Single Variable Calculus, Chapter 4, 4.3, Section 4.3, Problem 12

Suppose that $\displaystyle f(x) = \frac{x^2}{x^2 + 3}$

a.) Determine the intervals on which $f$ is increasing or decreasing.

If $\displaystyle f(x) = \frac{x^2}{x^2 + 3}$, then

By using Quotient Rule,


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{(x^2 + 3)(2x) - x^2 (2x)}{(x^2 + 3)^2}
\\
\\
f'(x) =& \frac{6x}{(x^2 + 3)^2}

\end{aligned}
\end{equation}
$


Again, by using Quotient Rule


$
\begin{equation}
\begin{aligned}

f''(x) =& \frac{(x^2 + 3)^2 (6) - 6x (2 (x + 3) (2x))}{((x^2 + 3)^2)}
\\
\\
f''(x) =& \frac{6 (x^2 + 3) [(x^2 + 3) - 4x^2]}{(x^2 + 3)^4}
\\
\\
f''(x) =& \frac{6x^2 + 18 - 24x^2}{(x^2 + 3)^3}
\\
\\
f''(x) =& \frac{-18x^2 + 18}{(x^2 + 3)^3}

\end{aligned}
\end{equation}
$


To find the critical numbers, we set $f'(x)$, so..


$
\begin{equation}
\begin{aligned}

0 =& \frac{6x}{(x^2 + 3)^2}
\\
\\
0 =& 6x

\end{aligned}
\end{equation}
$


The critical number is $x = 0$

Hence, we can divide the interval by

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
\hline\\
x < 0 & - & \text{decreasing on} (- \infty, 0) \\
\hline\\
x > 0 & + & \text{increasing on} (0, \infty)\\
\hline
\end{array}
$

These values are obtained by evaluating $f''(x)$ within the specified
interval. The concavity is upward when the sign of $f''(x)$ is positive. On the
other hand, the concavity is downward when the sign of $f''(x)$ is negative.

b.) Find the local maximum and minimum values of $f$.

We will use the Second Derivative Test to evaluate $f''(x)$ with the critical number:

So when $x = 0$,


$
\begin{equation}
\begin{aligned}

f''(0) =& \frac{-18(0)^2 + 18}{(0^2 + 3)^3}
\\
\\
f''(0) =& \frac{2}{3}

\end{aligned}
\end{equation}
$



Since $f'(0)$ and $f''(0) > 0, f(0) = 0$ is a local minimum.

c.) Find the intervals of concavity and the inflection points.

We set $f''(x) = 0$ to determine the point of inflection, so..


$
\begin{equation}
\begin{aligned}

f''(x) = 0 =& \frac{-18x^2 + 18}{(x^2 + 3)^3}
\\
\\
0 =& -18x^2 + 18
\\
\\
18x^2 =& 18

\end{aligned}
\end{equation}
$


The points of inflection are $x \pm 1$

Let's divide the interval to determine the concavity

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
\hline\\
x < -1 & - & \text{Downward} \\
\hline\\
-1 < x < 1 & + & \text{Upward} \\
\hline\\
x > 1 & - & \text{Downward}\\
\hline
\end{array}
$

These data obtained by substituting any values of $x$ to $f'(x)$ within the specified interval. Check its sign, if it's positive, it means that the curve is increasing on that interval. On the other hand, if the sign is negative, it means that the curve is decreasing on that interval.