Let M and m be respectively the absolute maximum and the absolute minimum value of the function,

$f\left(x\right)=2x^3-9x^2+12x+5$ in the interval $[0,3]$. Then M-m is equal to

$f\left(x\right)=2x^3-9x^2+12x+5,x\in [0,3]$

$\Rightarrow f^{\mathrm{\prime }}\left(x\right)=6x^2-18x+12=6\left(x^2-3x+2\right)$

$=6(x-1)(x-2)$

For $f^{\mathrm{\prime }}\left(x\right)=0,(x-1)(x-2)=0\Rightarrow x=1$ or x=2

Now, $f\left(1\right)=10,f\left(2\right)=9,f\left(0\right)=5,f\left(3\right)=14$

$\therefore M=14$ and $m=5$. So, $M-m=14-5=9$