Đáp án:
$\begin{array}{l}
f\left( x \right) = 4{\cos ^2}x + \sqrt 2 \sin \left( {2x - \frac{\pi }{4}} \right)\\
f\left( x \right) = 2\left( {2{{\cos }^2}x} \right) + \sin 2x - \cos 2x\\
f\left( x \right) = 2.\left( {1 + \cos 2x} \right) + \sin 2x - \cos 2x\\
f\left( x \right) = \sin 2x + \cos 2x + 2\\
f\left( x \right) - 2 = \sin 2x + \cos 2x\\
f\left( x \right) - 2 = \sqrt 2 \left( {\sin 2x + \frac{\pi }{4}} \right)\\
Do: - \sqrt 2 \le \sqrt 2 \left( {\sin 2x + \frac{\pi }{4}} \right) \le \sqrt 2 \forall x\\
\Rightarrow - \sqrt 2 \le f\left( x \right) - 2 \le \sqrt 2 \\
\Rightarrow 2 - \sqrt 2 \le f\left( x \right) \le 2 + \sqrt 2 \\
Vậy\,\left\{ \begin{array}{l}
GTNN:f\left( x \right) = 2 - \sqrt 2 \\
GTLN:f\left( x \right) = 2 + \sqrt 2
\end{array} \right.
\end{array}$
