How do you find the least value when an inverse trigonometric function is in the denominator? 
The function:
$f(x)=\frac{4\pi^2}{3\arccos{(x^4-2x^2)}}+\frac{5\pi}{3}$
If $B=\frac{m}{\pi}$
where $m$ is the minimum value which $f(x)$ can take. Find the value of $B$.
The choices given in my book are as follows:
$\begin{array}{ll}
1.&4\\
2.&3\\
3.&5\\
4.&2\\
\end{array}$

Gee, I'm confused on exactly what should I do here to solve this problem. Can someone help me here?. I think it has to be greater than $\frac{5\pi}{3}$ because that's a vertical shift for that function. But I don't know how to get the range here.

$x^4-2x^2=x^2(x^2-2) \ge -1$, and it takes value the value $-1$ when $x^2=1$.
As $x^2=1$, $\arccos(x^4-2x^2)$ has the largest value of $\pi$. Hence $$m=\frac{4\pi^2}{3\pi}+\frac{5\pi}{3}=\frac{9\pi}{3}=3\pi$$
Hence you are right, $$B=\frac{m}{\pi}=3.$$
