Why Rational-Function Concavity Needs Care
A rational function can change concavity at a point where its second derivative is zero, but that is not the only place to investigate. The function may also be undefined at a denominator zero, and an excluded input divides the domain into separate intervals. Keeping those facts visible is essential when deciding whether a graph has an inflection point.
The method in this workshop is to determine the domain first, calculate and factor the second derivative, then test its sign on every interval separated by a zero or an excluded input. Finally, check the function itself before calling any point an inflection point.
For a rational function \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) are polynomials, the domain consists of the real inputs for which \(Q(x)\ne 0\). A zero of \(Q\) is excluded from the domain even if an algebraic simplification later cancels a factor. Such excluded inputs must remain visible in the analysis.
A Sign Chart for the Second Derivative
The second derivative tells us how the first derivative changes: where \(f''\) is positive, \(f'\) is increasing; where \(f''\) is negative, \(f'\) is decreasing. This gives the link between the algebraic sign of \(f''\) and the graph’s concavity.
To make a sign chart, mark every real zero of \(f''\) and every point where the original function is undefined. These points split the domain into intervals. On each interval, determine the sign of \(f''\), either by factoring and reasoning about the factors or by evaluating \(f''\) at a convenient test input. A sign chart records what happens on the intervals; the final inflection-point decision must also use the function’s domain and continuity.
Why the Second-Derivative Test Works
Here is a justification for the positive-sign case. Suppose \(f''(x)>0\) on an interval \(I\). For any \(a<b\) in \(I\), the Mean Value Theorem applied to \(f'\) gives a number \(u\) between \(a\) and \(b\) such that
Both factors on the right are positive, so \(f'(b)>f'(a)\). Thus \(f'\) is increasing on \(I\). To see how this implies concavity upward, take any \(a<x<b\) in \(I\). Applying the Mean Value Theorem to \(f\) on \([a,x]\) and on \([x,b]\) gives numbers \(u\in(a,x)\) and \(v\in(x,b)\) such that
Since \(u<v\) and \(f'\) is increasing, \(f'(u)<f'(v)\). Therefore the slope from \(a\) to \(x\) is less than the slope from \(x\) to \(b\), the characteristic slope ordering for a concave-up graph. If \(f''<0\), the same reasoning reverses the inequalities and gives concavity downward.
Worked Example: Three Inflection Points
Worked Example: Three Inflection Points
Analyze the concavity and inflection points of \(f(x)=\frac{x}{x^2+1}\).
Step 1: Find the domain. The denominator satisfies \(x^2+1>0\) for every real \(x\), so the domain is all real numbers. There are no domain breaks.
Step 2: Calculate the derivatives. The quotient rule gives
Differentiate again, using the product rule on \((1-x^2)(x^2+1)^{-2}\):
The denominator is positive for all real \(x\). The zeros of \(f''\) are \(x=-\sqrt{3}\), \(x=0\), and \(x=\sqrt{3}\). These divide the real line into four intervals.
Step 3: Determine the signs. The factors \(2\) and \((x^2+1)^3\) are positive, so the sign comes from \(x(x^2-3)\). Testing one input in each interval confirms the signs:
| Interval | Sign of \(f''\) | Concavity |
|---|---|---|
| \((-\infty,-\sqrt{3})\) | Negative | Downward |
| \((-\sqrt{3},0)\) | Positive | Upward |
| \((0,\sqrt{3})\) | Negative | Downward |
| \((\sqrt{3},\infty)\) | Positive | Upward |
Step 4: Check the inflection claims. At each candidate, the function is defined and continuous, and the sign of \(f''\) changes across the input. The corresponding function values are
Thus the inflection points are \(\left(-\sqrt{3},-\frac{\sqrt{3}}{4}\right)\), \((0,0)\), and \(\left(\sqrt{3},\frac{\sqrt{3}}{4}\right)\). Substituting the inputs into the original function confirms the coordinates; the denominator is \(3+1=4\) at \(x=\pm\sqrt{3}\), and \(f(0)=0\).
Worked Example: Domain Breaks Are Not Inflection Points
Worked Example: Domain Breaks Are Not Inflection Points
Analyze the concavity of \(g(x)=\frac{1}{x^2-1}\) and decide whether either excluded input is an inflection point.
Step 1: Find the domain. The denominator is zero when \(x^2-1=0\), that is, at \(x=-1\) and \(x=1\). The domain is
Step 2: Compute the second derivative.
The numerator is always positive because \(3x^2+1>0\). The denominator has the sign of \(x^2-1\), since cubing preserves sign. Therefore \(g''>0\) when \(|x|>1\), and \(g''<0\) when \(|x|<1\).
For a direct check, \(g''(0)=2(1)/(-1)^3=-2<0\), while \(g''(2)=2(13)/3^3=26/27>0\). The function is concave downward on \((-1,1)\) and concave upward on \((-\infty,-1)\) and \((1,\infty)\). There are no real zeros of \(g''\), because its numerator is always positive.
Although the concavity differs on the two sides of each excluded input, neither \(x=-1\) nor \(x=1\) is in the domain. There is no point of the graph there, so neither input is an inflection point. The sign chart describes concavity on the separate domain intervals; it does not fill in missing points.
Worked Example: A Zero Without an Inflection Point
Worked Example: A Zero Without an Inflection Point
Analyze the concavity and inflection points of \(h(x)=\frac{x^4}{x^2+1}\).
Step 1: Find the domain and rewrite. Since \(x^2+1>0\), the domain is all real numbers. Polynomial division gives
because \((x^2+1)(x^2-1)+1=x^4-1+1=x^4\).
Step 2: Differentiate twice. Differentiating the rewritten expression,
The denominator is positive. Also, \(x^4+3x^2+6>0\) for every real \(x\), because each of \(x^4\) and \(3x^2\) is nonnegative and the constant term is positive. Thus \(h''(x)>0\) for every \(x\ne0\), while \(h''(0)=0\).
The sign chart has one candidate, \(x=0\), and \(h''\) is positive on both \((-\infty,0)\) and \((0,\infty)\). The function is concave upward on both intervals, so its concavity does not change at zero. In fact, \(h(0)=0\) and the function is continuous there, but \((0,0)\) is not an inflection point. Substitution into the factored second derivative confirms the zero: its numerator contains \(x^2\), while its denominator at zero is \(1\).
Reading the Evidence Correctly
These examples illustrate two distinct checks. First, a zero of \(f''\) does not by itself establish an inflection point: the sign of \(f''\) must change across the input, and the function must be defined and continuous there. Second, an excluded denominator zero is not a point on the graph, even if the concavity on one side differs from the concavity on the other.
A common incorrect conclusion is: “\(g''\) changes sign at \(x=1\), so \(x=1\) is an inflection point” for \(g(x)=1/(x^2-1)\). The sign analysis does show concavity downward just to the left of \(1\) and upward just to its right. But \(g(1)\) is undefined, so there is no graph point at input \(1\). The correction is to report the concavity on each domain interval and state that \(x=1\) is not an inflection point.
Likewise, a zero of \(f''\) deserves investigation rather than automatic classification. In the last example, \(h''(0)=0\), but its sign is positive on both sides; this rules out a change in concavity. Keep the sign chart tied to the intervals where the original function is defined, and check each proposed point against the function’s domain.
Find all real zeros of the original denominator. Keep excluded inputs marked throughout the analysis.
Differentiate carefully, simplify, and factor the numerator and denominator where possible.
Use the real zeros of \(f''\) and the domain breaks to form intervals, then determine the sign on each interval.
Confirm that the function is defined and continuous at the input and that concavity changes across it.
Check Your Understanding
Use the domain, the sign of the second derivative, and continuity to justify each conclusion.
- For \(f(x)=\frac{x}{x^2+1}\), why are the zeros of \(f''\) candidates rather than automatic inflection points?
- For \(g(x)=\frac{1}{x^2-1}\), list the three intervals in its domain and state its concavity on each.
- Why can an excluded denominator zero not be an inflection point, even if the concavity changes across it?
- For \(h(x)=\frac{x^4}{x^2+1}\), what is the sign of \(h''\) on either side of zero, and what does that imply?
- What two checks, beyond solving \(f''(x)=0\), are needed before identifying an inflection point?
