Given: f(x)=(x+3)2(x−2)3
⇒f′(x)=(x+3)2×(3)(x−2)2+(x−2)3(2)(x+3)
⇒f′(x)=(x+3)(x−2)2[(x+3)×(3)+(x−2)(2)]
⇒f′(x)=(x+3)(x−2)2[3x+9+2x−4]
⇒f′(x)=(x+3)(x−2)2(5x+5)
⇒f′(x)=5(x+3)(x−2)2(x+1)
For critical points, we put f′(x)=0.
⇒5(x+3)(x−2)2(x+1)=0
⇒x=−3,2,−1
⇒f(−3)=f(2)=0
⇒f(−1)=22(−3)3=−108
Also, x∈[−4,4]
⇒f(−4)=−216
⇒f(4)=49×8=392
⇒M=392,m=−216
⇒M−m=392+216
⇒M−m=608