In the previous article, we learned how to find extrema (maxima or minima) of a function. Now, let’s explore some related exercises involving basic and advanced extrema problems. These exercises often require us to find parameters *m* that satisfy certain conditions for the function to have extrema. We commonly encounter the following types:

**Review**: Methods for finding extrema of functions

**Type 1: Find**

*m*for the function $y = f(x)$ to have a maximum or minimum at ${x_0}$Method: We use the following conditions:

- If $\left\{ \begin{array}{l}f'({x_0}) = 0\\f”({x_0}) > 0\end{array} \right.$, then the function has a minimum at ${x_0}$.
- If $\left\{ \begin{array}{l}f'({x_0}) = 0\\f”({x_0}) < 0\end{array} \right.$, then the function has a maximum at ${x_0}$.

**Example 1**: Find *m* for the function $y = {\textstyle{1 \over 3}}{x^3} + \left( {{m^2} – m + 2} \right){x^2} + \left( {3{m^2} + 1} \right)x + m – 5$ to have a minimum at ${x = -2}.$

**Solution**:

$y’\left( x \right) = {x^2} + 2\left( {{m^2} – m + 2} \right)x + 3{m^2} + 1 \Rightarrow y”\left( x \right) = 2x + 2\left( {{m^2} – m + 2} \right)$

To have a minimum at ${x = -2},$ the necessary condition is $y’\left( { – 2} \right) = 0$:

$ \Leftrightarrow – {m^2} + 4m – 3 = 0 \Leftrightarrow \left[ \begin{array}{l}m = 1\\m = 3\end{array} \right.$

For $m = 3,$ we have $y\left( { – 2} \right) = 8 > 0,$ so the function has a minimum at ${x = -2}.$ Thus, $m = 3$ satisfies the requirement.

For $m = 1,$ we have $y = \dfrac{1}{3}{x^3} + 2{x^2} + 4x – 4.$ Using the sign table, we see that the function does not have extrema, so $m = 1$ does not satisfy the requirement.

Therefore, with $m = 3,$ the function has a minimum at ${x = -2}.$

**Note**: For $m = 1,$ $y\left( { – 2} \right) = 0,$ so we cannot conclude, and we need to use the sign table.

**Type 2: Find**

*m*for the function $y = f(x)$ to have extrema or no extrema.For this type of problem, we typically focus on two main types of functions:1. Cubic functions: $y = a{x^3} + b{x^2} + cx + d\,\left( {a \ne 0} \right)$

- The function has no extrema if the equation $y’ = 0$ has no or repeated roots, which is equivalent to $\Delta \le 0.$
- The function has two extrema if the equation $y’ = 0$ has two distinct roots, which is equivalent to $\Delta > 0.$

2. Quartic functions with a square term: $y = a{x^4} + b{x^2} + c\,\left( {a \ne 0} \right)$

- The function has one extremum if the equation $y’ = 0$ has a unique root, which is equivalent to \(a \cdot b \ge 0\).
- The function has three extrema if the equation $y’ = 0$ has three distinct roots, which is equivalent to \(a \cdot b < 0\).

**Example 2**: Given the function $y = {x^3} – 3(m + 1){x^2} + 9x – m$, with \(m\) as a real parameter. Determine \(m\) so that the given function has two extrema.

**Solution**:

We have $y’ = 3{x^2} – 6(m + 1)x + 9.$

To have two extrema, the equation \(y’ = 0\) must have two distinct roots.

$ \Leftrightarrow {x^2} – 2(m + 1)x + 3 = 0$ has two distinct roots.

$ \Leftrightarrow \Delta ‘ = {(m + 1)^2} – 3 > 0 \Leftrightarrow m \in \left( { – \infty ; – 1 – \sqrt 3 } \right) \cup \left( { – 1 + \sqrt 3 ; + \infty } \right)$

**Example 3**: Given the function $y = f(x) = m{x^3} + 3m{x^2} – \left( {m – 1} \right)x – 1$, where \(m\) is a parameter. Determine the values of \(m\) such that the function has no extrema.

**Solution**:

For \(m = 0 \Rightarrow y = x – 1\), and thus, the function has no extrema.

For \(m \ne 0\), we have \(y’ = 3m{x^2} + 6mx – \left( {m – 1} \right)\).

The function has no extrema if and only if the equation \(y’ = 0\) has no or repeated roots.

$ \Leftrightarrow \Delta ‘ = 9{m^2} + 3m\left( {m – 1} \right) = 12{m^2} – 3m \le 0 \Leftrightarrow 0 \le m \le \dfrac{1}{4}$

Therefore, for \(0 \le m \le \dfrac{1}{4},\) the function has no extrema.

**Type 3: Find parameters**

*m*for the function to have extrema satisfying certain conditions.This is an advanced type of problem that we often encounter in university and college entrance exams. To solve these problems, you need to be familiar with the methods for solving the problem types mentioned above and combine them with some additional knowledge of geometry, sequences, and more.

**Example 4**: Given the function $y = {x^4} – 2{m^2}{x^2} + 1\,\,\left( {{C_m}} \right)$. Find \(m\) such that the function has three extrema, and these extrema form the vertices of an isosceles right triangle.

**Solution**:

First, we apply the method from type 2 to find \(m\) such that the function has 3 extrema.

We have $y’ = 4{x^3} – 4{m^2}x = 4x\left( {{x^2} – {m^2}} \right)$

$y’ = 0 \Leftrightarrow 4x\left( {{x^2} – {m^2}} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\{x^2} = {m^2} (*)\end{array} \right.$

To have 3 extrema, the equation \(y’ = 0\) must have 3 distinct roots.

$ \Leftrightarrow $ Equation (*) must have 2 distinct roots, which means $m \ne 0$

So, for \(m \ne 0\), the function has 3 extrema.

Now, let’s find \(m\) such that these 3 extrema form the vertices of an isosceles right triangle.

We have \(y’ = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0 \Rightarrow y = 1\\x = m \Rightarrow y = 1 – {m^4}\\x = – m \Rightarrow y = 1 – {m^4}\end{array} \right.\)

Let the 3 extrema be: \(A(0;1), B( – m;1 – {m^4}),\) and \(C(m;1 – {m^4})\).

According to the properties of quartic functions, triangle ABC is isosceles at A. To make ABC an isosceles right triangle, we need \(AB\) to be perpendicular to \(AC \Leftrightarrow \overrightarrow{AB} \cdot \overrightarrow{AC} = 0\)

We have: \(\overrightarrow{AB} = (-m; – {m^4})\) and \(\overrightarrow{AC} = (m; – {m^4})\)

\(\overrightarrow{AB} \cdot \overrightarrow{AC} = 0 \Leftrightarrow – {m^2} + {m^8} = 0 \Leftrightarrow \left[ \begin{array}{l}m = 0\,\,(l)\\m = \pm 1\end{array} \right.\)

Therefore, \(m = -1\) and \(m = 1\) satisfy the problem’s requirements.

These are three common types of extremum (maxima or minima) problems involving functions that we often encounter. Among them, types 1 and 2 are fundamental, and you should master them before moving on to type 3.

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