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(