Question

Given x > 0, the value of $f\left(x\right)=–3\cos \sqrt{3+x+x^2}$ lie in the interval ____________.

Answer 1

Given x > 0,
$$\begin{gathered} f\left(x\right)=3\cos \sqrt{3+x+x^2} \\ \mathrm{Since}-1\le \cos \theta \le 1\mathrm{for}\mathrm{any}\theta \\ \mathrm{i}.\mathrm{e}-1\le \cos \left(\sqrt{3+x+x^2}\right)\le 1 \\ \mathrm{i}.\mathrm{e}-3\le 3\cos \left(\sqrt{3+x+x^2}\right)\le 3 \\ \mathrm{i}.\mathrm{e}3\cos \left(\sqrt{3+x+x^2}\right)\in \left[-3,3\right] \end{gathered}$$