Here, f(x)=2x3−9x2+12x+5 ⇒f′(x)=6x2−18x+12=0 For maxima or minima put f′(x)=0 ⇒x2−3x+2=0 ⇒x=1 or x=2 Now, f"(x)=12x−18 ⇒f"(1)=12(1)−18=−6<0 Hence, f(x) has maxima at x=1 ∴ maximum value =M=f(1)=2−9+12+5=10. And, f"(2)=12(2)−18=6>0. Hence, f(x) has minima at x=2. ∴ minimum value =m=f(2) =2(8)−9(4)+12(2)+5=9 ∴M−m=10−9=1