We have, f(x)=x2e−x2 f′(x)=2xe−x2−2x3e−x2 f′(x)=2xe−x2(1−x2) f′(x)=0,x=0,−1,1 f′(−h)<0,f′(h)>0, So minimum at x=0 f′(1−h)>0,f′(1+h)<0, So maximum at x=1 f′(−1−h)>0,f′(−1+h)<0, So maximum at x=−1 Minimum f(x)=f(0)=0 Maximum f(x)=e−1=
1
e
Since, f(x)≥0‌∀x So, difference between maximum and minimum values =