Question

\[ \text{Given } x>0, \]

\[ f(x)=-3\cos\sqrt{3+x+x^2} \]

\[ \text{lies in the interval} \]

Solution

We know that

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

Multiplying by \(-3\),

\[ -3\le-3\cos\theta\le3 \]

Now,

\[ 3+x+x^2>0 \]

for all \(x>0\).

Hence

\[ \sqrt{3+x+x^2} \]

is always real.

Therefore,

\[ -3\le f(x)\le3 \]

Answer

\[ \boxed{[-3,\,3]} \]