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

1 Answer
Aug 15, 2017

Answer:

#(1, 1/e)# is an absolute maximum in the given domain
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 #0# or is undefined. The derivative will never be undefined (because #e^(x^2)# and #x# are continuous functions and #e^(x^2) != 0# for any value of #x#.

So if #f'(x) = 0#:

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

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

As mentioned above #e^(x^2)# will never equal #0#, so our only two critical numbers will occur at the solution of

#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, #x = 1# is going to be a maximum (because #f(x)# converges to #0# as #x->+oo)#.

There will be no minimum

Hopefully this helps!