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

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

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To find the absolute minimum of the function $$ f(x) = e^{\sin x} $$ on the closed interval $$ [0, 2\pi] $$, we can analyze the behavior of the exponent $$ \sin x $$.

1. **Understanding $$ \sin x $$:**
   - The sine function, $$ \sin x $$, oscillates between -1 and 1.
   - Its minimum value is $$ -1 $$, which occurs at $$ x = \frac{3\pi}{2} $$ within the interval $$ [0, 2\pi] $$.

2. **Applying to $$ f(x) = e^{\sin x} $$:**
   - Since the exponential function $$ e^y $$ is strictly increasing, the minimum value of $$ e^{\sin x} $$ occurs when $$ \sin x $$ is at its minimum.
   - Therefore, the absolute minimum of $$ f(x) $$ is achieved when $$ \sin x = -1 $$, which is at $$ x = \frac{3\pi}{2} $$.

**Answer:**  
**Option C**: $$ \dfrac{3 \pi}{2} $$
The absolute minimum of $$ f(x) = e^{\sin x} $$ on $$ [0, 2\pi] $$ occurs at $$ \frac{3\pi}{2} $$.

On the closed interval , the absolute minimum of
occurs at

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On the closed interval , the absolute minimum of occurs at