The minimum and maximum values of $\cos x+3\sqrt{2}\sin (x+\frac{\pi }{4})+6$ are

Let $f\left(x\right)=\cos x+3\sqrt{2}(\sin x.\frac{1}{\sqrt{2}}+\cos x.\frac{1}{\sqrt{2}})+6$

$=\cos x+3\sin x+3\cos x+6$

$=4\cos x+3\sin x+6$

Minimum value =$C-\sqrt{a^2+b^2}=6-\sqrt{4^2+3^2}=6-5=1$

Maximum value $=C+\sqrt{a^2+b^2}=11$