From Derivative Signs to Function Behavior
A first-derivative sign chart organizes information about a function’s slope. On an interval where the derivative is positive, the function rises as the input increases. Where the derivative is negative, the function falls. The chart’s purpose is to identify those intervals carefully and record the behavior on each one.
The intervals are separated by relevant points: critical numbers, where the derivative is zero or does not exist, and breaks in the domain, where the function itself is not defined. These points divide the domain into pieces on which the derivative’s sign can be examined.
A sign chart describes increasing and decreasing behavior. It does not, by itself, classify extrema or describe concavity. Those are different conclusions that require further analysis.
Definitions and the Role of Partition Points
A critical number of \(f\) is a number \(c\) in the domain of \(f\) at which \(f'(c)=0\) or \(f'(c)\) does not exist. A domain break is a number not in the domain that separates parts of the domain—for example, a value that makes a denominator zero. A domain break is not a critical number, because the function is not defined there.
To construct a chart, first find the domain, then locate critical numbers and domain breaks. Together, these points partition the domain into intervals. On each interval, determine whether \(f'\) is positive or negative. A test value is useful because the derivative’s sign is constant between the partition points when the derivative is continuous there and has no zero there. For the examples below, the factored derivative makes that sign behavior explicit.
Why the Sign of the Derivative Determines Monotonicity
The connection between derivative sign and function behavior follows from the Mean Value Theorem. In the form needed here, the theorem says that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is a number \(c\) in \((a,b)\) such that
Suppose \(f'(x)>0\) throughout an interval \(I\). Choose any \(a,b\in I\) with \(a<b\). The function is continuous on \([a,b]\) and differentiable on \((a,b)\), since these points lie inside an interval where the derivative exists. The Mean Value Theorem gives a \(c\in(a,b)\) for which
Here \(f'(c)>0\) and \(b-a>0\), so \(f(b)>f(a)\). This holds for every such pair \(a<b\), which proves that \(f\) is increasing on \(I\). If \(f'(x)<0\) throughout \(I\), the same argument gives \(f(b)-f(a)<0\), so \(f\) is decreasing.
The strict-sign conclusions follow directly from the Mean Value Theorem proof. The non-strict versions follow by the same equation: a nonnegative derivative makes \(f(b)-f(a)\geq 0\), and a nonpositive derivative makes it at most zero.
A Practical Construction Process
Once the domain and derivative are known, use the following sequence. The key is to make a separate sign decision for each interval, rather than treating the whole domain as one piece.
List critical numbers and domain breaks in numerical order. Include domain endpoints as boundaries when relevant.
Use the partition points to divide the domain into intervals. Do not include a domain break as a point in any interval.
Factor \(f'\) when possible and use a test value from each interval to evaluate its sign.
A positive sign means increasing on that interval; a negative sign means decreasing there.
Worked Examples
Worked Example: A Polynomial with Two Critical Numbers
Construct a first-derivative sign chart for \(f(x)=x^3-3x\), and state the increasing and decreasing intervals.
The domain is all real numbers. Differentiate:
The derivative is zero at \(x=-1\) and \(x=1\). These critical numbers divide the real line into \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\). Test one value in each interval:
| Interval | Test value and derivative sign | Behavior |
|---|---|---|
| \((-\infty,-1)\) | \(f'(-2)=9>0\) | Increasing |
| \((-1,1)\) | \(f'(0)=-3<0\) | Decreasing |
| \((1,\infty)\) | \(f'(2)=9>0\) | Increasing |
As a check, the factorization \(f'(x)=3(x-1)(x+1)\) gives exactly the signs found by the test values: positive outside the two zeros and negative between them. Since \(f\) is a polynomial, it is differentiable on each interval. The Mean Value Theorem therefore justifies each increasing or decreasing conclusion.
Worked Example: A Domain Break Splits the Chart
Construct the sign chart for \(g(x)=\dfrac{x+1}{x-2}\), and state where \(g\) increases or decreases.
The denominator is zero at \(x=2\), so the domain is \((-\infty,2)\cup(2,\infty)\). Differentiate using the quotient rule:
For every \(x\neq 2\), the denominator \((x-2)^2\) is positive, so \(g'(x)<0\). For example, the test values \(x=0\) and \(x=3\) give
The sign chart has two separate intervals: \(g\) is decreasing on \((-\infty,2)\) and decreasing on \((2,\infty)\). The derivative is negative on both, but the function is undefined at \(2\). The Mean Value Theorem applies within either interval, not across the break. Thus the chart establishes decreasing behavior on each component of the domain; it does not justify a single monotonicity claim across the missing point.
Check the differentiation: the numerator simplifies to \(-3\), and the squared denominator is positive everywhere in the domain. Both facts agree with the negative signs recorded in the chart.
Worked Example: Zeros and a Domain Break Together
For \(h(x)=\dfrac{x^2}{x-1}\), find the intervals on which the function increases or decreases.
The domain excludes \(x=1\). Applying the quotient rule gives
The derivative is zero at \(x=0\) and \(x=2\), and the domain breaks at \(x=1\). The intervals to test are therefore \((-\infty,0)\), \((0,1)\), \((1,2)\), and \((2,\infty)\). The denominator is positive on all four intervals, so the sign is determined by \(x(x-2)\). Evaluate at one point in each:
| Interval | Derivative sign | Behavior |
|---|---|---|
| \((-\infty,0)\) | Positive | Increasing |
| \((0,1)\) | Negative | Decreasing |
| \((1,2)\) | Negative | Decreasing |
| \((2,\infty)\) | Positive | Increasing |
The factor \(x(x-2)\) is positive when both factors are negative or both are positive, and negative when their signs differ. The positive squared denominator never changes this result. This verifies the signs in the chart. Since \(h\) is differentiable on each listed interval, the Mean Value Theorem gives the stated behavior on each interval separately.
A Common Mistake: Joining Intervals Across a Break
A frequent error is to treat a derivative sign chart as one uninterrupted line, even when the function is undefined at a point. For \(g(x)=\dfrac{x+1}{x-2}\), one might see that \(g'\) is negative on both sides of \(2\), then claim that \(g\) is decreasing throughout the combined set \((-\infty,2)\cup(2,\infty)\). That conclusion does not follow: the Mean Value Theorem requires the function to be continuous on the closed interval between any two inputs being compared.
Indeed, \(g(0)=-\dfrac12\) and \(g(3)=4\). Since \(0<3\), a decreasing-function claim across the combined set would require \(g(0)>g(3)\), but \(-\dfrac12>4\) is false. The correction is to state the conclusion on each interval component: decreasing on \((-\infty,2)\) and decreasing on \((2,\infty)\). A shared derivative sign on both sides does not remove a domain break.
What a First-Derivative Sign Chart Establishes
A sign chart turns derivative information into interval-by-interval statements about increasing and decreasing behavior. Its conclusions depend on using the correct domain and partition points. In particular, a derivative zero splits the intervals to be checked, and a domain break prevents applying the Mean Value Theorem across that point.
When reporting the result, name the intervals explicitly and keep separate components separate. The chart records the sign of the first derivative and the resulting monotonicity on each interval. It is useful evidence for curve shape, but it does not by itself decide what happens at a critical number beyond the signs on either side. That question calls for a separate classification.
Check Your Understanding
For each question, identify the interval structure before drawing a conclusion about function behavior.
- If \(f'(x)>0\) at every point of an interval, what does the Mean Value Theorem imply about \(f\) there?
- Why must a point where the function is undefined be treated separately from a critical number?
- For \(p(x)=x^2-4\), find the critical number, test the derivative sign on either side, and state the increasing and decreasing intervals.
- For \(q(x)=1/x\), explain why the intervals on either side of \(0\) must be treated separately when making monotonicity claims.
- What does a negative sign in a first-derivative chart establish, and what does it not establish by itself?
