Discussion
The range of the function \(f(x) = \left(\frac{1}{3}\right)^{3 + sinx}\) is
Answer: D
The range is \(\left[\frac{1}{81}, \frac{1}{9}\right]\).
Since \(-1 \le \sin x \le 1\), the exponent \(3 + \sin x\) runs from 2 to 4. The base \(\frac{1}{3}\) is less than 1, so the function decreases as the exponent grows.
\[ \begin{aligned} \min f &= \left(\tfrac{1}{3}\right)^{4} = \tfrac{1}{81} \\ \max f &= \left(\tfrac{1}{3}\right)^{2} = \tfrac{1}{9} \end{aligned} \]
A positive base raised to any power is positive, so the options with negative endpoints are impossible. \(\frac{1}{3}\) would need the exponent 1, which \(3 + \sin x\) never reaches.
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