Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 66

Determine the intervals of increase or decrease, the intervals of concavity and the points of inflection of $\displaystyle f(x) = \frac{e^x}{x^2}$.


$
\begin{equation}
\begin{aligned}

\text{if } f(x) =& \frac{e^x}{x^2}, \text{then by using Quotient Rule..}
\\
\\
f'(x) =& \frac{x^2 (e^x) - e^x (2x)}{(x^2)^2} = \frac{xe^x (x - 2)}{x^4} = \frac{e^x (x - 2)}{x^3}

\end{aligned}
\end{equation}
$


Again, by using Quotient Rule as well as Product Rule..


$
\begin{equation}
\begin{aligned}

f''(x) =& \frac{x^3 [e^x (1) + e^x (x - 2)] - [e^x(x - 2)] (3x^2) }{(x^3)^2}
\\
\\
f''(x) =& \frac{x^2 [x^2 e^x - xe^x - 3xe^x + 6e^x]}{x^6}
\\
\\
f''(x) =& \frac{x^2 e^x - 4xe^x + 6ex}{x^4}
\\
\\
f''(x) =& \frac{e^x (x^2 - 4x + 6)}{x^4}

\end{aligned}
\end{equation}
$


Now, to determine the intervals of increase or decrease, we must get first the critical numbers by setting $f'(x) = 0$. So,

$\displaystyle f'(x) = \frac{e^x (x - 2)}{x^3}$

when $f'(x) = 0$

$\displaystyle 0 = \frac{e^x (x - 2)}{x^3}$

The real solution is..

$x = 2$

Hence, the interval of increase or decrease is..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
\hline\\
x < 2 & - & \text{decreasing on } (- \infty, 2 ) \\
\hline\\
x > 2 & + & \text{increasing on } (2, \infty)\\
\hline
\end{array}
$

Next to determine the inflection points, we set $f''(x) = 0$. So,

$\displaystyle 0 = \frac{e^x (x^2 - 4x + 6)}{x^4}$

It shows that we have no inflection point because we don't have real solution for the equation. Let's evaluate $f''(x)$ with interval..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
\hline\\
x < 0 & - & \text{Downward} \\
\hline\\
x > 0 & - & \text{Downward}\\
\hline
\end{array}
$

The function has downward concavity at $(- \infty,

0)$ and $(0, \infty)$.