Answer
$e$
Work Step by Step
Need to find the absolute minimum value of the function
$g(x) =\frac{e^{x}}{x} $ , $x>0$
Take first derivative of the function.
Apply quotient rule.
$g'(x) =\frac{xe^{x}-e^{x}}{x^{2}} $ , $x>0$
$x>1$ satisfies the given domain $x>0$
Thus, as per first derivative test the absolute minimum value is
$g(1) =\frac{e^{1}}{1} $ , $x>0$
Hence, $e$ is the absolute minimum value of the function.
