Monotone functions and derivative-sign conditions
Determine intervals of strict increase or decrease, retain domain breaks, and distinguish strict monotonicity from nondecrease.
Content updated: October 22, 2026
Monotonicity compares every pair of points on an interval. Derivative signs provide a way to make these comparisons, but each domain interval must be treated separately, and strict increase must be distinguished from nondecrease.
Objectives: find intervals of increase and decrease and parameter conditions; understand why a derivative may vanish at some points while the function remains strictly monotone.
Prerequisites: basic derivatives, polynomial signs, and continuity (English lesson).
On this page: Idea · Method · Examples · Practice · Further learning
1. Definitions on an interval
Let $f$ be defined on an interval $I$. It is strictly increasing on $I$ if, for every $x_1,x_2\in I$, $x_1<x_2$ implies $f(x_1)<f(x_2)$. Strict decrease uses $f(x_1)>f(x_2)$ instead. This lesson uses “increasing” and “decreasing” in these strict senses; nonstrict inequalities define nondecreasing and nonincreasing functions.
A graph rising from left to right illustrates increase, but a few plotted points do not prove the property for every pair of points. If the domain is interrupted at an undefined point, consider separate intervals; do not join them across the break into one monotonicity interval.
2. Derivative signs and the method
Suppose $f$ is differentiable on $I$. By the mean value theorem, $f^{\prime}\ge0$ on $I$ ensures nondecrease, and $f^{\prime}\le0$ ensures nonincrease. A strict derivative sign throughout the interval ensures strict monotonicity. With a nonstrict sign, strictness still follows if the derivative is not identically $0$ on any subinterval. A more convenient sufficient condition is that the derivative is $0$ at only finitely many points.
Indeed, if a nondecreasing function has $u<v$ with $f(u)=f(v)$, every intermediate value is equal to them. The function is then constant on $(u;v)$, so its derivative is $0$ on that subinterval. The stated condition excludes this possibility. The inequality $f^{\prime}\ge0$ alone, without excluding constant intervals, does not prove strict increase.
Method: find the domain, calculate the derivative, find the points where it is $0$ or does not exist, determine signs on each interval, and conclude. With parameters, check values causing a degree reduction or a denominator $0$ before applying polynomial criteria.
3. Five worked examples
Example 1. Using the definition directly
For $f(x)=2x+1$, if $x_1<x_2$, then $f(x_2)-f(x_1)=2(x_2-x_1)>0$. Thus the function is strictly increasing on $\mathbb R$.
Example 2. A quadratic function
Let $f(x)=x^2-2x+1$, with domain $\mathbb R$. Its derivative $f^{\prime}(x)=2x-2$ is negative for $x<1$, $0$ at $1$, and positive for $x>1$. Hence the function decreases on $(-\infty;1)$ and increases on $(1;+\infty)$.
Example 3. Two derivative roots
For $f(x)=x^3-x^2-x+2$, we have $f^{\prime}=3x^2-2x-1=(3x+1)(x-1)$. The signs are positive, negative, positive on the three intervals separated by $-1/3$ and $1$. Thus the function increases on $(-\infty;-1/3)$ and $(1;+\infty)$ and decreases on $(-1/3;1)$.
Example 4. A rational function with an undefined point
Let $f(x)=(2x+1)/(x-1)$, with domain $\mathbb R\setminus\{1\}$. We have $f^{\prime}=-3/(x-1)^2<0$ for $x\ne1$. Conclude decrease on each of $(-\infty;1)$ and $(1;+\infty)$. Their union is not a single interval of decrease: for example, $0<2$, but $f(0)=-1<5=f(2)$.
Example 5. A parameter for increase on R
Let $f(x)=x^3/3+(m+1)x^2+(m^2-1)x+3$. Its derivative is
If $m\le-1$, the derivative is nonnegative for every $x$ and has at most one zero, so the function is strictly increasing. If $m>-1$, its derivative at $x=-m-1$ is negative; a differentiable increasing function cannot have a negative derivative. Hence the necessary and sufficient condition is $m\le-1$. The boundary $m=-1$ is included, since then $f^{\prime}=x^2$, which creates no constant interval.
4. Practice exercises
- Determine the monotonicity of $f(x)=x^3$. Why does $f^{\prime}(0)=0$ not disprove increase?
- Consider $f(x)=-2x+1$ on $\mathbb R$.
- Find the monotonicity intervals of $(x+1)/(x-2)$.
Show answers and hints
- We have $f^{\prime}=3x^2\ge0$, equal to $0$ only at $0$, so the function is strictly increasing on $\mathbb R$.
- The derivative is $-2<0$, so the function is strictly decreasing on $\mathbb R$.
- The domain excludes $2$; the derivative $-3/(x-2)^2<0$, so the function decreases on each of $(-\infty;2)$ and $(2;+\infty)$.
5. Common mistakes
- Equating a nonnegative derivative with strict increase without excluding constant intervals.
- Joining intervals across an undefined point.
- Finding only derivative roots without determining signs.
- Discarding a boundary parameter value where the derivative has a double root.
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