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

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.