Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 16

Using the first and second derivative tests. Find the local maximum and minimum values of $\displaystyle f(x) = \frac{x}{x^2 + 4}$. State which method do you prefer.

$
\begin{equation}
\begin{aligned}
\text{if } f(x) &= \frac{x}{x^2 + 4} \quad \text{, then by using Quotient Rule}\\
\\
f'(x) &= \frac{(x^2+4) (1) - x (2x) }{(x^2+4)^2} = \frac{-x^2+4}{(x^2+4)^2}\\
\\
f''(x) &= \frac{(x^2+4)^2 ( -2x) - (-x^2+4)\left(2(x^2+4)(2x) \right)}{\left[ (x^2+4)^2 \right]^2}\\
\\
f''(x) &= \frac{2x^3-24x}{(x^2+4)^3}
\end{aligned}
\end{equation}
$


By using first derivative test, we set $f'(x) = 0$, then.

$
\begin{equation}
\begin{aligned}
f'(x) = 0 &= -x^2 +4\\
x^2 &= 4\\
\text{Solving for critical numbers, }\\
x &= \pm 2
\end{aligned}
\end{equation}
$

If we divide the interval by:

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

Since the function changes from positive to negative at $x=2$. It means that the local maximum is at $x = 2$. On the other hand, since the function changes from negative to positive at $x = -2$. It means that the local minimum is at $x = -2$

By using Second Derivative Test, we evaluate $f''(x)$ at these critical numbers:

$
\begin{equation}
\begin{aligned}
\text{so when } y &= 2, \\
f''(2) &= \frac{2(2)^3 - 24(2)}{(2^2+4)^3} = \frac{-1}{16} < 0\\
\\
\text{when } y &= -2, \\
f''(2) &= \frac{2(-2)^3 - 24(-2)}{((-2)^2+4)^3} = \frac{-1}{16} < 0
\end{aligned}
\end{equation}
$


We see that $f(2) < 0$, therefore, 2 is a local maximum. Also, $f(-2) > 0$, hence, -2 is a local minimum.
It's more easy to use the Second Derivative Test compare to the first one.