Question:
Given $x>0$, the value of $f(x)=-3 \cos \sqrt{3+x+x^{2}}$ lie in the interval ______________ .
Solution:
Given $x>0$,
$f(x)=3 \cos \sqrt{3+x+x^{2}}$
Since $-1 \leq \cos \theta \leq 1$ for any $\theta$
i. e $-1 \leq \cos \left(\sqrt{3+x+x^{2}}\right) \leq 1$
i. e $-3 \leq 3 \cos \left(\sqrt{3+x+x^{2}}\right) \leq 3$
i. e $3 \cos \left(\sqrt{3+x+x^{2}}\right) \in[-3,3]$
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