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:

$$x^2+x+1&gt;=0 \text { it is always true }$$

$$-1&lt;=\sqrt{x^2+x+1}&lt;=1$$
$$\sqrt{x^2+x+1}&gt;=-1 \text { and } \sqrt{x^2+x+1}&lt;=1$$
$$\sqrt{x^2+x+1}&lt;=1$$
$$x^2+x+1&lt;=1$$
$$x(x+1)&lt;=0$$
$$x\in[-1,0]$$

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.
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  $[\pi/3, \pi/2]$