Question

1.96 / 5.88 points
determine the open intervals on which the function is increasing, decreasing, or constant. (enter your answers using interval notation. if an answer does not exist, enter dne.)
$f(x)=-\frac{1}{2} x^{3}$

Explanation:

Step1: Find the derivative of the function

The function is \( f(x)=-\frac{1}{2}x^{3} \). Using the power rule \( (x^{n})^\prime = nx^{n - 1} \), the derivative \( f^\prime(x)=-\frac{1}{2}\times3x^{2}=-\frac{3}{2}x^{2} \).

Step2: Analyze the sign of the derivative

Since \( x^{2}\geq0 \) for all real \( x \), then \( -\frac{3}{2}x^{2}\leq0 \) for all real \( x \). The equality \( -\frac{3}{2}x^{2} = 0 \) holds when \( x = 0 \). For \( x
eq0 \), \( f^\prime(x)<0 \).

Answer:

increasing: DNE; decreasing: \( (-\infty,\infty) \); constant: DNE