## PROBLEM 2

Let the function $f: \mathbf{R} \rightarrow \mathbf{R}, f(x)=2 \sin 2x + 8(\sin x + \cos x) + 3$.

Determine the image of the function $f$.
## Solution:

Let $\sin x+\cos x=t \in\lfloor-\sqrt{2}, \sqrt{2}\rfloor$ and by squaring:

$1+\sin 2 x=t^{2}$, from which $\sin 2 x=t^{2}-1$.

It follows that $f(x)=g(t)=2\left(t^{2}-1\right)+8 t+3=2 t^{2}+8 t+1, t \in\lfloor-\sqrt{2}, \sqrt{2}\rfloor$

We have: $g(-\sqrt{2})=4-8 \sqrt{2}+1=5-8 \sqrt{2}$

$$
g(\sqrt{2})=4+8 \sqrt{2}+1=5+8 \sqrt{2}
$$

The function $h: \mathbf{R} \rightarrow \mathbf{R}, h(x)=2 x^{2}+8 x+1$ has a minimum $m=h\left(-\frac{8}{4}\right)=h(-2)=-7 \ldots 3$ points

but $-2 \notin\lfloor-\sqrt{2}, \sqrt{2}\rfloor$. Therefore, $\operatorname{Im} f=\lfloor 5-8 \sqrt{2}, 5+8 \sqrt{2}\rfloor$