Answer
$(-4,-2) \cup (2,4)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a Rational function.
$\displaystyle \frac{(x^2-16)}{(x^4-16)} \lt 0$
$f(x)=\displaystyle \frac{(x-4)(x+4)}{(x-2)(x+2)(x^2+4)} \lt 0$
2.The cut points are:
$\displaystyle \displaystyle \frac{(x-4)(x+4)}{(x-2)(x+2)(x^2+4)} = 0$
$x=-4$ or $x=-2$ or $x=2$ or $x=4$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \lt 0 ? \\
& & \displaystyle \displaystyle \frac{(a-4)(a+4)}{(a-2)(a+2)(a^2+4)} & \\
(-\infty,-4) & -5 & \frac{(-)(-)}{(-)(-)(+)}=(+) & F\\
(-4, -2) & -3 & \frac{(-)(+)}{(-)(-)(+)}=(-) & T\\
(-2, 2) & 0 & \frac{(-)(+)}{(-)(+)(+)}=(+) & F\\
(2, 4) & 3 & \frac{(-)(+)}{(+)(+)(+)}=(-) & T\\
(4,\infty) & 10 & \frac{(+)(+)}{(+)(+)(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-4,-2) \cup (2,4)$
