Question

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

Answer 1

`(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!