Why Derivative Signs Give Monotonicity Information
A derivative describes how a function changes locally. If the derivative is positive throughout an interval, the function rises as the input increases; if the derivative is negative throughout an interval, the function falls. A sign chart organizes this information and helps turn a derivative calculation into a precise claim about where the function is increasing or decreasing.
The important word is throughout. A derivative may change sign at a critical number, and a function may not even be defined at some inputs. Those points divide the domain into intervals that must be considered separately. To justify a claim, identify the intervals, establish the derivative’s sign on each one, and state the resulting monotonicity using accurate interval notation.
Some books use “increasing” and “decreasing” to include equality. Stating the strict or non-strict meaning avoids ambiguity. The derivative rules below give strict increase when \(f'(x)>0\) everywhere in the interval’s interior, and strict decrease when \(f'(x)<0\) there. A nonnegative or nonpositive derivative gives the corresponding non-strict conclusion.
The Mean Value Theorem Connects Derivatives to Change
The justification for using derivative signs is the Mean Value Theorem. In the form needed here, if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is at least one \(c\) in \((a,b)\) such that
The right-hand side is the average rate of change from \(a\) to \(b\). Since \(b-a>0\), its sign is the sign of \(f(b)-f(a)\). The theorem therefore links a derivative value inside an interval to the change between its endpoints.
Why the Derivative Sign Rule Works
Here is the proof for a positive derivative. Choose any two points \(x_1<x_2\) in the interval. The Mean Value Theorem applies on \([x_1,x_2]\), so for some \(c\) with \(x_1<c<x_2\),
The denominator is positive, and \(f'(c)>0\). Therefore the numerator must be positive:
Because the choice of \(x_1\) and \(x_2\) was arbitrary, \(f\) is increasing on the interval. If \(f'(x)<0\), the same equation makes the numerator negative, giving \(f(x_1)>f(x_2)\). Replacing \(>\) by \(\geq\), or \(<\) by \(\leq\), proves the non-strict versions. The continuity and differentiability conditions matter: without them, the Mean Value Theorem may not apply.
How to Build and Read a Sign Chart
A sign chart begins with the domain of the function and the derivative. A critical number is an input in the domain where \(f'(x)=0\) or \(f'(x)\) does not exist. Critical numbers, along with points excluded from the function’s domain, can divide the domain into intervals to investigate. The derivative is tested on each resulting interval, not at the dividing points alone.
Record any inputs excluded from the function, then calculate \(f'(x)\).
Locate derivative zeros and points where the derivative is undefined. Include domain breaks as separate divisions.
Use algebra or a valid test point, after verifying that no zero or break has been missed inside the interval.
Apply the derivative sign rule and name the intervals using the function’s domain.
A test point is a convenient way to determine a sign, but it is not a substitute for checking the interval. For example, if a factored derivative has no zero or undefined value inside an interval, its sign cannot switch there: derivatives have the intermediate value property, so a change from positive to negative would require a zero in between. In typical AP problems, factoring the derivative makes the relevant zeros visible. If the sign is not established on the whole interval, one test value alone is not enough.
Worked Examples
Worked Example: A Polynomial with Two Critical Numbers
Find the intervals where \(f(x)=x^3-3x\) is increasing or decreasing.
First differentiate and factor:
The derivative is zero at \(x=-1\) and \(x=1\). These points divide the real line into \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\). Test one input in each interval:
The factorization has no other zeros, and the derivative is defined everywhere, so the sign is constant on each of these intervals. Thus \(f\) is increasing on \((-\infty,-1)\), decreasing on \((-1,1)\), and increasing on \((1,\infty)\). For example, the sign on \((-1,1)\) is negative throughout, so the Mean Value Theorem justifies the decreasing claim for any two inputs in that interval.
Check the derivative and sign values: differentiating \(x^3-3x\) gives \(3x^2-3\), and substitution gives \(9,-3,9\), with signs \(+,-,+\) as recorded.
Worked Example: A Critical Number Where the Derivative Does Not Exist
Determine where \(f(x)=|x|\) is increasing or decreasing. For \(x<0\), \(f(x)=-x\), so \(f'(x)=-1\). For \(x>0\), \(f(x)=x\), so \(f'(x)=1\). At \(x=0\), the two one-sided slopes differ, so \(f'(0)\) does not exist. The function itself is defined at \(0\).
The derivative sign is negative on \((-\infty,0)\) and positive on \((0,\infty)\). The derivative sign rule shows that \(f\) is decreasing on \((-\infty,0)\) and increasing on \((0,\infty)\). The function is also continuous at \(0\), with \(f(0)=0\), so these claims extend to \((-\infty,0]\) and \([0,\infty)\), respectively.
To check the endpoint statements directly, if \(x_1<x_2\leq 0\), then \(-x_1>-x_2\), so \(f(x_1)>f(x_2)\). If \(0\leq x_1<x_2\), then \(f(x_1)=x_1<x_2=f(x_2)\). These verify decreasing and increasing behavior on the intervals that include \(0\), despite the derivative being undefined there.
Worked Example: A Domain Break Separates Two Intervals
Find where \(f(x)=\dfrac{1}{x-1}\) is increasing or decreasing.
The function is undefined at \(x=1\), so its domain is \((-\infty,1)\cup(1,\infty)\). Differentiate:
For every \(x\ne 1\), the square \((x-1)^2\) is positive. Therefore \(f'(x)<0\) on both \((-\infty,1)\) and \((1,\infty)\). The function is decreasing on each of those intervals.
Check by substitution: at \(x=0\), \(f'(0)=-1<0\); at \(x=2\), \(f'(2)=-1<0\). The factor \((x-1)^2\) never vanishes within either interval, so these signs persist there. The conclusion is decreasing on each domain interval separately, not on an interval that crosses \(1\), where the function is not defined.
Interval Endpoints and Critical Numbers
A sign chart usually displays open intervals between dividing points because the derivative sign is being tested between them. The resulting monotonicity claim may sometimes include an endpoint. If \(f\) is continuous at a finite endpoint and the derivative has the required sign in the interior, the Mean Value Theorem can be applied between that endpoint and any other point in the interval. The example \(f(x)=|x|\) illustrates this at \(0\).
Do not automatically include an endpoint where the function is undefined. Nor should a chart suggest monotonicity across a domain break: monotonicity compares function values for any two inputs in the claimed interval, and a function must be defined at those inputs. When a critical number has \(f'(c)=0\), it can be an endpoint of monotonicity intervals; whether the same monotonic behavior continues on both sides must be determined from the derivative signs there.
Common Sign-Chart Mistakes
A common error is to see \(f'(c)=0\) and immediately conclude that the function changes from increasing to decreasing, or vice versa. A zero marks a place to divide and investigate; it does not itself determine either neighboring sign.
Another error is writing a single interval that crosses a point where the function is undefined. The reciprocal-function example shows why domain breaks must remain separate. A final error is reporting only the derivative calculation without the requested conclusion. A complete justification identifies the sign on each interval, invokes the derivative sign rule, and states the corresponding intervals of increase or decrease.
Check Your Understanding
For each question, identify the relevant intervals and explain how the derivative sign supports your conclusion.
- For \(g(x)=x^2-4x\), find the intervals where \(g\) is increasing and decreasing.
- For \(h(x)=x^3\), what does the derivative sign chart show at \(x=0\), and why does a zero derivative there not by itself imply a sign change?
- For \(p(x)=1/(x+2)\), why must monotonicity claims be stated separately on either side of \(x=-2\)?
- What conditions allow the Mean Value Theorem to justify a monotonicity claim on an interval?
- Why is testing one point insufficient if the chosen interval may contain an unaccounted-for derivative zero or domain break?
