Finding the Inverse of a Function


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Finding the Inverse of a Function (page 3 of 7)

Sections: Definition / Inverting a graph, Is the inverse a function?, Finding inverses, Proving inverses

The customary method of finding the inverse is some variant of the method I'm going to use. Whatever method you use, make sure you do the exact same method in the exact same order every time, so you remember the steps when you get to the test.

Find the inverse of y = 3x – 2.

Here's how the process works:

Here's your original function:
Try to solve for "x =":
Try to solve for "x =":
Switch x and y; "y =" is the inverse.

Then the inverse is y = ^{(x + 2)} / ₃

By the way, if they ask you for the domain and range, you can get the domain from the equation and the range from the graph. Since the function is just a polynomial, x can be anything. Since the graph is a straight line with increasing slope, then y will be everything (eventually, if you go out far enough). That is, the domain is "all real numbers" and the range is "all real numbers". Since the x's and y's are flipped to get the inverse, then so are the domain and range. That is, the domain of the inverse is also "all real numbers", and the range of the inverse is also "all real numbers".

Find the inverse function of y = x² + 1, if it exists.

There will be times when they give you functions that don't have inverses.

From the graph, it's easy to see that this function can't possibly have an inverse, since it violates the Horizontal Line Test:

function without an inverse function

(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!) Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

So what does it look like when you try to find the inverse algebraically? Remember that the Vertical Line Test says that you 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:

Your original function:
Solve for "x =":

Well, I solved for "x =", but not a UNIQUE "x =". That is, I got that any given x corresponds to two different y's, one from the "plus" and one from the "minus". Any time you come up with a "±" sign, you can be pretty sure that the inverse isn't a function. So, while there is an inverse, this inverse is not a function.

Find the inverse function of y = x² + 1, x < 0.

The only difference between this function and the previous one is that the domain has been restricted to only the negative half of the x-axis. It looks like this:

y = x^2 + 1, x <= 0

With the domain restricted like this, the function does have an inverse that is also a function. Just about any time they give you a problem where they've taken the trouble to restrict the domain, you should take care with the algebra and draw a nice picture, because the inverse probably is a function, but it will probably take some extra effort to show this. In this case, since the domain is x < 0 and the range (from the graph) is 1 < y, then the inverse will have a domain of 1 < x and a range of y < 0. Here's how the algebra looks:

The original function:
Solve for "x =":


By figuring out the domain and range of the inverse, I know that I should choose the negative sign for the square root:
Switch the x and y; the "y =" is the inverse: