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To solve this problem, we need to find the minimum and maximum values of the function \( y = x^3 - 24 \ln(x) + 1 \) on the interval \( \left[ \frac{1}{2}, 3 \right] \). Here’s the step-by-step approach:

1. **Calculate the derivative** to find the critical points:
   \[
   y' = \frac{d}{dx} \left( x^3 - 24 \ln(x) + 1 \right).
   \]
   Using the power rule and the derivative of \( \ln(x) \):
   \[
   y' = 3x^2 - \frac{24}{x}.
   \]

2. **Set the derivative equal to zero** to find critical points:
   \[
   3x^2 - \frac{24}{x} = 0.
   \]
   Multiply through by \( x \) to clear the fraction:
   \[
   3x^3 - 24 = 0.
   \]
   Solving for \( x \):
   \[
   x^3 = 8 \Rightarrow x = 2.
   \]
   So, \( x = 2 \) is a critical point within the interval \( \left[ \frac{1}{2}, 3 \right] \).

3. **Evaluate the function at the endpoints and the critical point**:
   - At \( x = \frac{1}{2} \):
     \[
     y\left( \frac{1}{2} \right) = \left( \frac{1}{2} \right)^3 - 24 \ln\left( \frac{1}{2} \right) + 1.
     \]
     Simplifying this, we get:
     \[
     y\left( \frac{1}{2} \right) = \frac{1}{8} + 24 \ln(2) + 1.
     \]

   - At \( x = 2 \):
     \[
     y(2) = 2^3 - 24 \ln(2) + 1 = 8 - 24 \ln(2) + 1 = 9 - 24 \ln(2).
     \]

   - At \( x = 3 \):
     \[
     y(3) = 3^3 - 24 \ln(3) + 1 = 27 - 24 \ln(3) + 1 = 28 - 24 \ln(3).
     \]

4. **Compare the values** at \( x = \frac{1}{2} \), \( x = 2 \), and \( x = 3 \) to determine the minimum and maximum values.

- **Minimum**: The smallest of \( y\left(

Find the minimum and maximum values of the function y = x^3 - 24 ln(x) + 1 on the interval [1/2, 3] by comparing values at the critical points and endpoints.

Discover how to find the minimum and maximum values of the function y = x^3 - 24 ln(x) + 1 on the interval [1/2, 3]. This problem uses calculus to analyze critical points and interval endpoints, following the Extreme Value Theorem. Suitable for Grades 11-12.

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