Which one of the following is not bounded on | vedclass

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(B) To determine which function is not bounded,we analyze each option:$(A)$ For $f(x) = 2^{\frac{1}{x-1}}$ on $(0, 1)$:$\lim_{x 	o 0^+} f(x) = 2^{\fr

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$g(x) = x \cos \frac{1}{x}$ on $(-\infty, \infty)$

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(B) To determine which function is not bounded,we analyze each option:$(A)$ For $f(x) = 2^{\frac{1}{x-1}}$ on $(0, 1)$:$\lim_{x 	o 0^+} f(x) = 2^{\frac{1}{0-1}} = 2^{-1} = \frac{1}{2}$.$\lim_{x 	o 1^-} f(x) = 2^{-\infty} = 0$.Since the function is continuous and monotonic on $(0, 1)$,it is bounded in $(0, \frac{1}{2})$.$(B)$ For $g(x) = x \cos \frac{1}{x}$ on $(-\infty, \infty)$:As $x 	o \infty$,$g(x) = x \cos \frac{1}{x} \approx x(1) = x$,which approaches $\infty$.Thus,$g(x)$ is not bounded on $(-\infty, \infty)$.$(C)$ For $h(x) = x e^{-x}$ on $(0, \infty)$:$\lim_{x 	o 0^+} h(x) = 0$ and $\lim_{x 	o \infty} h(x) = 0$.The derivative $h'(x) = e^{-x} - x e^{-x} = e^{-x}(1-x)$ is zero at $x=1$.The maximum value is $h(1) = 1/e$. Thus,it is bounded.$(D)$ For $l(x) = 	an^{-1} 2^x$ on $(-\infty, \infty)$:As $x 	o -\infty$,$l(x) 	o 	an^{-1}(0) = 0$.As $x 	o \infty$,$l(x) 	o 	an^{-1}(\infty) = \frac{\pi}{2}$.Thus,it is bounded in $(0, \frac{\pi}{2})$.Therefore,the correct option is $(B)$.

Which one of the following is not bounded on the intervals as indicated?

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