)=0 x=0, or (x2−4)(x2−1)=0 x=0,x=±2,±1 Now, f′(x)=
2x(x+1)(x−1)(x+2)(x−2)
(ex2+2)
f′(x) changes sign from positive to negative at x=−1,1 So, number of local maximum points = 2 f′(x) changes sign from negative to positive at x=−2,0,2 So, number of local minimum points = 3 ∴m=2,n=3