solution:
E:IR
IR E'(x) = - xE(x) ,E(0)
a)
E'(x)
xE(x) = 0
x=0 (
E(x)
0
E(0) = 1 > .sp if E()=0
, which is contraduction thus E(x)>0
x
e IR.
b)
E',(x)= - xE(x)
Since E(x).o
x
so, if x <0
E'(x)>0
E is strically increasing
c)
since E is monotonically decreasing for E is
E (0) = s
E(x)
1
x>0
As E is monotonically increasing for x<0,
E(x) is bounded above
d)
As E decrease for x>0
so asx
E(x)
0
i. e
please give me thub up
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