Why Existence Needs Justification
A graph may appear to have a highest point and a lowest point, but a visual impression is not a proof. To justify that a function actually attains an absolute maximum and an absolute minimum, we need a theorem whose hypotheses match the function and its domain.
The relevant result is the Extreme Value Theorem. Its conditions are important: the function must be continuous on a closed, bounded interval. If those conditions hold, the theorem guarantees that both absolute extrema are attained somewhere in the interval. It does not tell us where they occur or what their values are.
Absolute Maximum and Minimum
Let \(f\) be a function defined on an interval \([a,b]\). A value \(f(c)\) is an absolute maximum value of \(f\) on \([a,b]\) if \(c\in[a,b]\) and
Similarly, \(f(d)\) is an absolute minimum value of \(f\) on \([a,b]\) if \(d\in[a,b]\) and
The values \(f(c)\) and \(f(d)\) are the absolute maximum and minimum values. The inputs \(c\) and \(d\) are points where those values are attained. More than one input may attain the same absolute extremum.
The Extreme Value Theorem
Each part of the hypothesis matters. The interval is closed, so it includes both endpoints, and bounded, so it has finite endpoints. Continuity must hold throughout the interval, including at those endpoints. The conclusion is about attained values: there really are inputs in the domain where the greatest and least values occur.
For an AP justification, state the conditions explicitly rather than simply naming the theorem. A useful structure is: identify the interval, establish continuity on that whole interval, invoke the theorem, and state the existence conclusion.
Worked Example: A Polynomial on a Closed Interval
Worked Example: A Polynomial on a Closed Interval
Justify that \(f(x)=x^2-4x+1\) attains an absolute maximum and minimum on \([-2,3]\).
Step 1: Identify the interval. The domain under consideration is \([-2,3]\). It is closed because it includes \(-2\) and \(3\), and bounded because both endpoints are finite.
Step 2: Verify continuity. The function \(f(x)=x^2-4x+1\) is a polynomial. Polynomials are continuous for every real input, so \(f\) is continuous at every point of \([-2,3]\), including both endpoints.
Step 3: Apply the theorem. Both hypotheses of the Extreme Value Theorem are satisfied. Therefore \(f\) attains an absolute maximum and an absolute minimum on \([-2,3]\).
Check: The function is defined and continuous throughout the specified closed, bounded interval. The conclusion asserts that the values are attained at inputs in \([-2,3]\); it does not claim which inputs or values they are.
Worked Example: A Rational Function
Worked Example: A Rational Function
Justify that \(g(x)=\dfrac{x+1}{x^2+1}\) attains an absolute maximum and minimum on \([-1,2]\).
Step 1: Check the interval. The interval \([-1,2]\) contains its endpoints and has finite endpoints, so it is closed and bounded.
Step 2: Check that the formula is defined throughout the interval. For every real \(x\), \(x^2\geq0\), so
In particular, the denominator is never zero on \([-1,2]\). Thus \(g\) is defined at every point of that interval.
Step 3: Verify continuity. The numerator \(x+1\) and denominator \(x^2+1\) are polynomials, hence continuous. A quotient of continuous functions is continuous wherever its denominator is nonzero. Since \(x^2+1>0\) throughout \([-1,2]\), \(g\) is continuous throughout the interval, including its endpoints.
Step 4: Apply the theorem. The interval is closed and bounded, and \(g\) is continuous on it. Therefore \(g\) attains an absolute maximum and an absolute minimum on \([-1,2]\).
Check: The denominator inequality verifies there is no point in the interval where the formula fails to be defined. Continuity and the interval conditions are both established, so the theorem applies.
Worked Example: A Square-Root Function
Worked Example: A Square-Root Function
Justify that \(h(x)=\sqrt{4-x^2}\) attains an absolute maximum and minimum on \([0,2]\).
Step 1: Check the interval. The interval \([0,2]\) includes both endpoints and is bounded, so it is a closed, bounded interval.
Step 2: Check the square-root input. If \(0\leq x\leq2\), then \(x^2\leq4\). Therefore
The square root is defined for every input in the interval.
Step 3: Verify continuity. The function \(4-x^2\) is a polynomial, so it is continuous everywhere. The square-root function is continuous for nonnegative inputs. Since \(4-x^2\geq0\) on \([0,2]\), their composition \(h(x)=\sqrt{4-x^2}\) is continuous throughout \([0,2]\), including at \(x=0\) and \(x=2\).
Step 4: Apply the theorem. Since \(h\) is continuous on the closed, bounded interval \([0,2]\), it attains an absolute maximum and an absolute minimum there.
Check: The radicand is nonnegative over the entire interval, so the function is defined and continuous there. The theorem guarantees attainment; this justification does not need to calculate the extremum values.
Why the Theorem Works
The theorem’s conclusion is stronger than saying that function values can get arbitrarily close to a greatest or least value. It guarantees that the function actually reaches those values. The following argument explains why continuity and a closed, bounded interval give that guarantee.
We use the Bolzano–Weierstrass property of a closed, bounded interval: every sequence of points in \([a,b]\) has a subsequence that converges to a point still in \([a,b]\). We also use the definition of continuity: if \(x_n\) approaches \(c\) and \(f\) is continuous at \(c\), then \(f(x_n)\) approaches \(f(c)\).
Suppose, to the contrary, that the values of \(f\) on \([a,b]\) are unbounded. Then for every positive integer \(n\), we can choose \(x_n\in[a,b]\) such that \(|f(x_n)|>n\). By the Bolzano–Weierstrass property, some subsequence \(x_{n_k}\) converges to a point \(c\in[a,b]\). Continuity at \(c\) implies \(f(x_{n_k})\to f(c)\), a finite number. But \(|f(x_{n_k})|>n_k\), and \(n_k\to\infty\), so this subsequence cannot converge to a finite number. This contradiction shows that \(f\) is bounded on \([a,b]\).
Because \(f\) is bounded above and its domain is nonempty, the set of its values has a finite supremum \(M\). By the defining property of the supremum, for every positive integer \(n\) there is an \(x_n\in[a,b]\) such that \(M-\frac{1}{n}<f(x_n)\leq M\).
Some subsequence \(x_{n_k}\) converges to a point \(c\in[a,b]\). Continuity gives \(f(x_{n_k})\to f(c)\). Since \(M-\frac{1}{n_k}<f(x_{n_k})\leq M\) and \(\frac{1}{n_k}\to0\), these function values also approach \(M\). Therefore \(f(c)=M\), so the supremum is an attained absolute maximum.
The function \(-f\) is also continuous on \([a,b]\). Applying the maximum argument to \(-f\) shows that \(-f\) attains a maximum at some point \(d\in[a,b]\). Equivalently, \(f\) attains a minimum at \(d\), since maximizing \(-f\) is the same as minimizing \(f\).
This argument uses continuity at the limits of convergent subsequences to turn limiting function values into actual function values. The fact that those limits remain in \([a,b]\) is essential: it is what allows continuity on the interval to apply at the limiting point.
What the Theorem Does—and Does Not—Say
The Extreme Value Theorem establishes existence, not location. From continuity on a closed, bounded interval, we may conclude that there are inputs where the absolute maximum and minimum are attained. The theorem alone does not identify those inputs, provide their function values, or show that either extremum occurs at an endpoint or at an interior point.
To answer a question asking only whether absolute extrema exist, stop after establishing the hypotheses and stating the existence conclusion. A question asking for the extrema themselves requires additional work. Keep that distinction clear: invoking the theorem is a justification of existence, not a calculation of the maximum or minimum.
Common Pitfall: Checking Only the Interval
A closed, bounded interval is not enough on its own. The function must also be continuous throughout that interval. For example, define \(q\) on \([0,1]\) by
The interval \([0,1]\) is closed and bounded, and \(q\) is defined at every point of it. But \(q\) is not continuous at \(0\). Moreover, for each positive integer \(n\), \(1/n\in[0,1]\) and
These values grow without bound, so \(q\) has no absolute maximum on \([0,1]\). The following attempted justification is therefore incorrect:
Incorrect: “The domain is the closed, bounded interval \([0,1]\), so the Extreme Value Theorem guarantees an absolute maximum and minimum.”
Correction: The theorem also requires continuity on the entire interval. Since \(q\) is discontinuous at \(0\), its hypotheses are not met, and the theorem provides no conclusion about the existence of extrema.
The converse also needs care: failing a hypothesis means the theorem cannot be applied, not that an extremum definitely does not exist. A separate argument would be needed to establish existence or nonexistence in that case.
A Clear Justification in an Exam Response
A concise response should connect each hypothesis to evidence and then state exactly what follows. For example:
The reasoning is complete because it identifies the interval, gives a reason for continuity, verifies that the interval is closed and bounded, names the applicable theorem, and states its conclusion. A response that only says “the graph has a highest and lowest point” does not provide this justification.
Check Your Understanding
For each question, focus on the theorem’s hypotheses and the precise scope of its conclusion.
- State the two main hypotheses of the Extreme Value Theorem for an interval.
- Why must continuity be checked at the endpoints as well as at interior points?
- A function is continuous on \([-3,4]\). What does the theorem guarantee, and what does it not determine?
- For \(r(x)=\dfrac{1}{x-2}\) on \([0,1]\), explain why the theorem applies or does not apply.
- If a function is discontinuous at one point of a closed, bounded interval, what can and cannot be concluded from the Extreme Value Theorem?
