Catch 22 situation involving inverting a function and finding the range of the function.

Let $f(x) = \sqrt{x+5} - \sqrt{x-5}$

Calculating the inverse:

$y = \sqrt{x+5} - \sqrt{x-5}$

$y + \sqrt{x-5} = \sqrt{x+5}$

$y^2 + x - 5 + 2y\sqrt{x-5} = x + 5$

$\frac{(10 - y^2)^2}{4y^2} + 5 = x$

Thus, $f^{-1}(x) = 5 + \frac{(10-x^2)^2}{4x^2}$

Now let's use the inverse function to calculate $f^{-1}(10)$. We get $f^{-1}(10) = 25.25$

Going back to the original function. $f(f^{-1}(10)) = f(25.25)$ should be equal to $10$. However, if we calculate, $f(25.25)$ comes out to be $1$.

I had originally wanted to find the range of $f(x)$. To do that, I wanted to take the inverse and calculate the inverse function's domain (which would be the original function's range).

Looking at the original function, I observed the following:

1) Taking the first derivative, it can be seen that the function is a monotonically decreasing function.

2) Domain of function is $[5, \infty)$

3) $f(5) = \sqrt{10}$

4) as $x \to \infty, f(x) \to 0$