Actividad 5. Máximos y mínimos. Criterio de la segunda derivada.
(Valor: 5 puntos)
(Valor 5%) Determina los puntos máximos y mínimos para la función   utiliza el criterio de la segunda derivada


To determine the maximum and minimum points of the function using the **second derivative test**, we will follow these steps:

1. **Find the first and second derivatives** of the function.
2. **Determine the critical points** where the first derivative is zero.
3. **Apply the second derivative test** to classify each critical point as a local maximum, local minimum, or a point of inflection.

Given the function:  
\[
f(x) = x^3 - 3x
\]

### Step-by-Step Solution

#### 1. Find the First and Second Derivatives

First, we already have the first derivative from the previous calculation:  
\[
f'(x) = 3x^2 - 3
\]

Now, we find the second derivative:  
\[
f''(x) = \frac{d}{dx}(3x^2 - 3) = 6x
\]

#### 2. Determine the Critical Points

The critical points are where the first derivative is zero. We previously found:  
\[
f'(x) = 3x^2 - 3 = 0 \implies x = \pm 1
\]

So, the critical points are \(x = -1\) and \(x = 1\).

#### 3. Apply the Second Derivative Test

The second derivative test states:

- If \(f''(x) > 0\) at a critical point, then \(f(x)\) has a **local minimum** at that point.
- If \(f''(x) < 0\) at a critical point, then \(f(x)\) has a **local maximum** at that point.
- If \(f''(x) = 0\), the test is inconclusive (the point could be an inflection point).

We evaluate the second derivative at the critical points:

- For \(x = -1\):
  \[
  f''(-1) = 6(-1) = -6
  \]
  Since \(f''(-1) < 0\), the function has a **local maximum** at \(x = -1\).

- For \(x = 1\):
  \[
  f''(1) = 6(1) = 6
  \]
  Since \(f''(1) > 0\), the function has a **local minimum** at \(x = 1\).

### Conclusion

- The function \(f(x) = x^3 - 3x\) has a **local maximum** at \(x = -1\) and a **local minimum** at \(x = 1\).

To find the coordinates of these points:

- At \(x = -1\):  
  \[
  f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2
  \]

- At \(x = 1\):  
  \[
  f(1) = (1)^3 - 3(1) = 1 - 3 = -2
  \]

Thus, the points are:

- **Local Maximum** at \((-1, 2)\)
- **Local Minimum** at \((1, -2)\)

Would you like more details or have any questions about this solution?

### Further Related Questions

1. What is the significance of the second derivative in understanding the shape of a graph?
2. How do you use the second derivative to find points of inflection?
3. Can a point where \(f''(x) = 0\) be a local maximum or minimum?
4. What are some practical applications of finding local maxima and minima?
5. How do the first and second derivative tests compare in terms of reliability and ease of use?

### Tip

The second derivative test is useful for quickly determining the concavity of a function at critical points, which helps to classify them as local maxima or minima. However, always check if \(f''(x) = 0\) because it could be a point of inflection rather than a maximum or minimum.

Learn how to determine maximum and minimum points of a function using the second derivative test. Example problem solved: f(x) = x^3 - 3x. Includes step-by-step solution, critical points analysis, and classification of local maxima and minima. Suitable for undergraduate level studies.

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