Increasing, Decreasing and Monotonic Functions

Introduction

1.A function is classified as increasing, decreasing, or monotonic according to how its output values change as the input grows.

2.Definition 1. Let $y = f(x)$ be a function defined on a domain $X \subseteq \mathbb{R}.$ We say that $f$ is strictly increasing on an interval $I \subseteq X$ if, for any two values $x_1, x_2 \in I$ with $x_1 < x_2,$ the following condition holds:

$$ f(x_1) < f(x_2) $$

3.As the input $x$ increases within the interval $I,$ the output $f(x)$ also strictly increases, without any flat or decreasing segments.

4.Definition 2. Let $y = f(x)$ be a function defined on a domain $X \subseteq \mathbb{R}.$ We say that $f$ is strictly decreasing on an interval $I \subseteq X$ if, for any two values $x_1, x_2 \in I$ with $x_1 < x_2,$ the following condition holds:

$$ f(x_1) > f(x_2) $$

5.As the input $x$ increases within the interval $I,$ the output $f(x)$ strictly decreases, with no flat or increasing sections.

6.A function with domain $X \subseteq \mathbb{R}$ is strictly monotonic on an interval $I \subseteq X$ if it is either strictly increasing or strictly decreasing throughout $I,$ with no change in direction or flat segments.

7.To summarize, let $X \subseteq \mathbb{R}$ and let $x_1, x_2 \in X$ with $x_1 < x_2.$ The function $f : X \to \mathbb{R}$ is said to be:

Derivatives and monotonic behavior

9.The derivative describes the shape of the graph of a function. In particular, the first derivative $f'(x)$ identifies the intervals where $f(x)$ increases and where it decreases. Given a function $y = f(x)$ that is continuous on an interval $I$ and differentiable at the interior points of $I$:

11.Applied to a position function $s(t),$ this criterion says that positive velocity makes the position increase with time, while negative velocity makes it decrease.

12.

A nonnegative derivative already gives monotonicity in the wide sense, since $f'(x) \geq 0$ on the interior of $I$ implies that $f$ is increasing. Strict monotonicity tolerates isolated points where the derivative vanishes, as $f(x) = x^3$ shows at $x = 0,$ where $f'(0) = 0$ while $f$ remains strictly increasing.

13.To prove these properties, we use Lagrange's theorem. Take two points $a, b \in I$ with $a < b,$ and consider a point $c$ in the open interval $(a, b).$ By Lagrange's theorem, there exists such a $c$ for which:

$$ f'(c) = \frac{f(b) - f(a)}{b - a} $$

14.Since $b - a > 0$ and $f'(c) > 0,$ it follows that $f(b) - f(a) > 0,$ which implies $f(b) > f(a).$ As $a$ and $b$ are arbitrary points of $I,$ the function is increasing on $I.$

15.In the opposite case, since $b - a > 0$ and $f'(c) < 0,$ it follows that $f(b) - f(a) < 0,$ which implies $f(b) < f(a).$ Again $a$ and $b$ are arbitrary, so the function is decreasing on $I.$

Example 1

16.Let us consider the function:

$$f(x) = \frac{x^4}{4} - \frac{x^2}{2}$$

17.We compute its derivative $f'(x) = x(x^2 - 1).$ To find the intervals where the derivative is positive, we study the sign of each factor:

18.$$x > 0$$ $$x^2 - 1 > 0 \implies x < -1 \text{ or } x > 1$$

19.By multiplying the signs of the two factors, we obtain the intervals where the derivative is positive:

21.Therefore, the derivative $x(x^2 - 1)$ is positive for $x \in (-1,0) \cup (1,+\infty).$

22.

The sign analysis of a product, as in this example, examines the sign of each factor and determines the overall sign on each interval by multiplying those signs.

23.Reading the sign of the derivative through the definition of monotonicity, $f$ is strictly increasing on each of the intervals $(-1, 0)$ and $(1, +\infty),$ where $f'(x) > 0,$ and strictly decreasing on each of the intervals $(-\infty, -1)$ and $(0, 1),$ where $f'(x) < 0.$

24.

The function is not increasing on the union $(-1, 0) \cup (1, +\infty)$ taken as a single set. Monotonicity must be read on each interval separately, since comparing a point of $(-1, 0)$ with a point of $(1, +\infty)$ crosses the decreasing stretch $(0, 1)$ that lies between them.

Strict monotonicity and injectivity

25.A strictly monotonic function is injective. Suppose $f$ is strictly increasing on $I$ and let $x_1, x_2 \in I$ be distinct. One of the two orderings holds, either $x_1 < x_2$ or $x_2 < x_1,$ and in each case strict monotonicity turns the strict inequality between the inputs into a strict inequality between the outputs, so $f(x_1) \neq f(x_2).$ The same reasoning applies to a strictly decreasing function.

26.A strictly monotonic function therefore admits an inverse on its image, and the inverse keeps the same direction of monotonicity.

27.

Wide-sense monotonicity is not enough for injectivity. A function that is increasing with the condition $f(x_1) \leq f(x_2)$ may remain constant on a subinterval, taking a single value at infinitely many points.

One-sided limits of a monotonic function

28.Monotonicity constrains the local behavior of a function strongly enough to force the existence of both one-sided limits at every interior point, even where the function is discontinuous.

29.Definition 3. Let $f$ be increasing on an open interval $(a, b).$ For every $x_0 \in (a, b)$ both one-sided limits exist and satisfy:

$$ \sup_{a < t < x_0} f(t) = f(x_0^-) \leq f(x_0) \leq f(x_0^+) = \inf_{x_0 < t < b} f(t) $$

30.To prove the claim for the left limit, consider the set of values $\{\ f(t) \mid a < t < x_0 \ \}.$ Since $f$ is increasing, every element of this set is bounded above by $f(x_0),$ so the set has a least upper bound:

$$ A := \sup_{a < t < x_0} f(t) $$

31.Fix $\varepsilon > 0.$ By definition of supremum there exists $t_0 \in (a, x_0)$ with $f(t_0) > A - \varepsilon.$ For every $t \in [t_0, x_0)$ monotonicity gives:

$$ A - \varepsilon < f(t_0) \leq f(t) \leq A $$

32.These inequalities state that $f(t)$ approaches $A$ as $t \to x_0^-,$ hence $f(x_0^-) = A.$ The proof that $f(x_0^+) = \inf_{x_0 < t < b} f(t)$ is symmetric, and the central inequality $f(x_0^-) \leq f(x_0) \leq f(x_0^+)$ follows from $f(t) \leq f(x_0) \leq f(s)$ whenever $t < x_0 < s.$ A decreasing function is handled by applying the same result to $-f,$ which is increasing.

33.The one-sided limits of an increasing function cannot overlap across distinct points. Let $x, y \in (a, b)$ with $x < y$ and choose any $z \in (x, y).$ Using the characterization above on both points:

$$ f(x^+) = \inf_{x < t < b} f(t) \leq f(z) \leq \sup_{a < t < y} f(t) = f(y^-) $$

34.The value approached from the right of $x$ never exceeds the value approached from the left of any later point $y.$

Discontinuities of a monotonic function

35.At an interior point $x_0$ both one-sided limits of an increasing function are finite and ordered as $f(x_0^-) \leq f(x_0) \leq f(x_0^+).$ Continuity at $x_0$ fails exactly when the outer terms differ, so the only discontinuity a monotonic function can have is a jump, and the nonnegative quantity

$$ f(x_0^+) - f(x_0^-) $$

36.measures its size. A removable discontinuity is excluded, because $f(x_0^-) = f(x_0^+)$ would already force equality with $f(x_0)$ and hence continuity. A monotonic function never has a discontinuity where a one-sided limit fails to exist.

37.The set of points where a monotonic function is discontinuous is at most countable. Take $f$ increasing on $(a, b).$ At each discontinuity $x$ we have $f(x^-) < f(x^+),$ so the open interval $(f(x^-), f(x^+))$ is nonempty and contains a rational number $r_x.$ If $x < y$ are two discontinuities, the bound $f(x^+) \leq f(y^-)$ keeps the intervals $(f(x^-), f(x^+))$ and $(f(y^-), f(y^+))$ disjoint, so $r_x \neq r_y.$ The map $x \mapsto r_x$ is an injection from the set of discontinuities into $\mathbb{Q},$ and any set that injects into $\mathbb{Q}$ is at most countable.

38.The floor function shows that this bound is attained. It is non-decreasing and has a jump at every integer.

39.

The discontinuities of a monotonic function need not be isolated. Given any countable set $E \subseteq (a, b),$ even a dense one such as $\mathbb{Q} \cap (a, b),$ one can build an increasing function that is discontinuous exactly at the points of $E$ and continuous everywhere else.