## Calculus 8th Edition

$e$
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.