Monotonicity (increasing or decreasing) is an important property of functions. This property is applied to solve many problems such as proving inequalities, solving equations, systems of equations, etc. In this article, we will explore the monotonicity of functions and the basic forms of problems that need to be mastered.

**Definition of Monotonicity of Functions**

For a function \(y = f(x)\) defined on a domain \(D\), we have:

\(f(x)\) is called

increasing(ornon-decreasing) on \(D\) if for all \(x_1, x_2 \in D\), if \(x_1 < x_2\), then \(f(x_1) \leq f(x_2)\).\(f(x)\) is called

decreasing(ornon-increasing) on \(D\) if for all \(x_1, x_2 \in D\), if \(x_1 < x_2\), then \(f(x_1) \geq f(x_2)\).

In simple terms, a function is increasing if both \(x\) and \(f(x)\) increase or decrease together, and a function is decreasing if \(x\) and \(f(x)\) change in opposite directions.

## Graph of Increasing and Decreasing Functions

If the function \(f(x)\) is **increasing** on the interval \((a, b)\), then the graph of \(f(x)\) on that interval is a line that **goes up from left to right**.

If the function \(f(x)\) is **decreasing** on the interval \((a, b)\), then the graph of \(f(x)\) on that interval is a line that **goes down from left to right**.

Observing the graph above, we can see that on the intervals \((-∞, -1)\) and \((1, +∞)\), the graph goes up from left to right, indicating that the function is increasing on these intervals. On the interval \((-1, 1)\), the graph goes down from left to right, indicating that the function is decreasing on this interval.

## Necessary and Sufficient Conditions for Monotonicity

Here we have an important theorem used to find the monotonic intervals of a function and apply it to some types of exercises.

For a function \(f(x)\) with a derivative on \(D\), we have:

\(f(x)\) is

increasingon \(D\) \( \Leftrightarrow f'(x) \geq 0 \, \forall x \in D\).\(f(x)\) is

decreasingon \(D\) \( \Leftrightarrow f'(x) \leq 0 \, \forall x \in D\).

Here, the condition is that \(f(x)\) is equal to 0 at only a finite number of points on \(D\). If \(f'(x) = 0 \, \forall x \in D\), then \(f(x)\) is a constant function, meaning it does not increase or decrease.

## Types of Exercises on Monotonicity of Functions

**Type 1**: Determine the Monotonicity of a Function

This means finding the intervals on which the function is increasing or decreasing. To solve this type of problem, follow these steps:

– Find the domain of the function.

– Calculate the derivative \(f'(x)\) and solve the equation \(f'(x) = 0\).

– Create a sign chart for \(f'(x)\) and use the theorem mentioned above to conclude (this is often referred to as creating a variation table because it also shows the variation of \(y\)).

**Example 1:** Determine the monotonicity of the function \(y = x^3 – 3x\).

**Solution:**

Domain: \(D = R\)

\(y’ = 3x^2 – 3\)

\(y’ = 0 \Rightarrow 3x^2 – 3 = 0 \Rightarrow x = \pm 1\)

Sign chart:

Conclusion:

The function is increasing on the intervals \((-∞, 1)\) and \((1, +∞)\).

The function is decreasing on the interval \((-1, 1)\).

**Note:** In the sign chart, if \(y’\) has a positive sign (+), it means the function is increasing, and if \(y’\) has a negative sign (-), it means the function is decreasing.

**Example 2:** Determine the monotonicity of the function \(y = \dfrac{x – 2}{2x + 4}\).

**Solution:**

Domain: \(D = R\setminus\{-2\}\)

\(y’ = \dfrac{8}{(2x + 4)^2} > 0 \, \forall x \in D\)

Sign chart:

Therefore, the function is increasing on the intervals \((-∞, -2)\) and \((-2, +∞)\).

**Observation:** In the example above, we can conclude that the function is increasing on all intervals of the domain without using the sign chart because \(y’ > 0 \, \forall x \in D\) (since both the numerator and denominator are positive). However, it’s recommended to create a sign chart to apply to different types of exercises later.

Note:The quick rule to find the derivative of a function \(y = \dfrac{ax + b}{cx + d}\) is \(y’ = \dfrac{ad – bc}{(cx + d)^2}\).

**Type 2**: Find Conditions for the Function to be Monotonic or Non-monotonic on domain

For this type of problem, you only need to use the theorem mentioned above and apply some knowledge you know about quadratic equations:

For \(f(x) = ax^2 + bx + c\) (\(a \neq 0\)), we have:

\(f(x)\) is

non-negativefor all \(x \in R\) \( \Leftrightarrow \left\{ \begin{array}{l}a > 0\\\Delta \leq 0\end{array} \right.\).\(f(x)\) is

non-positivefor all \(x \in R\) \( \Leftrightarrow \left\{ \begin{array}{l}a < 0\\\Delta \leq 0\end{array} \right.\).

**Example:** Find \(m\) such that the function \(y = x^3 – 3(2m + 1)x^2 + (12m + 5)x + 2\) is increasing on \(R\).

**Solution:**

Domain: \(D = R\)

\(y’ = 3x^2 – 6(2m + 1)x + (12m + 5)\)

\(\Delta’ = 9(2m + 1)^2 – 3(12m + 5) = 6(6m^2 – 1)\)

To make the function increasing on \(R\), we need \(y’ \geq 0\) for all \(x \in R\).

This condition is equivalent to \(\left\{ \begin{array}{l}3 > 0\\6(6m^2 – 1) \leq 0\end{array} \right.\), which simplifies to \(m \in \left[ -\dfrac{\sqrt{6}}{6}, \dfrac{\sqrt{6}}{6} \right]\).

So, for \(m \in \left[ -\dfrac{\sqrt{6}}{6}, \dfrac{\sqrt{6}}{6} \right]\), the function is increasing on \(R\).

**Note:** In the example above, we need to consider two cases: \(a = 0\) and \(a \neq 0\).

These are the two basic types of problems related to the monotonicity of functions that students need to master. In addition, there are more advanced types of problems related to monotonicity, such as finding conditions for a function to be monotonic on an interval, applying monotonicity to prove inequalities, solve equations, systems of equations, etc., which will be discussed in another article.

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