The difference between the greatest and least values of the function $f\left(x\right)=\cos x+\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3xis$

The given function is periodic, with period  $2\pi$.   So the  difference between the greatest and least values of the 
function is the difference between these values on the  interval  $[0,2\pi ]$ . 

We have 

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=-(\sin x+\sin 2x-\sin 3x)\\ =-4\sin x\sin (3x/2)\sin (x/2)\end{array}$

$\text{ Hence }x=0,2\pi /3,\pi \text{ and }2\pi \text{ are the critical points. Also, }$

$f\left(0\right)=1+1/2-1/3=7/6,f(2\pi /3)=-13/12,f\left(\pi \right)=-1/6andf\left(2\pi \right)=7/6.$

Hence the greatest value is 7/6 and the least value is  $-13/12$ . 

Thus the  difference is $\frac{7}{6}-(-\frac{13}{12})=\frac{27}{12}=\frac{9}{4}$