Why Domain Restrictions Matter
A rational function is a quotient of polynomials, and its denominator determines which inputs are allowed. Those restrictions remain important when the derivative is factored: a point excluded from the original function is not a possible local maximum or minimum. It can still divide the number line into intervals on which the derivative has different signs.
The method is to identify the domain first, differentiate and factor, then inspect the derivative’s sign on each interval separated by critical numbers and excluded points. A change from positive to negative indicates a local maximum; a change from negative to positive indicates a local minimum. No sign change means the derivative test gives no local extremum at that critical number.
Set Up the Sign Analysis Correctly
For a rational function \(f(x)=\frac{P(x)}{Q(x)}\), the domain consists of the real numbers for which \(Q(x)\ne 0\). Find this domain from the original expression before simplifying or differentiating. If a factor cancels, the corresponding excluded input remains excluded.
A critical number is a number \(c\) in the domain where \(f'(c)=0\) or where \(f'(c)\) does not exist. For rational functions, the derivative exists at every point in the domain, so critical numbers ordinarily come from zeros of the derivative. Points excluded from the domain are not critical numbers, but they must also be marked as breaks when forming sign intervals.
Once the derivative is factored, its sign is determined by the signs of its factors. A squared denominator is positive wherever it is defined, so it does not change the sign of the derivative. Other factors may change sign at their zeros. Check the sign on each interval in the domain; do not assign a sign at a point where the function is undefined.
Worked Example: Two Extrema Separated by a Pole
Worked Example: Two Extrema Separated by a Pole
Find and classify the local extrema of \(f(x)=\frac{x^2+1}{x-1}\).
Step 1: Find the domain. The denominator is zero when \(x=1\), so the domain is all real numbers except \(1\). Keep \(1\) marked as an excluded point throughout the sign analysis.
Step 2: Differentiate and factor. Using the quotient rule,
The numerator factors using its roots \(x=1\pm\sqrt{2}\):
The critical numbers are \(1-\sqrt{2}\) and \(1+\sqrt{2}\), both of which belong to the domain. The point \(1\) is excluded, not critical. Since the denominator is positive at every domain point, the numerator determines the derivative’s sign.
Step 3: Make the sign chart. In increasing order, the dividing points are \(1-\sqrt{2}\), \(1\), and \(1+\sqrt{2}\). The numerator is a quadratic opening upward, so it is positive outside its roots and negative between them. Therefore:
At \(1-\sqrt{2}\), the derivative changes from positive to negative, so there is a local maximum. At \(1+\sqrt{2}\), it changes from negative to positive, so there is a local minimum. There is no function value at \(x=1\), so that excluded point cannot be an extremum.
Step 4: Find the extremum values. At either critical number, write \(x=1+s\sqrt{2}\), where \(s\) is \(1\) or \(-1\). Then
Thus
Taking \(s=-1\) gives the local maximum \(f(1-\sqrt{2})=2-2\sqrt{2}\). Taking \(s=1\) gives the local minimum \(f(1+\sqrt{2})=2+2\sqrt{2}\).
Check: Substituting either critical number into the derivative’s numerator gives \(x^2-2x-1=0\), as required. The sign chart gives the maximum and minimum classifications, and the function values follow from direct substitution into the original rational function.
Why a Sign Change Gives an Extremum
The sign-change rule follows from the Mean Value Theorem and the relationship between derivative signs and monotonicity. Here is the reasoning for a local maximum. Suppose \(f\) is continuous at \(c\), differentiable on the nearby intervals on either side, and \(f'(x)>0\) immediately to the left of \(c\), while \(f'(x)<0\) immediately to the right.
For any nearby \(x<c\), the Mean Value Theorem on \([x,c]\) gives \(f(c)-f(x)=f'(\xi)(c-x)\) for some \(\xi\in(x,c)\). Since \(f'(\xi)>0\) and \(c-x>0\), \(f(c)-f(x)>0\), so \(f(x)<f(c)\).
For any nearby \(x>c\), the Mean Value Theorem on \([c,x]\) gives \(f(x)-f(c)=f'(\xi)(x-c)\) for some \(\xi\in(c,x)\). Since \(f'(\xi)<0\) and \(x-c>0\), \(f(x)-f(c)<0\), so \(f(x)<f(c)\).
Both nearby sides have function values below \(f(c)\), so \(f(c)\) is a local maximum. Reversing the derivative signs proves the corresponding local-minimum result.
For a rational function, a domain point \(c\) is not a denominator zero, so the function and its derivative are defined in a sufficiently small neighborhood of \(c\). This allows the reasoning to apply at its critical numbers. At an excluded point, the comparison cannot be made: there is no value \(f(c)\) to compare with nearby values.
Worked Example: A Cancelled Factor Does Not Restore a Point
Worked Example: A Cancelled Factor Does Not Restore a Point
Consider \(g(x)=\frac{x^2-1}{x-1}\). Determine whether the function has a local extremum at \(x=1\), and describe its behavior elsewhere.
Step 1: Preserve the original domain. The original denominator is zero at \(x=1\). Thus the domain is \(x\ne 1\), even though the numerator also has a factor \(x-1\).
Step 2: Differentiate on the domain. For \(x\ne 1\), the quotient rule gives
The derivative is positive everywhere in the domain, so \(g\) is increasing on each of the intervals \((-\infty,1)\) and \((1,\infty)\). There are no critical numbers in the domain. The point \(1\) is not a local extremum because it is not in the domain.
Check: Factoring the numerator gives \(x^2-1=(x-1)(x+1)\), so \(g(x)=x+1\) for \(x\ne 1\). Differentiating this simplified expression on the same restricted domain gives \(g'(x)=1\), matching the quotient-rule result. The value \(x+1=2\) at \(x=1\) is not a value of the original function.
Worked Example: A Stationary Point Without an Extremum
Worked Example: A Stationary Point Without an Extremum
Classify the critical numbers of \(h(x)=\frac{(x-2)^3}{x-1}\).
Step 1: Find the domain. The denominator vanishes at \(x=1\), so \(h\) is defined for \(x\ne 1\).
Step 2: Differentiate and factor. The quotient rule gives
The derivative is zero at \(x=\frac12\) and \(x=2\), and both numbers are in the domain. The point \(x=1\) is excluded. The squared factors are positive except where they are zero, so the sign is controlled by \(2x-1\) wherever the derivative is defined and nonzero.
Step 3: Inspect signs on every interval. The ordered dividing points are \(\frac12\), \(1\), and \(2\). The signs are
The derivative changes from negative to positive at \(x=\frac12\), so \(h\) has a local minimum there. Its value is
At \(x=2\), the derivative is zero but remains positive on both sides. Therefore \(x=2\) is not a local extremum. Its function value is \(h(2)=0\).
Check: The derivative formula gives \(h'(\frac12)=0\) because \(2(\frac12)-1=0\), and \(h'(2)=0\) because of the factor \((x-2)^2\). The sign chart confirms that only the first zero changes the derivative’s sign. Direct substitution confirms the minimum value \(27/4\).
A Common Mistake: Calling a Pole an Extremum
A common error is to treat a denominator zero as a critical number and then label it a maximum or minimum because the derivative is undefined there. For example, for \(g(x)=\frac{x^2-1}{x-1}\), one might simplify to \(x+1\), calculate \(g(1)=2\), and use that invented value to include \(x=1\) in the analysis. This is incorrect: the original formula is undefined at \(1\), so the function has no value there.
The correction is to record the original domain first. Cancellation can simplify calculations at points that are already in the domain, but it cannot fill a hole in the original function. Likewise, a vertical asymptote is a domain break, not a point where a local extremum can occur.
A Reliable Procedure
For rational-function extrema, use the following sequence. It keeps derivative zeros, excluded inputs, and actual function values distinct.
- Find the domain from the original denominator, before cancelling any factors.
- Differentiate using the quotient rule or a valid simplification on the domain.
- Factor the derivative fully and list its zeros that belong to the domain.
- Include excluded points as breaks in the sign chart, but do not call them critical numbers or extrema.
- Determine the derivative’s sign on each interval and classify each domain critical number by its signs on either side.
- Substitute each extremum point into the original function to find its value.
The method matters because rational functions may have several separate domain intervals, and derivative signs can be different on opposite sides of a denominator zero. Factoring reveals the sign changes that determine local behavior, while keeping the original domain intact prevents conclusions about points where the function does not exist.
Check Your Understanding
For each question, keep track of both derivative zeros and excluded domain points.
- For \(f(x)=\frac{x^2}{x^2-1}\), what is the domain, and which points must divide a derivative sign chart?
- Why can a denominator zero not be a local maximum or minimum of the original rational function?
- If a factored derivative has the form \(\frac{(x-3)^2(x+1)}{(x-2)^2}\), what sign behavior should you check at \(x=3\) and \(x=-1\), assuming those points are in the domain?
- What derivative sign changes indicate a local maximum and a local minimum, respectively?
- Why does cancelling a common factor not automatically make its zero part of the original function’s domain?
