find the maximum value of f(x)=xe^(-0.5x)^2

The maximum value of the function could be calculated
using the first derivative test.

The roots of derivative
represents the critical values of the function, where the function has an extreme
value.

We'll determine the first derivative using product
rule:

f'(x) = x'*e^(-0.5x)^2 +
x*[e^(-0.5x)^2]'

e^(-0.5x)^2 =
e^(0.25x^2)

[e^(-0.5x)^2]' =
0.5x* e^(0.25x^2)

f'(x) = 1*e^(0.25x^2) +
x* 0.5x* e^(0.25x^2)

We'll factorize by
e^(0.25x^2):

f'(x) = e^(0.25x^2)(1 +
0.5x^2)

We'll put f'(x) =
0.

e^(0.25x^2)(1 + 0.5x^2) =
0

Since the factor e^(0.25x^2) cannot be zero for any real
value of x, we'll impose to the 2nd factor to be zero.

(1 +
0.5x^2) = 0

0.5x^2 = -1

x^2 =
-1/0.5

Again, neither real value of x, raised to square,
can give a negative value, such as
-1/0.5.

So, the function has not a maximum
value. The function is continuously and increasing all over real set
number.