Question

\[ f(x)=2\sin\sqrt{x^2+x+1} \]

\[ \text{The values of } f(x) \text{ lie in the interval} \]

Solution

Since

\[ -1\le\sin\theta\le1 \]

therefore,

\[ -2\le2\sin\theta\le2 \]

Now,

\[ x^2+x+1 = \left(x+\frac12\right)^2+\frac34 \]

\[ x^2+x+1>0 \]

So

\[ \sqrt{x^2+x+1} \]

is always real.

Hence,

\[ -2\le f(x)\le2 \]

Answer

\[ \boxed{[-2,\,2]} \]