**Finding Extrema of Functions**

The problem of finding extrema (maxima or minima) of a function is a common topic in calculus, especially in 12th-grade mathematics. Students need to grasp various methods for finding extrema of functions to apply them when analyzing the behavior of functions and solving related problems.

**Learn more**: Some Basic and Advanced Types of Extrema Problems in Functions

A fundamental problem that students often encounter is finding extrema for a function \(y = f(x)\). There are two methods to solve this problem:

### Method 1: Finding Extrema Using the Sign Table

The steps for creating the sign table, which you may already know from the method of examining the monotonicity of functions, are as follows:

**Step 1:** Determine the domain of the function \(f(x)\).

**Step 2:** Calculate \(y’\) and solve the equation \(y’ = 0\).

**Step 3:** Create a sign table and conclude:

- If \(y’\) changes from negative to positive as it passes through the point \({x_0}\) (from left to right), then the function has a local minimum at \({x_0}\).
- If \(y’\) changes from positive to negative as it passes through the point \({x_0}\) (from left to right), then the function has a local maximum at \({x_0}\).

**Example 1**: Find the extrema of the function \(y = \dfrac{1}{3}x^3 – \dfrac{1}{2}x^2 – 2x + 2\).

**Solution**:

Domain: \(D = \mathbb{R}\)

\(y’ = x^2 – x – 2\)

\(y’ = 0 \Leftrightarrow x^2 – x – 2 = 0 \Leftrightarrow \left[ \begin{array}{l}x = – 1\\x = 2\end{array} \right.\)

Sign table:

Therefore, the function has a local maximum at \(x = -1\) with a maximum value of \({y_{\text{CD}}} = y\left( -1 \right) = \dfrac{19}{6}\).

The function has a local minimum at \(x = 2\) with a minimum value of \({y_{\text{CT}}} = y\left( 2 \right) = \dfrac{-4}{3}\).

**Example 2**: Find the extrema of the function \(y = \dfrac{x + 3}{2x – 1}\).

**Solution**:

Domain: \(D = \mathbb{R} \setminus \left\{\dfrac{1}{2}\right\}\)

\(y’ = \dfrac{-7}{(2x – 1)^2} < 0\) for all \(x \in D\).

Therefore, the function has no extrema.

### Method 2: Finding Extrema Using the Second Derivative

This method is often used for functions where creating a sign table is relatively difficult. Follow these steps:

**Step 1:** Determine the domain.

**Step 2:** Calculate \(y’\), solve the equation \(y’ = 0\), and denote \({x_i}\) (\(i = 1, 2, \ldots\)) as the solutions.

**Step 3:** Calculate \({f”}(x)\) and \({f”}({x_i})\) and conclude:

- If \({f”}({x_i}) < 0\), then the function has a local maximum at \({x_i}\).
- If \({f”}({x_i}) > 0\), then the function has a local minimum at \({x_i}\).

**Example 3**: Find the extrema of the function \(y = \cos x + \dfrac{1}{2}\cos 2x – 1\).

**Solution**:

Domain: \(D = \mathbb{R}\)

\(y’ = -\sin x – \sin 2x\)

\(y’ = 0 \Leftrightarrow -\sin x(1 + 2\cos x) = 0 \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\cos x = – \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \pm \dfrac{2\pi}{3} + k2\pi \end{array} \right.\)

\({f”}(x) = -\cos x – 2\cos 2x\)

For \(x = k\pi\), \({f”}(k\pi) = -\cos(k\pi) – 2\cos(2k\pi) = \pm 1 – 2 < 0\) Therefore, the function has a local maximum at \(x = k\pi\) (\(k \in \mathbb{Z}\)). For \(x = \pm \dfrac{2\pi}{3} + k2\pi\), \({f”}\left(\pm \dfrac{2\pi}{3} + k2\pi\right) = -\cos\left(\pm \dfrac{2\pi}{3}\right) – 2\cos\left(\pm \dfrac{4\pi}{3}\right) = \dfrac{1}{2} + 1 = \dfrac{3}{2} > 0\)

Therefore, the function has a local minimum at \(x = \pm \dfrac{2\pi}{3} + k2\pi\) (\(k \in \mathbb{Z}\)).

These are the two methods for finding extrema of functions that students must master. Extrema problems in functions can also involve finding parameters to make a function have extrema or not, or finding parameters that satisfy certain conditions. These types of problems will be discussed in a future article.

**Read the article**: Some Basic and Advanced Types of Extrema Problems in Functions

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