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

Combining all $4$ points, we get the range of the function to be $(0, \sqrt{10}]$ and so, the domain of the inverse function is $(0, \sqrt{10}]$. That means that when we put $x = 10$ in $f^{-1}(x), we were outside the domain of the function.

This seems to be a catch 22 situation. I wanted to take the inverse of the function $f(x)$ to calculate the range, however to be able to use the inverse function properly, I needed to know the range of the function $f(x)$ in the first place. How does one solve for the range of the function in such cases?

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    $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – Jessica B Nov 16 '14 at 9:15
  • $\begingroup$ The domain of $f$ is $[5,\infty)$. If you can prove that $f$ is decreasing, then you know the range is $(0,\sqrt{10}]$. $\endgroup$ – Gerry Myerson Nov 16 '14 at 9:34
  • $\begingroup$ @GerryMyerson, this is a special case where the function is decreasing monotonically. Thus, I could find out the range using that fact. In the more general case, this might not be the case. I was interested in getting the range from inverting the function. What's of interest in this example is that the inverse function does not reveal anything about the range of $f(x)$ $\endgroup$ – Chaos Nov 16 '14 at 9:36
  • $\begingroup$ @JessicaB, thanks for the links. :) Peter Brown, thanks for the edits. :) $\endgroup$ – Chaos Nov 16 '14 at 9:42
  • $\begingroup$ I found a similar question asking about the general method of finding the range here However, the general method proposed by the OP doesn't seem to be effective for my example. I am not able to comment on his answer due to a lack of rep points. $\endgroup$ – Chaos Nov 16 '14 at 10:16

First of all, the domain is $[5, \infty)$ since we want to restrict ourselves to real numbered outputs and since $\sqrt{x-5}$ only gives real outputs when $x > 5$. You are right that $$f'(x) = \frac{1}{2 \sqrt{x+5}} - \frac{1}{2 \sqrt{x-5}}$$ is always negative for $x > 5$ so the function is decreasing and since $f(0) = \sqrt{10}$, we know $f(x) < \sqrt{10}$ for all $x > 5$. Consequently it doesn't make sense to plug in $10$ to our inverse function. $10$ isn't in the range of $f$ so we shouldn't expect a reasonable answer. With that said, I think you did a little of the plugging in incorrectly. The symbolic manipulation looks right to me. With that said, finding an inverse isn't the most efficient to find the range. In fact, for many polynomials of degree $5$ and higher no neat formula exists for finding inverses. The following techniques are pretty common.

  1. Intermediate value theorem: if a function is continuous, we only need to compute a few very small or very large values to know that it hits everything in between.
  2. Mean value theorem: we need to only check where the derivative of the function is $0$ and at the end points. In our example, we checked that the derivative was never $0$ so the only possible maxima and minima are at the end points $5$ and $\infty$.

Honorable mention goes to the extereme value theorem which tells us that for a domain of the form $[a, b]$, both maxima and minima exist. Without calculus, the mean value theorem is off the table. However, if the class you are taking is not proof based, it is frequently easier to just look at the graph and guess the range. By the way, in your case the range is $(0, \sqrt{10}]$.


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