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

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Answer 1 · 2015-11-12T04:01:18

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

$1^x = 1$ 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 $0$ .

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