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: [2024]
(4)
Given, f(x)=(x+3)2(x-2)3 ...(i)
Differentiate (i), w.r.t. x, we get
f'(x)=2(x+3)(x-2)3+3(x+3)2(x-2)2
=(x+3)(x-2)2[2(x-2)+(x+3)×3]
=(x+3)(x-2)2(5x+5)
For maxima / minima, f'(x)=0
⇒(x+3)(x-2)2(5x+5)=0⇒x=-3,-1,2
Now find the value of (i) at x=-4,-3,-1,2,4
f(-4)=(-4+3)2(-4-2)3=1×(-216)=-216
f(-3)=0
f(-1)=(-1+3)2(-1-2)3=4×(-27)=-108
f(2)=0
f(4)=(4+3)2(4-2)3=49×8=392
∴ Maximum value, M=392
Minimum value, m=-216
Hence, M-m=392+216=608