Verify by calculation that a root lies between 2 and 3 (which equation do I put this in to, and what am I looking for?) I am given
$$ y = 4cos(\frac{1}{2}x)$$
$$y = \frac{1}{4-x}$$
$$0 <= x < 4$$
When x = a, the tangents to the curves are perpendicular (see image for reference):

I have successfully done the first step and obtained:
$$ a = 4 - \sqrt{2sin(\frac{1}{2}x)}$$
I also have the derivative functions of each of the above functions.
Now I am to verify by calculation that a lies between 2 and 3.
I am not sure which equation to put my 2 and 3 in to find a sign change (that's what we have to do, right?). Also, I tried putting it in each of the equations (both of the curve equations and the derivative equations), but I couldn't find a sign-change anywhere. Where am I going wrong, and why?
EDIT: Let me add the answer key, since this question seems to be a tad confusing (and I'm confused too, so can't clarify):

 A: So if I understand the question correctly, I assume that the solution of:
$x = 4-\sqrt{2sin(\frac{1}{2}x)}$
is your a value.
So you want
$4-\sqrt{2sin(\frac{1}{2}x)} -x = 0$
Look at the function
$f(x) = 4-\sqrt{2sin(\frac{1}{2}x)} -x$
It is a continuous function from $2 \le x \le 3$
$f(2) = 0.7027$
$f(3) = -0.4124$
By the intermediate value theorem $f(x)$ takes on every value in the interval $[-0.4124,0.7027]$ at some x in $[2,3]$. So it hits 0 at some x in this interval.
A: $$f(x)=4 \cos \left(\frac{x}{2}\right);\;g(x)=\frac{1}{4-x}$$
Derivatives are
$$f'(x)=-2 \sin \left(\frac{x}{2}\right);\;g'(x)=\frac{1}{(4-x)^2}$$
Tangents are perpendicular at $x\in[0,4]$ means that the product of their slope is $-1$.
$$f'(x)g'(x)=-1\to 2 \sin \left(\frac{x}{2}\right)-(4-x)^2=0$$
Let $h(x)=2 \sin \left(\frac{x}{2}\right)-(4-x)^2$
We have $h(2)\approx -2.3;\;h(3)\approx 0.99$ therefore in the interval $(2,3)$ there is at least one root.
This root is about $x=2.61$.
There is another root, actually, at $x=5.07$, out of the given interval.
The points where the tangent are perpendicular are actually two

