From Critical Numbers to Absolute Extrema
A derivative can identify where a function has a horizontal tangent, but that information alone does not determine the greatest or least value on an interval. An absolute maximum or minimum may occur at an endpoint, or at an interior critical number. To find absolute extrema reliably, list every possible candidate, evaluate the function at each one, and compare the resulting values.
For polynomials and rational functions, this process is systematic. The key conditions are that the interval is closed and bounded, and that the function is defined and continuous throughout it. Rational functions require particular care: their denominators must not be zero anywhere in the interval.
The Candidate-Point Method
A critical number is an interior number \(c\) in the domain of \(f\) where \(f'(c)=0\) or \(f'(c)\) does not exist. For polynomials and rational functions, the derivative exists at every point where the function itself is defined. Thus, for these functions, interior critical numbers are found by solving \(f'(x)=0\) and retaining only solutions that lie in the interval and in the function’s domain.
On a closed interval \([a,b]\), include both endpoints \(a\) and \(b\) in the candidate list, whether or not the derivative is zero there. The candidate list consists of the endpoints together with all critical numbers strictly between them.
A polynomial is continuous and differentiable everywhere. A rational function \(f(x)=\frac{P(x)}{Q(x)}\), where \(P\) and \(Q\) are polynomials, is continuous and differentiable wherever \(Q(x)\neq 0\). Therefore, the theorem applies to a rational function on \([a,b]\) only when its denominator is nonzero throughout that interval.
Confirm that the interval is closed and bounded and that the function is defined and continuous at every point in it. For a rational function, check its denominator throughout the interval.
Differentiate, solve \(f'(x)=0\), and keep only solutions strictly inside the interval and in the function’s domain.
Find the function values at every retained critical number and at both endpoints.
The largest candidate value is the absolute maximum; the smallest is the absolute minimum. Report every candidate location attaining either value.
Why the Candidate List Works
The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval actually attains its greatest and least values. It remains to explain why an interior location of an extremum must appear on the critical-number list.
Suppose \(c\) is an interior location of an absolute maximum and \(f\) is differentiable at \(c\). For sufficiently small positive \(h\), the point \(c+h\) is in the interval, and \(f(c+h)\leq f(c)\). Hence \(\frac{f(c+h)-f(c)}{h}\leq 0\). For sufficiently small negative \(h\), the numerator is still nonpositive but the denominator is negative, so \(\frac{f(c+h)-f(c)}{h}\geq 0\). If the derivative exists, the two-sided limit of these difference quotients must satisfy both inequalities. It follows that \(f'(c)=0\). At an interior minimum the inequalities reverse, and the same reasoning again gives \(f'(c)=0\). So an interior extremum of a differentiable function must be a critical number; the only other candidates are the endpoints.
Worked Examples
Worked Example: A Polynomial with Interior Critical Numbers
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 the entire closed interval. Differentiate:
Set the derivative equal to zero:
Both critical numbers lie in \([-2,2]\). The complete candidate list is \(-2,-1,1,2\). Evaluate the function at each:
The largest value is \(2\), attained at \(x=-1\) and \(x=2\). The smallest value is \(-2\), attained at \(x=-2\) and \(x=1\). As a check, substituting each reported location into \(x^3-3x\) gives the stated value; the derivative equation also confirms that the two interior locations are critical numbers.
Worked Example: A Rational Function with One Critical Number
Find the absolute maximum and minimum of \(f(x)=\frac{x+1}{x^2+1}\) on \([-1,2]\).
The denominator \(x^2+1\) is positive for every real \(x\), so the function is defined and continuous on the entire interval. Apply the quotient rule:
The denominator of \(f'(x)\) is positive, so \(f'(x)=0\) precisely when \(1-2x-x^2=0\):
The value \(-1-\sqrt{2}\) is less than \(-1\), so it is outside the interval. The value \(c=-1+\sqrt{2}\) is inside \([-1,2]\). Evaluate at the endpoints and at \(c\):
At \(c=\sqrt{2}-1\), the numerator is \(c+1=\sqrt{2}\), and
Therefore,
The candidates give \(0\), \(\frac{3}{5}\), and \(\frac{\sqrt{2}+1}{2}\). Since \(\frac{\sqrt{2}+1}{2}>\frac{3}{5}>0\), the absolute maximum is \(\frac{\sqrt{2}+1}{2}\) at \(x=-1+\sqrt{2}\), and the absolute minimum is \(0\) at \(x=-1\). Substituting \(c\) into the original function gives the same value calculated above, and the derivative numerator vanishes at \(c\), checking both calculations.
Worked Example: Comparing a Rational Critical Value with Both Endpoints
Find the absolute extrema of \(g(x)=\frac{x^2+1}{x+2}\) on \([-1,1]\).
The denominator \(x+2\) is positive on this interval: its smallest value is \(1\), at \(x=-1\). Thus \(g\) is continuous there. Differentiate using the quotient rule:
Since the denominator is nonzero, critical numbers satisfy \(x^2+4x-1=0\):
Only \(c=-2+\sqrt{5}\) lies in \([-1,1]\). The endpoint values are
To evaluate at \(c\), use \(c^2+4c-1=0\), so \(c^2=1-4c\). Then \(c^2+1=2-4c\), and
This critical value is positive because \(\sqrt{5}>2\). It is less than \(\frac{2}{3}\), since \(2\sqrt{5}-4<\frac{2}{3}\) is equivalent to \(\sqrt{5}<\frac{7}{3}\), which follows by squaring the positive quantities: \(5<\frac{49}{9}\). Thus the absolute maximum is \(2\) at \(x=-1\), and the absolute minimum is \(2\sqrt{5}-4\) at \(x=-2+\sqrt{5}\). Substitution into the original formula verifies the endpoint values, and the critical-number equation verifies the interior location.
Common Mistakes and Domain Checks
A frequent error is to find critical numbers but omit one or both endpoints. For example, for \(p(x)=x^2\) on \([-2,1]\), the only interior critical number is \(x=0\), and \(p(0)=0\). If the endpoints are ignored, one might incorrectly report \(0\) as the maximum as well as the minimum. But \(p(-2)=4\) and \(p(1)=1\), so the actual maximum is \(4\), attained at \(-2\), and the minimum is \(0\), attained at \(0\).
Another important error is applying the method when a rational function is not defined throughout the interval. Consider \(r(x)=\frac{1}{x-1}\) on the proposed interval \([0,2]\). A mistaken endpoint-only calculation gives \(r(0)=-1\) and \(r(2)=1\), and might claim these are the minimum and maximum. But \(r\) is undefined at \(x=1\), so it is not continuous on the whole interval and the Extreme Value Theorem does not apply. In fact, \(r(x)\) decreases without bound as \(x\) approaches \(1\) from the left, and increases without bound as \(x\) approaches \(1\) from the right. It has neither an absolute minimum nor an absolute maximum on its domain within \([0,2]\).
Check Your Understanding
For each question, make a complete candidate list before deciding which values are absolute extrema.
- Find the absolute maximum and minimum of \(f(x)=x^2-4x\) on \([0,5]\).
- For \(h(x)=\frac{x}{x^2+1}\) on \([0,2]\), find the critical number and compare its value with both endpoint values.
- Why must a rational function’s denominator be nonzero throughout the closed interval for this method to apply?
- Can an absolute maximum occur at an endpoint even when the derivative is nonzero there? Explain briefly.
- For \(q(x)=\frac{1}{x+1}\) on \([-2,0]\), explain why the candidate-point theorem cannot be used on the stated interval.
