M-\frac{1}{n}<f(x_n)\le M\). Otherwise \(M-\frac{1}{n}\) would be an upper bound smaller than \(M\).
By Step 1, a subsequence \(x_{n_k}\) converges to some \(c\in[a,b]\). The inequalities \(M-\frac{1}{n_k}<f(x_{n_k})\le M\), together with \(n_k\to\infty\), imply that \(f(x_{n_k})\to M\). Continuity at \(c\) also implies \(f(x_{n_k})\to f(c)\). Limits are unique, so \(f(c)=M\). The supremum is therefore an actual function value: \(f\) attains its absolute maximum.
The function \(-f\) is continuous on \([a,b]\), so the argument above shows that \(-f\) attains an absolute maximum there. If that maximum occurs at \(d\), then \(-f(d)\ge -f(x)\) for every \(x\in[a,b]\). Reversing signs gives \(f(d)\le f(x)\) for every \(x\in[a,b]\). Thus \(f\) attains an absolute minimum as well.
This proof explains the roles of the hypotheses. Boundedness ensures that there are finite upper and lower bounds to consider. The closed interval ensures that a limit point reached by a subsequence remains in the domain. Continuity ensures that function values along inputs approaching that point converge to the value at the point. Together, these facts turn a limiting bound into an attained value.
Worked Example: Absolute Extrema of an Absolute-Value Function
Worked Example: Absolute Extrema of an Absolute-Value Function
Let \(g(x)=|x-1|\) on \([0,3]\). For \(x\in[0,3]\), we have \(-1\le x-1\le2\). Hence \(0\le|x-1|\le2\). The lower bound is attained at \(x=1\):
The upper bound is attained at \(x=3\):
Therefore, the absolute minimum value is \(0\), and the absolute maximum value is \(2\). For an additional check at the other endpoint, \(g(0)=|0-1|=1\), which lies between \(0\) and \(2\). The theorem guarantees that extrema exist because \(g\) is continuous on \([0,3]\); the bounds and substitutions identify and verify them in this example.
Worked Example: Extrema of a Cubic on an Interval
Worked Example: Extrema of a Cubic on an Interval
Consider \(h(x)=x^3\) on \([-2,1]\). To bound its values directly, first note that \(x+2\ge0\) on this interval and
Thus \((x+2)(x^2-2x+4)\ge0\). Since this product equals \(x^3+8\), it follows that \(x^3\ge-8\). For the upper bound, \(1-x\ge0\), and
Therefore, \((1-x)(x^2+x+1)\ge0\). This product equals \(1-x^3\), so \(x^3\le1\). We have proved \(-8\le h(x)\le1\) throughout the interval. At the endpoints,
The bounds are both attained: the absolute minimum is \(-8\) at \(x=-2\), and the absolute maximum is \(1\) at \(x=1\). Substitution into the function confirms each value.
What the Theorem Does Not Tell You
A common mistake is to treat the existence guarantee as a rule about where an extremum must occur. For example, consider \(p(x)=4-(x-1)^2\) on \([0,2]\). A mistaken conclusion might be, “The interval has endpoints \(0\) and \(2\), so the maximum must be at one of them.” But direct substitution gives
At the interior input \(x=1\), however,
Indeed, \((x-1)^2\ge0\) implies \(p(x)\le4\), with equality at \(x=1\). The absolute maximum is \(4\), not the endpoint value \(3\). The correction is simple: the Extreme Value Theorem guarantees that a maximum and minimum are attained somewhere in the interval. It does not say they occur at the endpoints.
The distinction matters in applications. If a continuous quantity, such as a modeled height or cost, is considered over a fixed closed interval of inputs, the theorem guarantees that its highest and lowest values are reached. To identify those values or the inputs where they occur, additional reasoning is needed.
Check Your Understanding
Use the theorem’s hypotheses and the meaning of “attains” to answer these questions.
- State the Extreme Value Theorem in your own words, including its hypotheses.
- What does it mean for \(f\) to attain an absolute maximum at \(c\)?
- Why does continuity matter in the proof when a sequence of inputs converges to a point in the interval?
- Does the theorem guarantee that an absolute maximum is unique? Explain.
- Does the theorem guarantee that an absolute maximum occurs at an endpoint? Explain.
