https://r2.erweima.ai/mathbot/image/1725425784008chb9i6fo.jpg

resuelve los ejercicos 7,8,9 con sus respectivos incisos

Let's review the exercises and solve them based on the image provided. The exercises involve finding derivatives, solving optimization problems, and interpreting results.

### Exercise 7
The function provided is:
\[ f(x) = x^3 - ax^2 + bx - 2 \]

**(a) ¿Qué valores deben tener \(a\) y \(b\) para que \(f\) tenga un máximo local igual a 3 en \(x = 1\)?**

To solve this, we need to ensure the function has a maximum at \(x = 1\) with a value of 3. 

1. **Find the first derivative of \(f(x)\)** to locate critical points:
   \[
   f'(x) = 3x^2 - 2ax + b
   \]
   For a maximum at \(x = 1\), \(f'(1) = 0\).

2. **Apply the condition for a critical point:**
   \[
   f'(1) = 3(1)^2 - 2a(1) + b = 3 - 2a + b = 0 \quad \text{(Equation 1)}
   \]

3. **Apply the condition that \(f(1) = 3\):**
   \[
   f(1) = (1)^3 - a(1)^2 + b(1) - 2 = 1 - a + b - 2 = -a + b - 1 = 3 \quad \text{(Equation 2)}
   \]

From Equation 2, \(b = a + 4\). Substitute this into Equation 1:
\[
3 - 2a + (a + 4) = 0 \implies 3 - a + 4 = 0 \implies -a + 7 = 0 \implies a = 7
\]
\[
b = 7 + 4 = 11
\]
So, \(a = 7\) and \(b = 11\).

**Answer:**
- \(a = 7\)
- \(b = 11\)

**(b) Hay un segundo valor de \(x\) y de \(y\) del inciso (a). Encuentra este valor.**

To find the other value of \(x\) where the function also has a local maximum or minimum:

We substitute \(a = 7\) and \(b = 11\) back into the derivative:
\[
f'(x) = 3x^2 - 14x + 11
\]
Set \(f'(x) = 0\) to find the other critical point:
\[
3x^2 - 14x + 11 = 0
\]
Solve this quadratic equation using the quadratic formula:
\[
x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(3)(11)}}{2(3)} = \frac{14 \pm \sqrt{196 - 132}}{6} = \frac{14 \pm \sqrt{64}}{6} = \frac{14 \pm 8}{6}
\]
This gives two solutions:
\[
x_1 = \frac{22}{6} = \frac{11}{3} \quad \text{and} \quad x_2 = 1 \quad (\text{already known})
\]

Now we check the function value at \(x = \frac{11}{3}\):
\[
f\left(\frac{11}{3}\right) = \left(\frac{11}{3}\right)^3 - 7\left(\frac{11}{3}\right)^2 + 11\left(\frac{11}{3}\right) - 2
\]

### Exercise 8
\[ f(x) = x \ln(x) \]

**(a) ¿En qué punto la pendiente de la recta tangente a la función \( f(x) = x \ln(x) \) es igual a \( 2 \)?**

To solve this:
1. **Find the derivative of \(f(x)\):**
   \[
   f'(x) = \ln(x) + 1
   \]
2. **Set the derivative equal to 2 to find the point:**
   \[
   \ln(x) + 1 = 2 \implies \ln(x) = 1 \implies x = e
   \]

**Answer:**
- The point is \(x = e\).

### Exercise 9

**(a) Si se cuenta con 1200 m\(^2\) de cartón para hacer una caja sin tapa con base cuadrada y se desea maximizar el volumen, ¿cuáles son las dimensiones de la caja de esa condición?**

Let the side length of the base be \(x\) and the height be \(h\).

1. **Set up the surface area constraint:**
   \[
   x^2 + 4xh = 1200
   \]
2. **Volume of the box:**
   \[
   V = x^2h
   \]
   Express \(h\) in terms of \(x\):
   \[
   h = \frac{1200 - x^2}{4x}
   \]
   Substitute into the volume equation:
   \[
   V(x) = x^2 \left(\frac{1200 - x^2}{4x}\right) = \frac{1200x - x^3}{4}
   \]
3. **Maximize \(V(x)\)** by taking the derivative and setting it to zero:
   \[
   V'(x) = \frac{d}{dx}\left(\frac{1200x - x^3}{4}\right) = \frac{1200 - 3x^2}{4}
   \]
   \[
   1200 - 3x^2 = 0 \implies x^2 = 400 \implies x = 20
   \]
   Substitute back to find \(h\):
   \[
   h = \frac{1200 - 20^2}{4(20)} = 10
   \]

**Answer:**
- Side of the base \(x = 20\) meters.
- Height \(h = 10\) meters.

**(b) El observador de la figura en clase constató que las dimensiones que se obtienen al querer maximizar el ángulo \(\theta\). ¿Qué valor de \(\theta\) maximiza el volumen del observador?**

This question seems to involve additional information, possibly related to a diagram or specific geometrical setup. Without this context, it's difficult to provide an accurate solution.

If you provide more information about this part or clarify the scenario, I can assist further.

Would you like more details or have any specific questions? Here are five related questions you might consider:

1. How do changes in parameters \(a\) and \(b\) affect the shape of the cubic function \(f(x)\)?
2. What are the physical implications of finding the point where the slope of a tangent to \(f(x) = x \ln(x)\) is equal to 2?
3. How does the method used to find the dimensions of the box relate to real-world packaging optimization problems?
4. How would the results change if the box was required to have a top?
5. What other optimization problems could be solved using similar techniques?

**Tip:** In optimization problems, always check if your critical points correspond to maxima, minima, or saddle points by considering the second derivative or using other tests.

Explore the detailed solutions for algebraic problems involving derivatives, optimization, and quadratic equations. These exercises are suitable for students in Grades 10-12 and cover key concepts like finding maxima and minima, solving quadratic equations, and applying optimization techniques to real-world scenarios.

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