<s>[INST] Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$ Find the domain and range of $f(x)=\sin^{-1}(\sqrt{x^2+x+1})$
Finding the domain:
$$x^2+x+1>=0 \text { it is always true }$$
$$-1<=\sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}>=-1 \text { and } \sqrt{x^2+x+1}<=1$$
$$\sqrt{x^2+x+1}<=1$$
$$x^2+x+1<=1$$
$$x(x+1)<=0$$
$$x\in[-1,0]$$
Finding the range
$$f(x)=\sin^{-1}\left(\sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left([-1,0]+\frac{1}{2}\right)^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{-1}{2},\frac{1}{2}\right]^2+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[0,\frac{1}{4}\right]+\frac{3}
{4}}\right)$$
$$f(x)=\sin^{-1}\left(\sqrt{\left[\frac{3}{4},1\right]}\right)$$
$$f(x)=\sin^{-1}\left({\left[\frac{\sqrt{3}}{2},1\right]}\right)$$
$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$
but actual answer is $\left[\dfrac{\pi}{3},\dfrac{\pi}{2}\right]$
 [/INST] The domain of $$f(x)=\sin^{-1}(\sqrt{x^2+x+1})$$ is where you get $$0\le x^2+x+1\le 1$$ which is the interval $[-1,0]$
Note that $f(-1/2)=\pi/3$ is the minimum value and $f(-1)=f(0)= \pi/2$ is the maximum.
Thus the range is the interval  $[\pi/3, \pi/2]$
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