How do I find the length attained by a bullet when fired by a rifle? The problem is as follows:

The horizontal range of a bullet fired by a rifle from a certain height above sea level is given by the minimum value of the function presented below:
$$f(x)=2\sec\left(\pi x-\frac{\pi}{4}\right)+1$$
in kilometers when, $-\frac{1}{12}\leq x \leq \frac{5}{12}$
With the given information, find the length attained by the bullet.

The alternatives given in my workbook are as follows:
$\begin{array}{ll}
1.&\textrm{4 km}\\
2.&\textrm{1 km}\\
3.&\textrm{3 km}\\
4.&\textrm{2 km}\\
\end{array}$
What I attempted to do in order to solve this problem was to use the given domain to reconstruct the function and hence having the range, and with that minimum value I can get what it is being asked. In other words the horizontal range of that bullet.
Since the domain is in this:
$-\frac{1}{12}\leq x \leq \frac{5}{12}$
Then:
$-\frac{\pi}{12}\leq \pi x \leq \frac{5\pi}{12}$
$-\frac{\pi}{12}\leq \pi x \leq \frac{5\pi}{12}$
$-\frac{\pi}{12}-\frac{\pi}{4}\leq \pi x - \frac{\pi}{4} \leq \frac{5\pi}{12}-\frac{\pi}{4}$
$-\frac{4\pi}{12}\leq \pi x - \frac{\pi}{4} \leq \frac{2\pi}{12}$
$-\frac{\pi}{3}\leq \pi x - \frac{\pi}{4} \leq \frac{\pi}{6}$
Now here's where it comes the source of controversy.
I'm assuming that I can "take the secant function" to that interval in order to get the range.
Which would yield.
$\sec\left(-\frac{\pi}{3}\right)\leq \sec \left(\pi x - \frac{\pi}{4}\right) \leq \sec \left(\frac{\pi}{6}\right)$
Since the negative sign is absorved by the secant function this will generate:
$2 \leq \sec \left(\pi x - \frac{\pi}{4}\right) \leq \frac{2}{\sqrt{3}}$
$4 \leq 2\sec \left(\pi x - \frac{\pi}{4}\right) \leq 2\frac{2}{\sqrt{3}}$
$4 +1  \leq 2\sec \left(\pi x - \frac{\pi}{4}\right)+1 \leq \frac{2 \cdot 2}{\sqrt{3}}+1$
The finally:
$5  \leq 2\sec \left(\pi x - \frac{\pi}{4}\right)+1 \leq \frac{4}{\sqrt{3}}+1$
But rewritting this doing the rationalization it would be:
$5 \leq f(x) \geq \frac{4\sqrt{3}+3}{3}$
Now:
$\frac{4\sqrt{3}+3}{3} \approx 3.3094$
But looking at the orientation of the signs it doesn't make sense. How it can possibly be that this result is greater than $4$? What went wrong here?.
Can someone help me here?. Or is it just that I did not got the right picture?.
I can spot that $4\,km$ appears in one of the choices But I have no idea if whether I got to the right answer or some critical concept I missunderstood here. Thus I require help to settle down this contradiction. Help please?. I'd also like to know that since I got this problem in my precalculus workbook I'd like that the answer would follow the similar approach given here and not use derivatives.
 A: The function you are given determines the following domain for the secant function:
$$
\left[-\frac1{12}\pi-\frac\pi4,\frac5{12}\pi-\frac\pi4\right]=\left[-\frac{\pi}{3},\frac\pi{6}\right]
$$
On this interval the function $\sec (t)$ is positive and takes on the minimal value $1$ at $t=0$. To obtain the answer simply substitute the value into the equation.
A: Hint: a simpler way for solving this problem is taking the derivative of the function and after taking the value of $x$ such that $y$ of the line is mininum.
A: Combining some of the comments ...
Here is one conceptual error:

I'm assuming that I can "take the secant function" to that interval in order to get the range.

It is true that for any function $f,$ if $a=b=c$ then $f(a)=f(b)=f(c).$
But $\leq$ does not work the same as $=$.
From $a \leq b \leq c$ you get $f(a) \leq f(b) \leq f(c)$
if $f$ is a monotonically increasing function,
but not for functions $f$ in general.
In particular, the function $\sec(x)$ is sometimes decreasing and sometimes increasing.
The second conceptual error is unclear, but somehow you lost track of what the question asked. The question simply requires you to find the minimum value of
$$ f(x)=2\sec\left(\pi x-\frac{\pi}{4}\right)+1 $$
where $-\frac1{12} \leq x \leq \frac5{12}.$
You made some progress: you found that the input of the secant function in this formula,
$x - \frac\pi4$, is between $-\frac\pi3$ and $\frac\pi6.$
Now you need to find the minimum output value the secant function can have if its input is between $-\frac\pi3$ and $\frac\pi6.$
If you sketch the secant function over this interval, you should find that it is decreasing at the start and increasing at the end; the final value is less than the initial value, but the function's minimum value occurs at an input somewhere in between the two ends of the interval.
Once you identify that minimum value, multiply by $2$ and add $1.$
By the way, this question really has nothing to do with the range of a bullet or a rifle;
you would never use a function like this to compute such a thing.
