2. On the closed interval \( [0,2 \pi] \), the absolute minimum of \( f(x)=e^{\sin x} \) occurs at \( \begin{array}{llll}\text { (A) } 0 & \text { (B) } \frac{\pi}{2} & \text { (C) } \frac{3 \pi}{2} & \text { (D) } 2 \pi\end{array} \)

To find the absolute minimum of \( f(x)=e^{\sin x} \) on the interval \( [0, 2\pi] \), we need to evaluate \( f(x) \) at the endpoints and any critical points inside the interval. The function \( \sin x \) reaches its minimum value of -1 at \( \frac{3\pi}{2} \), resulting in \( f\left(\frac{3\pi}{2}\right) = e^{-1} \approx 0.3679 \). For the endpoints, \( f(0) = e^0 = 1 \) and \( f(2\pi) = e^0 = 1 \). Thus, the absolute minimum occurs at \( \frac{3\pi}{2} \). Additionally, considering the behavior of the sine function can add to our understanding. The sine function oscillates between -1 and 1, and since the function \( e^x \) is monotonically increasing, the minimum of \( e^{\sin x} \) directly aligns with where \( \sin x \) hits its minimum. It's a delightful way to see how one function's behavior can dictate outcomes for another!