The maximum and minimum values of the function f : ℝ → ℝ defined by f x = 5 cos x + 3 cos…

Given, $f\left(x\right)=5\cos x+3\cos \left(x+\frac{\pi }{3}\right)+8$

$=5\cos x+3\left\{\cos x\cos \left(\frac{\pi }{3}\right)-\sin x\sin \left(\frac{\mathrm{\pi }}{3}\right)\right\}+8$

$=5\cos x+3\left\{\cos x\left(\frac{1}{2}\right)-\sin x\left(\frac{\sqrt{3}}{2}\right)\right\}+8$

$=5\cos x+\frac{3}{2}\cos x-\left(\frac{3\sqrt{3}}{2}\right)\sin x+8$

$=\frac{13}{2}\cos x-\left(\frac{3\sqrt{3}}{2}\right)\sin x+8$

Now, we know that

$A\sin x+B\cos x\in \left[-\sqrt{A^2+B^2},\sqrt{A^2+B^2}\right]$

Here, we have

$A=\frac{13}{2}$ and $B=\frac{3\sqrt{3}}{2}$

Therefore,

$-\sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2}+8\le f\left(x\right)\le \sqrt{{\left(\frac{13}{2}\right)}^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2}+8$

$\Rightarrow 1\le f\left(x\right)\le 15$

Minimum value of $f\left(x\right)=1$

Maximum value of $f\left(x\right)=15$