Analyzing Function Decreasing Intervals

The problem asks us to find the set S, which represents the interval where the function \(f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 6x + 7\) decreases.

Derivative Calculation

To find where a function decreases, we need to analyze the sign of its first derivative, \(f'(x)\).

1. Calculate the derivative of \(f(x)\): \(f'(x) = \frac{d}{dx} \left( \frac{x^3}{3} - \frac{5x^2}{2} + 6x + 7 \right)\)
2. Applying the power rule for differentiation: \(f'(x) = \frac{3x^2}{3} - \frac{5(2x)}{2} + 6\) \(f'(x) = x^2 - 5x + 6\)

Finding Decreasing Intervals

A function decreases where its derivative is negative, i.e., \(f'(x) < 0\).

1. Set the derivative less than zero: \(x^2 - 5x + 6 < 0\)
2. Find the roots of the quadratic equation \(x^2 - 5x + 6 = 0\). Factoring the quadratic gives: \((x - 2)(x - 3) = 0\) The roots are \(x = 2\) and \(x = 3\).
3. Determine the sign of \(f'(x)\) in the intervals defined by the roots \((-\infty, 2)\), (2, 3), and \((3, \infty)\). The quadratic \(x^2 - 5x + 6\) represents a parabola opening upwards. Therefore, it is negative between its roots.
   • For \(x < 2\), \(f'(x) > 0\) (function increases).
   • For \(2 < x < 3\), \(f'(x) < 0\) (function decreases).
   • For \(x > 3\), \(f'(x) > 0\) (function increases).
4. The set S, where the function decreases, corresponds to the interval where \(f'(x) < 0\). \(S = \{x \mid 2 < x < 3\}\)

This interval is represented as (2, 3).

Conclusion

The set S where the function \(f(x)\) decreases is the open interval (2, 3).