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


$$f(x)\in \left[2m\pi+\dfrac{\pi}{3},2m\pi+\dfrac{2\pi}{3}\right] \text { where m is integer }$$

This isn't true because the range of the inverse sine function is from $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Let $$\theta=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$$
then
$$\sin{\theta}=\frac{\sqrt{3}}{2}$$
with the restriction that $\theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. The possible triangles with this ratio of $y$ to hypotenuse (i.e. $\sin$) are shown below

where only $\pi/ 3$ falls within the restricted range for the inverse sine function. Similarly, letting
$$\theta=\sin^{-1}\left(1\right)$$
means
$$\sin{\theta}=1$$
where only $ \pi/ 2$ falls within the restricted range for the inverse sine function.
