Maxima and MinimaHard
Question
A minimum value of 
dt is-
dt is-Options
A.1
B.2
C.0
D.3
Solution
f(x) = 
dt
f′(x) = xe-x2 = 0
x = 0
f″(x) = e-x2 (1 - 2x2)
f″ (0) = 1 > 0
Minimum value = f(0) = 0
dtf′(x) = xe-x2 = 0
x = 0
f″(x) = e-x2 (1 - 2x2)
f″ (0) = 1 > 0
Minimum value = f(0) = 0
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