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### Equation Given:
[tex]\[ y = x^3 - x^2 - 24x - 36 \][/tex]
A. Finding the Roots of the Equation \( y = 0 \):
1. To determine the values of \( x \) where \( y \) equals zero (\( y = 0 \)), solve the equation:
[tex]\[ x^3 - x^2 - 24x - 36 = 0 \][/tex]
The solutions (roots) are:
[tex]\[ x = -3, \, x = -2, \, x = 6 \][/tex]
So, the roots of the equation are:
[tex]\[ x = -3, -2, 6 \][/tex]
B. Determining the Critical Points:
2. First, find the first derivative \( y' \) of the given function \( y = x^3 - x^2 - 24x - 36 \):
[tex]\[ y' = \frac{d}{dx}(x^3 - x^2 - 24x - 36) \][/tex]
The derivative is:
[tex]\[ y' = 3x^2 - 2x - 24 \][/tex]
3. To obtain the critical points, set the first derivative equal to zero and solve for \( x \):
[tex]\[ 3x^2 - 2x - 24 = 0 \][/tex]
Solving this quadratic equation gives:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{73}}{3} \text{ and } x = \frac{1}{3} + \frac{\sqrt{73}}{3} \][/tex]
The critical points are:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{73}}{3} \text{ and } x = \frac{1}{3} + \frac{\sqrt{73}}{3} \][/tex]
C. Classifying the Critical Points Using the Second Derivative:
4. Find the second derivative \( y'' \) of the given function:
[tex]\[ y'' = \frac{d}{dx}(3x^2 - 2x - 24) \][/tex]
The second derivative is:
[tex]\[ y'' = 6x - 2 \][/tex]
5. Evaluate the second derivative at each of the critical points to determine whether each is a local maximum, local minimum, or an inflection point.
For \( x = \frac{1}{3} - \frac{\sqrt{73}}{3} \):
[tex]\[ y'' \left( \frac{1}{3} - \frac{\sqrt{73}}{3} \right) = 6 \left( \frac{1}{3} - \frac{\sqrt{73}}{3} \right) - 2 = -2\sqrt{73} \][/tex]
For \( x = \frac{1}{3} + \frac{\sqrt{73}}{3} \):
[tex]\[ y'' \left( \frac{1}{3} + \frac{\sqrt{73}}{3} \right) = 6 \left( \frac{1}{3} + \frac{\sqrt{73}}{3} \right) - 2 = 2\sqrt{73} \][/tex]
Since:
[tex]\[ y'' \left( \frac{1}{3} - \frac{\sqrt{73}}{3} \right) = -2\sqrt{73} \quad (\text{Negative, indicating a local maximum}) \][/tex]
[tex]\[ y'' \left( \frac{1}{3} + \frac{\sqrt{73}}{3} \right) = 2\sqrt{73} \quad (Positive, indicating a local minimum) \][/tex]
Summary:
- The roots of the equation \( y = x^3 - x^2 - 24x - 36 \) are \( x = -3, -2, 6 \).
- The critical points of the function are at \( x = \frac{1}{3} - \frac{\sqrt{73}}{3} \) and \( x = \frac{1}{3} + \frac{\sqrt{73}}{3} \).
- At \( x = \frac{1}{3} - \frac{\sqrt{73}}{3} \), there is a local maximum since the second derivative is negative.
- At [tex]\( x = \frac{1}{3} + \frac{\sqrt{73}}{3} \)[/tex], there is a local minimum since the second derivative is positive.
### Equation Given:
[tex]\[ y = x^3 - x^2 - 24x - 36 \][/tex]
A. Finding the Roots of the Equation \( y = 0 \):
1. To determine the values of \( x \) where \( y \) equals zero (\( y = 0 \)), solve the equation:
[tex]\[ x^3 - x^2 - 24x - 36 = 0 \][/tex]
The solutions (roots) are:
[tex]\[ x = -3, \, x = -2, \, x = 6 \][/tex]
So, the roots of the equation are:
[tex]\[ x = -3, -2, 6 \][/tex]
B. Determining the Critical Points:
2. First, find the first derivative \( y' \) of the given function \( y = x^3 - x^2 - 24x - 36 \):
[tex]\[ y' = \frac{d}{dx}(x^3 - x^2 - 24x - 36) \][/tex]
The derivative is:
[tex]\[ y' = 3x^2 - 2x - 24 \][/tex]
3. To obtain the critical points, set the first derivative equal to zero and solve for \( x \):
[tex]\[ 3x^2 - 2x - 24 = 0 \][/tex]
Solving this quadratic equation gives:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{73}}{3} \text{ and } x = \frac{1}{3} + \frac{\sqrt{73}}{3} \][/tex]
The critical points are:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{73}}{3} \text{ and } x = \frac{1}{3} + \frac{\sqrt{73}}{3} \][/tex]
C. Classifying the Critical Points Using the Second Derivative:
4. Find the second derivative \( y'' \) of the given function:
[tex]\[ y'' = \frac{d}{dx}(3x^2 - 2x - 24) \][/tex]
The second derivative is:
[tex]\[ y'' = 6x - 2 \][/tex]
5. Evaluate the second derivative at each of the critical points to determine whether each is a local maximum, local minimum, or an inflection point.
For \( x = \frac{1}{3} - \frac{\sqrt{73}}{3} \):
[tex]\[ y'' \left( \frac{1}{3} - \frac{\sqrt{73}}{3} \right) = 6 \left( \frac{1}{3} - \frac{\sqrt{73}}{3} \right) - 2 = -2\sqrt{73} \][/tex]
For \( x = \frac{1}{3} + \frac{\sqrt{73}}{3} \):
[tex]\[ y'' \left( \frac{1}{3} + \frac{\sqrt{73}}{3} \right) = 6 \left( \frac{1}{3} + \frac{\sqrt{73}}{3} \right) - 2 = 2\sqrt{73} \][/tex]
Since:
[tex]\[ y'' \left( \frac{1}{3} - \frac{\sqrt{73}}{3} \right) = -2\sqrt{73} \quad (\text{Negative, indicating a local maximum}) \][/tex]
[tex]\[ y'' \left( \frac{1}{3} + \frac{\sqrt{73}}{3} \right) = 2\sqrt{73} \quad (Positive, indicating a local minimum) \][/tex]
Summary:
- The roots of the equation \( y = x^3 - x^2 - 24x - 36 \) are \( x = -3, -2, 6 \).
- The critical points of the function are at \( x = \frac{1}{3} - \frac{\sqrt{73}}{3} \) and \( x = \frac{1}{3} + \frac{\sqrt{73}}{3} \).
- At \( x = \frac{1}{3} - \frac{\sqrt{73}}{3} \), there is a local maximum since the second derivative is negative.
- At [tex]\( x = \frac{1}{3} + \frac{\sqrt{73}}{3} \)[/tex], there is a local minimum since the second derivative is positive.
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