Why Endpoints Matter
A function can reach its greatest or least value at an interior point, but it can also reach either value at an endpoint. To classify absolute extrema on a closed interval, you must check both kinds of locations.
The method is systematic: first identify the endpoints and all interior critical points, then evaluate the function at every candidate and compare the resulting values. The largest candidate value is the absolute maximum; the smallest is the absolute minimum.
Definitions and Conditions
Let \(f\) be a real-valued function defined on the closed interval \([a,b]\), where \(a<b\). An absolute maximum occurs at a point \(x=M\) if \(f(M)\geq f(x)\) for every \(x\in[a,b]\). An absolute minimum occurs at \(x=m\) if \(f(m)\leq f(x)\) for every \(x\in[a,b]\). The maximum and minimum are the corresponding function values, \(f(M)\) and \(f(m)\).
A point \(c\) in the interior \((a,b)\) is a critical point of \(f\) if \(f'(c)=0\), or if \(f'(c)\) does not exist. In this procedure, endpoints are listed separately, whether or not a derivative exists there.
Continuity on the entire interval is essential to the existence guarantee. If \(f\) is discontinuous, an absolute maximum or minimum might not exist. A point where \(f'\) does not exist must not be ignored: it may be where an extreme occurs, as happens at a corner.
The Candidate-Value Method
For a specific problem, use the following sequence. Include each relevant location once, evaluate the original function there, and compare the values—not merely the locations.
Check that the function is defined and continuous throughout the closed interval, so the existence guarantee applies.
On \((a,b)\), solve \(f'(x)=0\) and identify points where the derivative does not exist.
Include \(a\), \(b\), and every interior critical point.
Find \(f(x)\) at every candidate. The greatest value is the absolute maximum and the least is the absolute minimum.
A critical point is only a candidate; it is not automatically an absolute maximum or minimum. It is the function values that determine the classification. If several candidates share the greatest or least value, the absolute extremum occurs at each of those locations.
Why the Candidate List Is Complete
The method rests on two results. The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval attains both an absolute maximum and an absolute minimum. Fermat’s Theorem explains why a differentiable interior location of an extremum must have derivative zero.
Here is the reasoning behind Fermat’s Theorem. Suppose \(f\) has a local maximum at \(c\) and is differentiable there. For sufficiently small \(h\), \(f(c+h)-f(c)\leq 0\). If \(h>0\), dividing by \(h\) gives a difference quotient at most zero. If \(h<0\), division reverses the inequality, so the quotient is at least zero. Because the derivative exists, both one-sided limits must equal the same number \(f'(c)\). That number must be both at most zero and at least zero, hence it is zero. At a local minimum, the inequalities reverse, and the same conclusion follows.
To prove the closed-interval procedure, begin with a function continuous on \([a,b]\). By the Extreme Value Theorem, it attains an absolute maximum and minimum. Consider either one. If it occurs at \(a\) or \(b\), it is an endpoint candidate. If it occurs at an interior point where the derivative does not exist, it is a critical point by definition. If it occurs at an interior point where the derivative exists, Fermat’s Theorem gives \(f'(c)=0\), so it is also a critical point. Thus each absolute extremum appears on the candidate list. Comparing the values on that list therefore finds both extrema.
Worked Examples
Worked Example: A Polynomial with Tied Absolute Extrema
Find the absolute maximum and minimum of \(f(x)=x^3-3x\) on \([-2,2]\).
The function is a polynomial, so it is continuous on \([-2,2]\). Differentiate and solve for interior critical points:
Thus \(f'(x)=0\) at \(x=-1\) and \(x=1\), both in \((-2,2)\). The complete candidate list is \(-2,-1,1,2\). Evaluate the original function at each:
The greatest candidate value is \(2\), attained at \(x=-1\) and \(x=2\). The least is \(-2\), attained at \(x=-2\) and \(x=1\). As a check, substituting each reported location into \(x^3-3x\) gives exactly the listed value; the comparisons include both endpoints and both interior critical points.
Worked Example: A Corner as a Critical Point
Find the absolute extrema of \(g(x)=|x-1|\) on \([0,3]\).
The absolute-value function is continuous on the interval. For \(x<1\), \(g(x)=1-x\), so \(g'(x)=-1\). For \(x>1\), \(g(x)=x-1\), so \(g'(x)=1\). At \(x=1\), the left and right derivatives differ, so \(g'(1)\) does not exist. Therefore \(x=1\) is an interior critical point. The candidate list is \(0,1,3\).
The absolute minimum is \(0\) at \(x=1\), and the absolute maximum is \(2\) at \(x=3\). Substitution into the original absolute-value formula confirms all three candidate values. This example shows why checking only solutions of \(g'(x)=0\) would be incomplete: the derivative is never zero, but the corner is where the minimum occurs.
Worked Example: Comparing Several Interior and Endpoint Values
Find the absolute maximum and minimum of \(h(x)=x^3-6x^2+9x+1\) on \([0,4]\).
This polynomial is continuous on \([0,4]\). Its derivative is
The interior critical points are \(x=1\) and \(x=3\). Include the endpoints to obtain the candidate list \(0,1,3,4\). Now evaluate:
The greatest value is \(5\), attained at \(x=1\) and \(x=4\). The least is \(1\), attained at \(x=0\) and \(x=3\). Direct substitution in the defining polynomial verifies each candidate value, and comparing all four values confirms both classifications, including the ties.
Common Pitfall: Omitting the Endpoints
A frequent error is to find the interior critical points and treat them as the entire candidate list. For example, consider \(p(x)=x^2-2x\) on \([0,3]\). Its derivative is \(p'(x)=2x-2\), so the only interior critical point is \(x=1\). An incomplete calculation might evaluate only \(p(1)=-1\) and claim that this is both the absolute minimum and maximum because it is the only value checked. That conclusion is not justified.
The correct list includes \(0,1,3\). The function values are
Thus the absolute minimum is \(-1\) at \(x=1\), while the absolute maximum is \(3\) at \(x=3\). Substitution confirms the endpoint values. The correction is simple but essential: endpoints are always candidates on a closed interval, even when the derivative has no critical point there.
Writing a Complete Conclusion
A complete response should show enough work to make the candidate list and comparison clear. State the interval, identify interior critical points, evaluate the function at every candidate, and name both the extreme value and where it occurs. If an extreme is attained at multiple points, report all of them.
For example, “The absolute maximum is \(5\) at \(x=1\) and \(x=4\); the absolute minimum is \(1\) at \(x=0\) and \(x=3\)” communicates both values and locations. Reporting only the \(x\)-coordinates, or only the function values, leaves the classification incomplete.
Check Your Understanding
For each question, use the closed-interval candidate method and explain your comparison.
- For a continuous function on \([a,b]\), which locations must be included in the candidate list?
- Why can an interior point where the derivative does not exist be important?
- Find the absolute extrema of \(q(x)=x^2-4x\) on \([0,5]\), showing the candidate values.
- A differentiable function has an absolute maximum at an interior point \(c\). What does Fermat’s Theorem imply about \(f'(c)\), and why?
- What should you report if two different candidates give the same greatest value?
