Why Combine the Information?
A formula can have several features that matter at once: points where it is not defined, critical numbers, changes in concavity, and behavior near the edges of its domain. A combined sign chart gives these features a common structure. Instead of considering each one in isolation, it divides the real line into intervals on which the function’s domain and derivative signs are settled.
The chart does not replace checking limits, function values, or continuity. It organizes the work so that those checks happen at the right points and are not confused with one another.
Which Points Belong in the Chart?
Begin with the domain \(D\) of \(f\). A point outside \(D\) is not a point on the graph, even if a simplified formula appears to have a value there. Next identify points in \(D\) where \(f'=0\) or \(f'\) does not exist. An interior point with either property is a critical number. Also identify points where \(f''=0\) or \(f''\) does not exist; these are inflection candidates, not automatically inflection points.
Put all these relevant points in increasing order. They divide the real line into open intervals. On each interval, determine whether \(f\) is defined and, when the derivatives exist throughout that interval, determine the signs of \(f'\) and \(f''\). A point excluded from the domain still splits the intervals, but it is not a critical number of \(f\), because critical numbers must belong to the domain.
Use factorization and known signs to determine derivative signs. A test value from each interval can help, provided the derivative expression has no zero or undefined point inside that interval. Record zero and undefined values at the points themselves separately from the signs on the open intervals.
What the Derivative Rows Tell You
If \(f'>0\) throughout an interval, \(f\) is increasing there; if \(f'<0\), it is decreasing. This follows from the Mean Value Theorem: for any \(x_1<x_2\) in the interval, there is a \(c\) between them such that \(f(x_2)-f(x_1)=f'(c)(x_2-x_1)\). Since \(x_2-x_1>0\), the sign of the difference is the sign of \(f'(c)\).
The sign of \(f''\) describes concavity on intervals where \(f''\) exists: positive \(f''\) means concave up, and negative \(f''\) means concave down. To establish an inflection point, check that the function is defined and continuous at the candidate and that its concavity changes across it. A zero or undefined value of \(f''\) alone does not establish an inflection point.
For the positive-to-negative case, the Mean Value Theorem shows that \(f\) increases on the interval to the left and decreases on the interval to the right. Continuity at \(c\) makes the limiting values from both sides equal to \(f(c)\). Thus nearby function values are no greater than \(f(c)\), which is a local maximum. Reversing the signs gives a local minimum. If \(f\) is differentiable at \(c\), it is continuous there, so the continuity condition is satisfied.
Worked Example: A Rational Function with Two Turning Points
Worked Example: Charting a Rational Function
Analyze \(f(x)=\frac{x^2+1}{x-1}\) using its domain, derivative signs, concavity, and asymptote behavior.
The denominator is zero at \(x=1\), so the domain is \(x\ne1\). Division gives a useful equivalent form on that domain:
Differentiate each term and then differentiate again:
The numerator of \(f'\) is zero when \((x-1)^2=2\), giving \(a=1-\sqrt2\) and \(b=1+\sqrt2\). The denominator of \(f'\) and \(f''\) is zero at \(1\), where \(f\) is not defined. These points divide the line into four intervals:
| Interval or point | Domain | Sign of \(f'\) | Sign of \(f''\) |
|---|---|---|---|
| \((-\infty,a)\) | Defined | Positive | Negative |
| \(a\) | Defined; \(f'=0\) | Zero | Negative |
| \((a,1)\) | Defined | Negative | Negative |
| \(1\) | Not defined | Undefined | Undefined |
| \((1,b)\) | Defined | Negative | Positive |
| \(b\) | Defined; \(f'=0\) | Zero | Positive |
| \((b,\infty)\) | Defined | Positive | Positive |
To check the derivative signs, the numerator \((x-1)^2-2\) is positive outside \((a,b)\) and negative inside it. The denominator \((x-1)^2\) is positive wherever it is defined. The sign of \(f''\) is the sign of \((x-1)^3\), so it is negative to the left of \(1\) and positive to the right.
At \(a\), \(f'\) changes from positive to negative, and \(f\) is continuous there because its denominator is nonzero. Thus \(a\) is a local maximum. Its value is
At \(b\), \(f'\) changes from negative to positive, so \(b\) is a local minimum. Its value is
These values can be checked by substitution into the original quotient. For \(a=1-\sqrt2\), the numerator is \((1-\sqrt2)^2+1=4-2\sqrt2\) and the denominator is \(-\sqrt2\), giving \(2-2\sqrt2\). For \(b=1+\sqrt2\), the numerator is \((1+\sqrt2)^2+1=4+2\sqrt2\) and the denominator is \(\sqrt2\), giving \(2+2\sqrt2\).
The concavity changes across \(x=1\), but that does not make \(1\) an inflection point: \(f\) is not defined there. The same decomposition shows a vertical asymptote at \(x=1\), since the term \(2/(x-1)\) tends to negative infinity from the left and positive infinity from the right. It also shows the slant asymptote \(y=x+1\), because the difference \(f(x)-(x+1)=2/(x-1)\) tends to zero as \(x\) tends to either infinity. The chart locates where to check these behaviors; the limits and difference establish them.
Worked Example: A Polynomial with an Inflection Point
Worked Example: Charting a Cubic
Let \(g(x)=x^3-3x\). Its domain is all real numbers. Differentiating gives
The breakpoints are \(-1\) and \(1\), where \(g'=0\), and \(0\), where \(g''=0\). The signs are:
| Interval | Sign of \(g'\) | Sign of \(g''\) | Behavior |
|---|---|---|---|
| \((-\infty,-1)\) | Positive | Negative | Increasing; concave down |
| \((-1,0)\) | Negative | Negative | Decreasing; concave down |
| \((0,1)\) | Negative | Positive | Decreasing; concave up |
| \((1,\infty)\) | Positive | Positive | Increasing; concave up |
The signs follow from the factors of \(g'\) and the sign of \(6x\). At \(-1\), \(g'\) changes from positive to negative, so there is a local maximum. At \(1\), it changes from negative to positive, so there is a local minimum. Substitution verifies their values:
At \(0\), \(g''\) changes from negative to positive. Since \(g\) is a polynomial, it is continuous there; hence \((0,g(0))=(0,0)\) is an inflection point. The checks agree with the derivatives: \(g'(0)=-3\), while \(g''(x)=6x\) is negative for \(x<0\) and positive for \(x>0\).
A Pitfall: A Sign Change Does Not Override Discontinuity
A chart reports derivative signs on intervals, but a sign change does not by itself settle the function value at the dividing point. Consider the function
Worked Example: Checking Continuity Before Using the Sign Test
The domain is all real numbers. On the two open intervals, \(h'(x)=1\) for \(x<0\) and \(h'(x)=-1\) for \(x>0\). On both intervals, \(h''(x)=0\). At \(0\), \(h'\) and \(h''\) do not exist. Thus the derivative signs change from positive to negative at \(0\), but \(h\) is discontinuous there: the values from either side approach \(0\), whereas \(h(0)=-1\).
A wrong calculation would be: “\(h'\) changes from positive to negative at \(0\), so \(h\) has a local maximum at \(0\).” This omits the continuity condition. In fact, for \(0<|x|<\frac12\), the values satisfy \(h(x)>-\frac12>-1=h(0)\): on the left \(h(x)=x>-\frac12\), and on the right \(h(x)=-x>-\frac12\). Therefore \(0\) is a local minimum, not a local maximum. The correction is to check continuity before applying the first derivative test.
A Reliable Chart-Building Process
Record excluded points and domain endpoints. These divide the line even when no derivative is involved.
Locate zeros and undefined points of \(f'\) and \(f''\), retaining whether each point belongs to the domain.
Determine the signs of \(f'\) and \(f''\) on every interval where the function and derivatives are defined.
Use \(f'\) for monotonicity and, with continuity at an interior point, the first derivative test. Use \(f''\) for concavity; verify a concavity change and continuity before identifying an inflection point.
Evaluate the function at relevant points in its domain. At excluded points and domain boundaries, examine limits when needed to assess asymptotes or endpoint behavior.
Why the Combined Chart Matters
A combined chart prevents several common confusions. A point where a derivative is undefined may still be in the domain and be a critical number; an excluded point is not. A zero of \(f''\) is only an inflection candidate until concavity is checked on both sides. And a change in the sign of \(f'\) yields a local extremum through the first derivative test only when the necessary continuity condition holds.
The intervals are also useful for organizing a sketch: they specify where the curve rises or falls and where it bends upward or downward. They do not supply every feature of the graph. Function values, intercepts, domain boundaries, and limits still need attention. In particular, a sign chart cannot establish a vertical or slant asymptote without the corresponding limit or difference check.
Check Your Understanding
Use the definitions and conditions in this tutorial when answering.
- Why is a point where \(f\) is undefined not a critical number, even if \(f'\) is also undefined there?
- What sign change in \(f'\) indicates a local minimum, and what additional condition is required at the point?
- If \(f''\) changes sign across a point excluded from the domain, is that point an inflection point? Explain.
- For \(f(x)=x^3-3x\), list the intervals where the function is increasing and where it is concave up.
- What extra check is needed to conclude that a domain exclusion gives a vertical asymptote?
