# What are the absolute extrema of #f(x)=x / e^(x^2) in[1,oo]#?

##### 1 Answer

#### Answer:

There is no minimum

#### Explanation:

The derivative is given by

#f'(x) = (1(e^(x^2)) - x(2x)e^(x^2))/(e^(x^2))^2#

#f'(x) = (e^(x^2) - 2x^2e^(x^2))/(e^(x^2))^2#

Critical values will occur when the derivative equals

So if

#0 = e^(x^2) - 2x^2e^(x^2)#

#0 = e^(x^2)(1 - 2x^2)#

As mentioned above

#0 = 1 -2x^2#

#2x^2 = 1#

#x^2 = 1/2#

#x = +- sqrt(1/2) = +- 1/sqrt(2)#

But neither of these lie in our given domain. Therefore,

There will be no minimum

Hopefully this helps!