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

Let's analyze and solve the exercises 7, 8, and 9 from the image.

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

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

1. First, apply \( x = 1 \) in the first derivative:
\[
f'(1) = 3(1)^2 - 2a(1) + b = 3 - 2a + b = 0 \quad \text{(Equation 1)}
\]

2. Apply \( x = 1 \) in the original function and set \( f(1) = 3 \):
\[
f(1) = (1)^3 - a(1)^2 + b(1) - 2 = 1 - a + b - 2 = -a + b - 1 = 3 \quad \text{(Equation 2)}
\]
Thus, we have two equations:
\[
\text{(1) } 3 - 2a + b = 0
\]
\[
\text{(2) } -a + b = 4
\]
From Equation (2), \( b = a + 4 \). Substitute this into Equation (1):
\[
3 - 2a + (a + 4) = 0 \implies 3 - a + 4 = 0 \implies -a + 7 = 0 \implies a = 7
\]
Substitute \( a = 7 \) into Equation (2):
\[
b = 7 + 4 = 11
\]
Therefore, \( a = 7 \) and \( b = 11 \).

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

### Exercise 8
The problem states:
\[ f(x) = x \ln(x) \]

- **(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, find the derivative of \( f(x) \) to get the slope of the tangent line:
\[
f'(x) = \frac{d}{dx} [x \ln(x)] = \ln(x) + 1
\]
Set this equal to 2 and solve for \( x \):
\[
\ln(x) + 1 = 2 \implies \ln(x) = 1 \implies x = e
\]
Therefore, the slope of the tangent line is 2 when \( x = e \).

**Answer:**
- \( x = e \)

### Exercise 9
The problem contains two parts:

- **(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 square base be \( x \), and the height be \( h \).

The surface area constraint (without the top) is:
\[
x^2 + 4xh = 1200
\]
The volume of the box is:
\[
V = x^2h
\]
Express \( h \) in terms of \( x \) from the surface area equation:
\[
h = \frac{1200 - x^2}{4x}
\]
Substitute this into the volume equation:
\[
V(x) = x^2 \left( \frac{1200 - x^2}{4x} \right) = \frac{1200x - x^3}{4}
\]
To maximize \( V(x) \), take the derivative and set it equal to zero:
\[
V'(x) = \frac{d}{dx} \left( \frac{1200x - x^3}{4} \right) = \frac{1200 - 3x^2}{4}
\]
Set \( V'(x) = 0 \):
\[
1200 - 3x^2 = 0 \implies 3x^2 = 1200 \implies x^2 = 400 \implies x = 20
\]
Substitute \( x = 20 \) back into the equation for \( h \):
\[
h = \frac{1200 - 20^2}{4(20)} = \frac{1200 - 400}{80} = \frac{800}{80} = 10
\]
Thus, the dimensions of the box are:
- Side of the base \( x = 20 \) meters
- Height \( h = 10 \) meters

**Answer:**
- Side length of 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 part of the problem requires information not visible in the image, so I can't provide an exact answer. It seems to involve maximizing some function related to an angle, possibly involving a geometric setup. Please provide more details or clarify this part.

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.

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

This page provides detailed solutions for calculus and optimization problems involving cubic functions, logarithmic functions, and maximizing the volume of a box. Learn how to find the maximum and minimum points of a function, calculate slopes of tangent lines, and solve real-world optimization problems. Suitable for students in Grades 11-12.

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