Why Rational Functions Need a Domain Check
A rational function is a quotient of polynomials. Its derivative can be found algebraically, and the usual closed-interval strategy is to compare endpoint values with values at interior critical points. But before using that strategy, one important question comes first: is the function defined and continuous throughout the entire interval?
The denominator of a rational function can vanish. If that happens at a point of the interval, the function may have a pole there and fail to be defined on the whole interval. The Extreme Value Theorem then does not guarantee absolute extrema. If the denominator never vanishes on the interval, the rational function is continuous there, and the closed-interval method applies.
The Closed-Interval Method
Suppose \(f(x)=P(x)/Q(x)\) on a closed interval \([a,b]\), and \(Q(x)\ne 0\) for every \(x\in[a,b]\). The quotient of polynomials is continuous wherever its denominator is nonzero, so \(f\) is continuous on the whole interval. By the Extreme Value Theorem, \(f\) attains both an absolute maximum and an absolute minimum on \([a,b]\).
To find them, include the endpoints and any interior points where the derivative is zero. The quotient rule gives
On the interval, the denominator \([Q(x)]^2\) is positive. Therefore, the interior points where \(f'(x)=0\) are exactly the solutions of \(P'(x)Q(x)-P(x)Q'(x)=0\). Evaluate \(f\) at those points and at \(a\) and \(b\); the largest and smallest resulting values are the absolute maximum and minimum.
In the usual nonconstant case, the derivative numerator is a nonzero polynomial, so it has only finitely many roots. If that numerator is identically zero, the derivative is zero throughout the interval; the Mean Value Theorem then implies that \(f\) is constant there, so its endpoint values give its common value.
Why the Candidate List Is Complete
The Extreme Value Theorem guarantees that a continuous function on a closed, bounded interval attains its absolute extrema. The remaining question is why an interior absolute extremum of a differentiable rational function must appear among the solutions of \(f'(x)=0\).
Here is the needed reasoning. Suppose \(c\) is an interior local maximum and \(f\) is differentiable at \(c\). For sufficiently small positive \(h\), \(f(c+h)\le f(c)\), so \([f(c+h)-f(c)]/h\le 0\). For sufficiently small negative \(h\), \(f(c+h)\le f(c)\), but now dividing by the negative \(h\) gives \([f(c+h)-f(c)]/h\ge 0\). Since the two-sided limit of these difference quotients exists and equals \(f'(c)\), that limit must be both nonpositive and nonnegative. Thus \(f'(c)=0\). At an interior local minimum, the inequalities reverse and the same conclusion follows.
Worked Examples
Worked Example: A Rational Function with One Interior Candidate
Find the absolute extrema of \(f(x)=(x^2+1)/(x+1)\) on \([0,2]\).
The denominator \(x+1\) is positive on \([0,2]\), so \(f\) is defined and continuous throughout the interval. By the quotient rule,
The denominator of \(f'\) is positive, so solve \(x^2+2x-1=0\). The quadratic formula gives \(x=-1\pm\sqrt{2}\). Of these roots, only \(r=-1+\sqrt{2}\) lies in \([0,2]\). The candidates are \(0\), \(r\), and \(2\).
Evaluate each value:
For \(r=\sqrt{2}-1\), we have \(r+1=\sqrt{2}\) and \(r^2=3-2\sqrt{2}\). Therefore,
The critical-point equation checks directly: \(r^2+2r-1=(3-2\sqrt{2})+(2\sqrt{2}-2)-1=0\). Also, \(2\sqrt{2}-2<1\), since \(\sqrt{2}<3/2\). Thus the three values are ordered \(2\sqrt{2}-2<1<5/3\). The absolute minimum is \(2\sqrt{2}-2\) at \(x=\sqrt{2}-1\), and the absolute maximum is \(5/3\) at \(x=2\).
As a check, substituting the minimizing input into the original function gives \((4-2\sqrt{2})/\sqrt{2}=2\sqrt{2}-2\), and substituting \(x=2\) gives \((4+1)/(2+1)=5/3\), as used in the comparison.
Worked Example: Equal Absolute Maximum Values at Both Endpoints
Find the absolute extrema of \(g(x)=x^2/(x^2+1)\) on \([-2,2]\).
Since \(x^2+1>0\) for every real \(x\), the function is continuous on the interval. Differentiate:
The denominator is positive, so \(g'(x)=0\) exactly when \(x=0\). The candidates are \(-2\), \(0\), and \(2\). Their function values are
The least value is \(0\), attained at \(x=0\). The greatest value is \(4/5\), attained at both \(x=-2\) and \(x=2\). To check the interior candidate, \(g'(0)=0\); direct substitution gives \(g(0)=0^2/(0^2+1)=0\). The endpoint substitutions both give \(4/(4+1)=4/5\).
What Changes When the Interval Contains a Pole?
The closed-interval method requires a function that is continuous on the entire closed interval. A pole at any point of \([a,b]\), including an endpoint, prevents the rational function from meeting that requirement: it is not defined there. The Extreme Value Theorem therefore cannot be used to guarantee absolute extrema on that interval.
This does not mean that every function on a punctured interval lacks local extrema, or that checking derivative zeros is never useful. It means that the closed-interval theorem and its guaranteed candidate procedure do not apply to a rational function with a pole in the interval. In particular, a critical-point calculation cannot replace the missing continuity condition.
Worked Example: A Pole Inside the Interval
Consider \(h(x)=1/(x-1)\) on the proposed interval \([0,2]\). The denominator is zero at \(x=1\), so \(h\) is undefined there and is not a continuous function on the whole interval. The Extreme Value Theorem does not apply.
The values also show that \(h\) is unbounded in both directions. For every integer \(n\ge 2\), the points \(1-1/n\) and \(1+1/n\) lie in \([0,2]\), and neither is the pole. Substitution gives
As \(n\) increases, these values become arbitrarily negative and arbitrarily positive. Hence there is no absolute minimum and no absolute maximum on the function’s domain within \([0,2]\). The calculation confirms both the failure of the theorem’s continuity hypothesis and the lack of extrema in this example.
A Common Mistake: Checking Derivative Zeros Before Checking the Domain
A tempting but incorrect approach is to differentiate a rational expression, find its critical points, and assume that comparing those values with the endpoints settles the problem. This skips the condition that the function be defined and continuous throughout the interval.
For example, with \(h(x)=1/(x-1)\) on \([0,2]\), the derivative is
This derivative is never zero where it is defined. Concluding that there are no extrema simply because there are no derivative-zero candidates would be an invalid use of the closed-interval method: \(h\) is undefined at \(x=1\), so the method’s hypotheses fail. The direct values \(-n\) and \(n\) above establish the correct conclusion for this example. The reliable order is to check the denominator on the whole interval first, then use the candidate method only if the function is continuous there.
Practical Checklist
Determine whether \(Q(x)\) is nonzero at every point of the closed interval.
If there is no denominator zero, the rational function is continuous on the interval and attains an absolute minimum and maximum.
Solve \(P'Q-PQ'=0\) for points inside the interval.
Compute the function values at every interior candidate and both endpoints; identify the largest and smallest values.
The key distinction is between a rational function that is continuous on the whole closed interval and one with a pole there. In the first case, the Extreme Value Theorem guarantees absolute extrema, and the derivative-based candidate list makes the search finite in the nonconstant case. In the second, the theorem’s hypotheses do not hold, so the closed-interval guarantee is unavailable.
Check Your Understanding
For each question, explain the role of the denominator, the candidate points, or the theorem’s hypotheses.
- Why must you check the denominator on the entire closed interval before applying the closed-interval method?
- For \(f(x)=P(x)/Q(x)\), what equation gives the interior points where \(f'(x)=0\), provided \(Q(x)\ne0\)?
- Why is an interior absolute extremum of a differentiable rational function a critical point?
- What is wrong with using only the derivative-zero candidates for \(1/(x-1)\) on \([0,2]\)?
- If a rational function is continuous on a closed interval, which points must be included in the absolute-extrema comparison?
