Let f(x)=(x+3)^2 (x-2)^3,x∈ [-4,4]. If M and m are the maximum and minimum values of f, respectively in [-4,4], then the value of M-m is :

Given: $f(x)={(x+3)}^{2}{(x-2)}^{3}$f(x)=(x+3)2(x−2)3

$$\Rightarrow {f}^{'}(x)={(x+3)}^{2}\times (3){(x-2)}^{2}+{(x-2)}^{3}(2)(x+3)$$⇒f′(x)=(x+3)2×(3)(x−2)2+(x−2)3(2)(x+3)

$$\Rightarrow {f}^{'}(x)=(x+3){(x-2)}^{2}[(x+3)\times (3)+(x-2)(2)]$$⇒f′(x)=(x+3)(x−2)2[(x+3)×(3)+(x−2)(2)]

$$\Rightarrow {f}^{'}(x)=(x+3){(x-2)}^{2}[3x+9+2x-4]$$⇒f′(x)=(x+3)(x−2)2[3x+9+2x−4]

$$\Rightarrow {f}^{'}(x)=(x+3){(x-2)}^{2}(5x+5)$$⇒f′(x)=(x+3)(x−2)2(5x+5)

$$\Rightarrow {f}^{'}(x)=5(x+3){(x-2)}^{2}(x+1)$$⇒f′(x)=5(x+3)(x−2)2(x+1)

For critical points, we put ${f}^{'}(x)=0$f′(x)=0.

$$\Rightarrow 5(x+3){(x-2)}^{2}(x+1)=0$$⇒5(x+3)(x−2)2(x+1)=0

$$\Rightarrow x=-3,2,-1$$⇒x=−3,2,−1

$$\Rightarrow f(-3)=f(2)=0$$⇒f(−3)=f(2)=0

$$\Rightarrow f(-1)={2}^{2}{(-3)}^{3}=-108$$⇒f(−1)=22(−3)3=−108

Also, $x\in [-4,4]$x∈[−4,4]

$$\Rightarrow f(-4)=-216$$⇒f(−4)=−216

$$\Rightarrow f(4)=49\times 8=392$$⇒f(4)=49×8=392

$$\Rightarrow M=392,m=-216$$⇒M=392,m=−216

$$\Rightarrow M-m=392+216$$⇒M−m=392+216

$$\Rightarrow M-m=608$$⇒M−m=608