Derivative Signs Describe Direction of Change
A derivative measures a function’s instantaneous rate of change. If that rate is positive throughout an interval, the function rises as its input increases there; if it is negative throughout an interval, the function falls. The key phrase is throughout an interval: a derivative value at just one point cannot establish the function’s behavior over a range of inputs.
To use derivative signs carefully, we need precise meanings for increasing and decreasing. We also need to distinguish strict change from behavior that may include flat portions.
Some texts use “increasing” and “decreasing” to include equality. The terms nondecreasing and nonincreasing remove that ambiguity. Here, “increasing” and “decreasing” mean strict change.
The Derivative-Sign Criteria
For the criteria below, \(I\) is an interval and \(f\) is differentiable at every point in \(I\). Differentiability supplies the derivative values whose signs we examine. These are sufficient conditions: they let us conclude monotonicity from derivative signs. They do not say that every increasing function must have a positive derivative at every point.
- If \(f'(x)>0\) for every \(x\in I\), then \(f\) is strictly increasing on \(I\).
- If \(f'(x)<0\) for every \(x\in I\), then \(f\) is strictly decreasing on \(I\).
- If \(f'(x)\ge 0\) for every \(x\in I\), then \(f\) is nondecreasing on \(I\).
- If \(f'(x)\le 0\) for every \(x\in I\), then \(f\) is nonincreasing on \(I\).
The strict inequalities give strict monotonicity, while the weak inequalities give non-strict monotonicity. In particular, a derivative that is nonnegative can be zero at some points. That condition alone guarantees nondecreasing behavior, not strict increase. A function may nevertheless be strictly increasing even if its derivative is zero at some points.
The interval condition matters just as much as the sign condition. The theorem applies on an interval where the derivative has the stated sign at every point. If the sign changes, the conclusion may differ on the intervals to either side. If the domain itself is disconnected, conclusions on its separate intervals do not automatically compare function values across the gap.
Worked Example: A Quadratic with Opposite Behaviors
Worked Example: A Quadratic with Opposite Behaviors
Determine where \(f(x)=x^2-4x+1\) is increasing or decreasing.
First differentiate:
If \(x<2\), then \(x-2<0\), so \(f'(x)<0\). Therefore, \(f\) is strictly decreasing on \((-\infty,2)\). If \(x>2\), then \(x-2>0\), so \(f'(x)>0\). Therefore, \(f\) is strictly increasing on \((2,\infty)\).
As a numerical check on the first interval, \(f(0)=1\) and \(f(1)=1-4+1=-2\), so the function’s values fall as the input moves from \(0\) to \(1\). On the second interval, \(f(3)=9-12+1=-2\) and \(f(4)=16-16+1=1\), so the values rise from \(3\) to \(4\). The derivative signs give the conclusions over the full intervals, not just at these sample inputs: \(f'(0)=-4<0\), \(f'(1)=-2<0\), \(f'(3)=2>0\), and \(f'(4)=4>0\).
At \(x=2\), \(f'(2)=0\). That point is the boundary between the two intervals considered here; the sign conclusions above concern inputs strictly to its left or right.
Worked Example: A Derivative with Two Sign Changes
Worked Example: A Derivative with Two Sign Changes
Find intervals of increase and decrease for \(g(x)=x^3-3x\).
Differentiate and factor:
For \(x<-1\), both \(x-1\) and \(x+1\) are negative, so their product is positive and \(g'(x)>0\). Thus \(g\) is strictly increasing on \((-\infty,-1)\).
For \(-1<x<1\), \(x-1<0\) while \(x+1>0\). Their product is negative, so \(g'(x)<0\), and \(g\) is strictly decreasing on \((-1,1)\).
For \(x>1\), both factors are positive, so \(g'(x)>0\). Therefore, \(g\) is strictly increasing on \((1,\infty)\).
Check representative values: \(g(-2)=-8+6=-2\) and \(g(-1)=-1+3=2\); \(g(0)=0\) and \(g(1)=1-3=-2\); \(g(2)=8-6=2\). These comparisons agree with the derivative-sign conclusions on the corresponding intervals. Direct substitution also confirms the derivative signs: \(g'(-2)=9>0\), \(g'(0)=-3<0\), and \(g'(2)=9>0\).
Worked Example: A Zero Derivative Does Not Prevent Strict Increase
Worked Example: A Zero Derivative Does Not Prevent Strict Increase
Consider \(h(x)=(x-1)^3\). Its derivative is
for every real \(x\), with \(h'(1)=0\). The nonnegative-derivative criterion shows that \(h\) is nondecreasing on \(\mathbb{R}\). In this example, we can also verify directly that it is strictly increasing.
Take any real numbers \(a<b\). Then
The first factor \(b-a\) is positive. To check the second factor, set \(u=b-1\) and \(v=a-1\). Then
This expression is nonnegative, and it can be zero only if \(u=v=0\). That would mean \(a=b=1\), contradicting \(a<b\). Hence the second factor is positive, so \(h(b)-h(a)>0\). Thus \(h(a)<h(b)\) whenever \(a<b\), proving that \(h\) is strictly increasing.
In particular, \(h'(1)=0\) does not mean that \(h\) is constant or that it stops being strictly increasing. The derivative is zero at one point, while remaining nonnegative everywhere.
Why the Interval Must Be Checked
A derivative can have different signs on different parts of a function’s domain. Each monotonicity conclusion must name an interval where its required sign condition holds. Do not combine adjacent intervals into one conclusion if the derivative changes sign between them.
For example, the quadratic in the first worked example decreases to the left of \(2\) and increases to the right. It would be incorrect to say it is increasing on all of \(\mathbb{R}\), or decreasing on all of \(\mathbb{R}\). The derivative is negative on one interval and positive on the other.
Likewise, if a function is defined on two separated intervals, a derivative-sign conclusion on each interval describes behavior within that interval. It does not compare values at inputs on opposite sides of a missing point. The definition of monotonicity compares every pair \(a<b\) in the interval under consideration, so the set being considered must genuinely be an interval.
Common Mistakes
Mistake: “The derivative is zero, so the function is constant.” A zero derivative at one point says only that the instantaneous rate of change is zero at that point. For instance, \(h'(1)=0\) above, but \(h\) is strictly increasing on \(\mathbb{R}\). A function with derivative zero at every point on an interval is a different situation; one zero value at a single point does not establish that condition.
Mistake: “The derivative is positive at one point, so the function is increasing.” A sign at one point is not the sign condition on an interval that the theorem requires. For \(f(x)=x^2-4x+1\), we have \(f'(3)=2>0\), but \(f\) is not increasing on all of \(\mathbb{R}\): it is strictly decreasing on \((-\infty,2)\). Check the derivative throughout the interval before making a monotonicity claim.
Mistake: Using a strict conclusion from a weak sign condition. The condition \(f'(x)\ge0\) guarantees nondecreasing behavior. It does not, by itself, give the theorem’s strict-increase conclusion. For a strict claim, either establish \(f'(x)>0\) throughout the interval or supply additional reasoning, as in the direct comparison for \(h(x)=(x-1)^3\).
Using the Criteria Carefully
State where you intend to make the monotonicity claim, and make sure the function is defined there.
The derivative-sign criteria require differentiability at every point in the interval.
Use the derivative expression and the conditions on the interval; a sample value alone is not enough.
A strictly positive or negative derivative gives strict monotonicity; a nonnegative or nonpositive derivative gives the corresponding non-strict conclusion.
These criteria turn derivative information into statements about the function’s behavior over intervals. Their careful use depends on checking the whole interval, distinguishing strict from non-strict conclusions, and not treating a zero derivative at one point as a complete description of the function.
Check Your Understanding
For each question, state the interval and the precise monotonicity conclusion supported by the information given.
- If \(f\) is differentiable on \((2,5)\) and \(f'(x)>0\) throughout that interval, what can you conclude?
- If \(g'(x)\le0\) on an interval, which type of decreasing behavior is guaranteed?
- Why does knowing \(p'(4)>0\) alone not establish that \(p\) is increasing on \((3,5)\)?
- For \(h(x)=(x-1)^3\), what is \(h'(1)\), and does \(h\) remain strictly increasing on \(\mathbb{R}\)?
- What additional issue arises when trying to make a monotonicity claim on a disconnected domain?
