The range of the real valued function f(x) = | vedclass

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(C) Let $g(x) = x^2+x+1$. We can write this as $g(x) = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}$.Since $\left(x+\frac{1}{2}\right)^2 \ge 0$,the mini

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$\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$

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(C) Let $g(x) = x^2+x+1$. We can write this as $g(x) = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}$.Since $\left(x+\frac{1}{2}\right)^2 \ge 0$,the minimum value of $g(x)$ is $\frac{3}{4}$.Thus,the range of $\sqrt{g(x)}$ is $\left[\sqrt{\frac{3}{4}}, \infty\right) = \left[\frac{\sqrt{3}}{2}, \infty\right)$.However,the domain of $\operatorname{Sin}^{-1}(u)$ is $u \in [-1, 1]$.Therefore,we must have $\frac{\sqrt{3}}{2} \le \sqrt{x^2+x+1} \le 1$.Squaring the inequality,we get $\frac{3}{4} \le x^2+x+1 \le 1$.Since the function $f(u) = \operatorname{Sin}^{-1}(u)$ is an increasing function,the range of $f(x)$ is $\left[\operatorname{Sin}^{-1}\left(\frac{\sqrt{3}}{2}\right), \operatorname{Sin}^{-1}(1)\right]$.This simplifies to $\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$.

The range of the real valued function $f(x) = \operatorname{Sin}^{-1}\left(\sqrt{x^2+x+1}\right)$ is

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