(b) \( y=\frac{3 x^{2}}{2 \sin x} \)

To analyze the function \( y = \frac{3x^2}{2 \sin x} \), we can explore various aspects such as its domain, behavior, and any critical points. ### Domain The function is defined wherever the denominator is not zero. Therefore, we need to find where \( \sin x = 0 \). The sine function is zero at integer multiples of \( \pi \): \[ x = n\pi, \quad n \in \mathbb{Z} \] Thus, the domain of \( y \) is: \[ x \in \mathbb{R} \setminus \{ n\pi \mid n \in \mathbb{Z} \} \] ### Behavior as \( x \to n\pi \) As \( x \) approaches any integer multiple of \( \pi \), \( \sin x \) approaches 0, which means \( y \) will approach \( \pm \infty \) depending on the direction from which \( x \) approaches \( n\pi \). ### Critical Points To find critical points, we can take the derivative of \( y \) with respect to \( x \) and set it to zero. Using the quotient rule: \[ y' = \frac{(2 \sin x)(6x) - (3x^2)(2 \cos x)}{(2 \sin x)^2} \] Simplifying the numerator: \[ y' = \frac{12x \sin x - 6x^2 \cos x}{(2 \sin x)^2} \] Setting the numerator equal to zero: \[ 12x \sin x - 6x^2 \cos x = 0 \] Factoring out \( 6x \): \[ 6x(2 \sin x - x \cos x) = 0 \] This gives us two cases: 1. \( 6x = 0 \) which gives \( x = 0 \). 2. \( 2 \sin x - x \cos x = 0 \). The second equation \( 2 \sin x = x \cos x \) may require numerical methods or graphical analysis to find solutions. ### Asymptotic Behavior As \( x \to \infty \) or \( x \to -\infty \), the behavior of \( y \) can be analyzed: - Since \( \sin x \) oscillates between -1 and 1, \( y \) will oscillate as well, but the overall trend will depend on the growth of \( x^2 \) in the numerator. ### Summary - **Domain**: \( x \in \mathbb{R} \setminus \{ n\pi \mid n \in \mathbb{Z} \} \) - **Critical Points**: \( x = 0 \) and solutions to \( 2 \sin x = x \cos x \). - **Behavior**: \( y \to \pm \infty \) as \( x \to n\pi \) and oscillates as \( x \to \pm \infty \). If you need further analysis or specific evaluations, please let me know!