Question

What are the local extrema of `f(x)= e^xln1^x`?

Answer 1

I assume that either there is an error or this is a 'trick' question.

Explanation:

`1^x = 1` for all `x`, so `ln1^1 = ln1 = 0`

Therefore, `f(x) =e^xln1^x = e^x*0 = 0` for all `x`.

`f` is a constant. The minimum and maximum of `f` are both `0`.