The least integer in the range of f(x) = sqrt | vedclass

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(C) The domain of $f(x)$ is determined by the conditions: $(x + 4)(1 - x) \ge 0$ and $x > 0$.This implies $x \in [-4, 1]$ and $x > 0$,so the domain is

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$0$

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(C) The domain of $f(x)$ is determined by the conditions: $(x + 4)(1 - x) \ge 0$ and $x > 0$.This implies $x \in [-4, 1]$ and $x > 0$,so the domain is $x \in (0, 1]$.We have $f(x) = \sqrt{-x^2 - 3x + 4} + \log_{1/2} x$.Since $\sqrt{-x^2 - 3x + 4}$ is a decreasing function on $(0, 1]$ and $\log_{1/2} x$ is also a decreasing function on $(0, 1]$,the sum $f(x)$ is a strictly decreasing function.As $x 	o 0^+$,$f(x) 	o \infty$.At $x = 1$,$f(1) = \sqrt{(1 + 4)(1 - 1)} - \log_2 1 = 0 - 0 = 0$.Since the function is strictly decreasing on $(0, 1]$,the range of $f(x)$ is $[0, \infty)$.The least integer in the range $[0, \infty)$ is $0$.

The least integer in the range of $f(x) = \sqrt{(x + 4)(1 - x)} - \log_2 x$ is

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