The range of the function f(x) = (sin x)^{sin | vedclass

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(D) Let $y = f(x) = (\sin x)^{\sin x}$ for $x \in (0, \pi)$.Taking the natural logarithm on both sides,we get $\ln y = \sin x \ln(\sin x)$.Let $u = \s

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$[e^{-1/e}, 1]$

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(D) Let $y = f(x) = (\sin x)^{\sin x}$ for $x \in (0, \pi)$.Taking the natural logarithm on both sides,we get $\ln y = \sin x \ln(\sin x)$.Let $u = \sin x$. Since $x \in (0, \pi)$,the range of $u$ is $(0, 1]$.We define $g(u) = u^{\ln u}$ for $u \in (0, 1]$.To find the range,we differentiate $g(u)$ with respect to $u$:$g'(u) = u^u (1 + \ln u)$.Setting $g'(u) = 0$,we get $1 + \ln u = 0$,which implies $\ln u = -1$,so $u = e^{-1} = \frac{1}{e}$.At $u = \frac{1}{e}$,$g(\frac{1}{e}) = (e^{-1})^{e^{-1}} = e^{-1/e}$.As $u 	o 0^+$,$g(u) = u^u 	o 1$ (using the limit $\lim_{u 	o 0^+} u^u = 1$).At $u = 1$,$g(1) = 1^1 = 1$.Since $e^{-1/e} \approx 0.692$,the minimum value is $e^{-1/e}$ and the maximum value is $1$.Thus,the range is $[e^{-1/e}, 1]$.

The range of the function $f(x) = (\sin x)^{\sin x}$ defined on $(0, \pi)$ is

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