What a Derivative Sign Chart Records
A derivative sign chart organizes information about a function’s derivative. It shows where the derivative is positive, negative, zero, or undefined, and connects those signs to the function’s behavior on intervals.
The chart is not a shortcut for finding critical numbers: those must first be identified. Nor does a chart by itself tell us the function’s values. Its purpose is to make the interval-by-interval reasoning visible and systematic.
After finding the critical numbers and relevant domain breaks, place them in order on a number line. They divide the domain into intervals. On each interval, determine the sign of \(f'(x)\), then record the corresponding increase or decrease behavior.
A Reliable Chart-Building Process
A sign chart is only as complete as the points used to divide the domain. Include every critical number and every domain break that separates the function’s domain. Then test the derivative on each resulting interval—not at the dividing points, where the derivative may be zero or undefined.
Identify where the function is defined. Mark any breaks that separate parts of its domain.
Solve \(f'(x)=0\), and find points in the domain where \(f'\) does not exist.
Arrange the critical numbers and domain breaks from least to greatest. These points divide the domain into intervals.
Choose a convenient point in each interval and evaluate \(f'\), or determine its sign from a factored expression.
Write \(+\) or \(-\) for the derivative and “increasing” or “decreasing” for the function on each interval.
For the usual AP functions, \(f'\) is continuous on each interval of the chart. If all zeros and undefined points of \(f'\) have been included as dividing points, then \(f'\) has no zero or discontinuity inside any such interval. Its sign therefore cannot switch within that interval: a continuous function that changed from positive to negative would have to equal zero somewhere in between, by the Intermediate Value Theorem.
Worked Example: A Polynomial with Two Critical Numbers
Worked Example: A Polynomial with Two Critical Numbers
Build a derivative sign chart for \(f(x)=x^3-3x^2-9x+2\).
1. Find the domain and derivative. A polynomial is defined for every real number, so the domain is \(\mathbb{R}\). Differentiate term by term:
Factor the derivative:
Expanding the factored form checks the factorization: \(3(x-3)(x+1)=3(x^2-2x-3)=3x^2-6x-9\).
2. Find critical numbers. Set the derivative equal to zero:
The derivative is defined everywhere, so the critical numbers are \(-1\) and \(3\). They partition the real line into \((-\infty,-1)\), \((-1,3)\), and \((3,\infty)\).
3. Test the sign on each interval. Use \(x=-2\), \(x=0\), and \(x=4\), respectively:
The sign chart is:
| Location | \((-\infty,-1)\) | \(-1\) | \((-1,3)\) | \(3\) | \((3,\infty)\) |
|---|---|---|---|---|---|
| Sign of \(f'(x)\) | + | 0 | − | 0 | + |
| Behavior of \(f\) | Increasing | — | Decreasing | — | Increasing |
Thus, the chart records that \(f\) is increasing on \((-\infty,-1)\), decreasing on \((-1,3)\), and increasing on \((3,\infty)\). As a check, the tested derivative values agree with the signs of the factors \(3(x-3)(x+1)\) on their respective intervals.
Why the Sign Chart Gives Monotonicity
The connection between derivative signs and increase or decrease follows from the Mean Value Theorem. Here is the complete argument for a positive derivative; the negative case follows by reversing the inequality.
Let \(x_1,x_2\in[a,b]\) with \(x_1<x_2\). The function is continuous on \([x_1,x_2]\) and differentiable on \((x_1,x_2)\), so the Mean Value Theorem applies.
There is some \(c\in(x_1,x_2)\) such that \(f(x_2)-f(x_1)=f'(c)(x_2-x_1)\).
Since \(f'(c)>0\) and \(x_2-x_1>0\), their product is positive. Thus \(f(x_2)-f(x_1)>0\), so \(f(x_2)>f(x_1)\).
Because this holds for every pair \(x_1<x_2\) in the interval, \(f\) is strictly increasing there. If instead \(f'(x)<0\) throughout, then \(f'(c)(x_2-x_1)<0\), giving \(f(x_2)<f(x_1)\); the function is strictly decreasing. This explains why each derivative sign in a chart has a precise interval behavior attached to it.
Worked Example: A Domain Break Splits the Chart
Worked Example: A Domain Break Splits the Chart
Build a sign chart for \(g(x)=\dfrac{x+1}{x-2}\).
1. Find the domain. The denominator cannot be zero, so \(x\ne 2\). The domain is \((-\infty,2)\cup(2,\infty)\). The point \(2\) is not a critical number because \(g(2)\) is not defined; it is a domain break that separates the chart.
2. Differentiate and find critical numbers. By the quotient rule,
The numerator is never zero, so \(g'\) has no zeros. It is undefined at \(x=2\), where \(g\) itself is also undefined. Thus there are no critical numbers in the domain.
3. Test both domain intervals. At \(x=0\),
At \(x=3\),
The denominator is a positive square for every \(x\ne2\), so the derivative is negative on both domain intervals. The chart is:
| Location | \((-\infty,2)\) | \(2\) | \((2,\infty)\) |
|---|---|---|---|
| Domain status | In domain | Excluded | In domain |
| Sign of \(g'(x)\) | − | Undefined | − |
| Behavior of \(g\) | Decreasing | — | Decreasing |
The conclusion is that \(g\) is decreasing separately on \((-\infty,2)\) and \((2,\infty)\). The domain break matters: the chart does not claim that \(g\) is decreasing across \(2\), because \(g\) is not defined there.
Worked Example: A Critical Number Where the Derivative Is Undefined
Worked Example: A Critical Number Where the Derivative Is Undefined
Build a sign chart for \(h(x)=|x-2|\).
1. Find the domain and derivative. The absolute value is defined for every real \(x\). Writing the function piecewise gives
Therefore \(h'(x)=-1\) for \(x<2\), while \(h'(x)=1\) for \(x>2\). At \(x=2\), the left-hand slope is \(-1\) and the right-hand slope is \(1\), so the derivative does not exist there. Since \(h(2)\) is defined, \(2\) is a critical number.
2. Partition and record the signs. The critical number divides the domain into \((-\infty,2)\) and \((2,\infty)\). For example, \(h'(1)=-1<0\) and \(h'(3)=1>0\). The chart is:
| Location | \((-\infty,2)\) | \(2\) | \((2,\infty)\) |
|---|---|---|---|
| Sign of \(h'(x)\) | − | Undefined | + |
| Behavior of \(h\) | Decreasing | — | Increasing |
As a check, the piecewise formulas confirm the derivative values: differentiating \(2-x\) gives \(-1\), and differentiating \(x-2\) gives \(1\). The chart records the behavior on either side without assigning a derivative value at \(2\).
Common Pitfall: Treating a Domain Break as a Critical Number
A point where the formula for \(f'\) is undefined is not automatically a critical number. The function itself must be defined there. In the rational example, \(g'(2)\) does not exist, but \(g(2)\) does not exist either, so \(2\) is not a critical number.
A related mistake is to test only once and then combine intervals separated by a domain break. For \(g(x)=\dfrac{x+1}{x-2}\), one might calculate \(g'(0)=-\dfrac34<0\) and incorrectly conclude that \(g\) is decreasing on all of \(\mathbb{R}\). That conclusion is invalid: \(g\) is not defined at \(2\), and the domain is not all of \(\mathbb{R}\). The correction is to split the chart at \(2\), test each domain interval, and state that \(g\) is decreasing on \((-\infty,2)\) and on \((2,\infty)\).
Using the Chart Carefully
Before interpreting a sign chart, check that it contains every relevant dividing point. A missed zero of \(f'\) can leave an interval on which the derivative does not have one consistent sign. Also check that each test value lies inside the interval it represents. A test value at a critical number or domain break cannot determine the sign on either neighboring interval.
Finally, keep the chart’s conclusion at the right level: it records the derivative’s sign and the resulting increase or decrease on each interval. A critical number’s role in a chart is to divide intervals for testing; the chart-building task does not require classifying what happens at that number.
Check Your Understanding
For each question, focus on the domain, the intervals, and the derivative signs.
- What two conditions make a number a critical number of a function?
- Why must a point where the function is undefined still be marked when it separates the domain?
- For \(p(x)=x^2-4\), find the critical number and state the derivative sign and function behavior on each resulting interval.
- For \(q(x)=\dfrac{1}{x+1}\), explain why \(x=-1\) is a domain break rather than a critical number.
- What theorem explains why a positive derivative throughout an interval implies that the function is increasing there?
