Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 62

Determine the absolute minimum value of the function $\displaystyle f(x) = \frac{e^x}{x}, x > 0$

To determine the absolute minimum value, we set $f'(x) = 0$, so..


$
\begin{equation}
\begin{aligned}

\text{if } f(x) =& \frac{e^x}{x}, \text{ then by using Quotient Rule..}
\\
\\
f'(x) =& \frac{x(e^x) - e^x(1)}{x^2}
\\
\\
f'(x) =& \frac{e^x (x - 1)}{x^2}

\end{aligned}
\end{equation}
$


When $f'(x) = 0$, we have..


$
\begin{equation}
\begin{aligned}

0 =& \frac{e^x (x - 1)}{x^2}
\\
\\
e^x =& 0 \text{ and } x -1 = 0

\end{aligned}
\end{equation}
$


The real solution and the critical number is..

$x = 1$

If we divide the interval and evaluate $f'(x)$, we can determine if the critical number we got is an absolute maximum or absolute minimum..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f' \\
\hline\\
x < 1 & - & \text{decreasing on } (- \infty, 1) \\
\hline\\
x > 1 & + & \text{increasing on } (1, \infty)\\
\hline
\end{array}
$

Since $f'(x)$ changes from negative to positive at $x = 1, f(1) = e^1$ is an absolute minimum.