Finding the Inverse of a Function

Here's how the inverse-finding process works: I start with my original function:

Now I'll try to solve for "x =":

Once I have "x=", I'll switch x and y; the new "y=" is the inverse.

Then the inverse is:

Since the variable is in the denominator, this is a rational function. Also, because I can't have division by zero, the domain of this function is x ≠ 5. A fraction is zero when its numerator is zero, which this numerator never will be, so the range at least contains the restriction that y ≠ 0. From what I know about rational functions and vertical asymptotes (of which, this function has one), I know that the graph will go forever upward and forever downward, so the range is indeed everything other than y = 0. I'll use this to find the domain and range of my inverse.

Here's the algebra for finding that inverse, starting with the original function:

I multiply the denominator up to the left-hand side of the equation:

I take the y through the parentheses:

I get the x-stuff by itself on one side of the "equals" sign:

Then I solve for x=:

And then switch the x's and y's:

So I've found the inverse, and it's another rational function; in particular, the inverse is indeed a function.

The domain and range of this inverse function are the reverses of the original function's domain and range. So my answer is:

There will be times when they give you functions that don't have inverses, and, from the graph of the given function, it's easy to see that this function can't possibly have an inverse, since it violates the Horizontal Line Test:

It is usually considered acceptable to draw the above graph, draw a horizontal line across it that crosses the graph twice, and then say something like "The inverse of this function is not itself a function, because of the Horizontal Line Test". But some teachers want to see the algebra anyway. Be sure to check with your teacher and verify what will be an acceptable answer -- and do this before the test!

What will this look like when I try to find the inverse algebraically? The Vertical Line Test says that I can't have two y's that share an x-value. That is, each x has to have a UNIQUE corresponding y value. But look at what happens when I try to solve for "x=":

My original function:

First, I'll swap out the f (x) for the simpler y:

Now I try to solve for "x=":

Well, I found the inverse, but is it a function? No; because, for any input x in the inverse, but I won't get a UNIQUE output y. Instead, I've shown that any given x-value will actually correspond to two different y-values, one from the "plus" on the square root and the other from the "minus". Because of this, my answer is: