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## College Algebra

### Course: College Algebra>Unit 8

Lesson 1: Rational equations

# Finding inverses of rational functions

The inverse of a function ƒ is a function that maps every output in ƒ's range to its corresponding input in ƒ's domain. We can find an expression for the inverse of ƒ by solving the equation 𝘹=ƒ(𝘺) for the variable 𝘺. See how it's done with a rational function.

## Want to join the conversation?

• Is the 1 not negative? The answer isn't...

g^-1(x) = (-1 - 3x)/(x - 2) ?
• You are correct... At time stamp , there is a box that pops up in the lower part of the video that says "Sal made a mistake. This is suppose to be g^-1(x) = (-1 - 3x)/(x - 2)"
• At in the video, it should be -1? Correct?
• If you look without full screen you'll see that it says that he made a mistake. Whenever you think he made a mistake, disable full screen and see if he did.
• I am just curious about how come the inverse of something like 1/2 is 2/1 verses when we deal with rational functions why do we end up doing what is done in the video? I understand that it is a rule we follow in math, therefore we do it, I was only wondering if anyone knows why we do what we do in this scenario. Also sorry if that sounds convoluted. Thanks for any ideas/answers!
• The word "inverse" has a different meaning for every operation. In the case of 1/2 and 2, we call it the "multiplicative inverse," since multiplying 1/2 and 2 gives the multiplicative identity, 1. Multiplicative inverses are also called "reciprocals".

We also have inverses for addition, called "additive inverses." These are number pairs that add up to 0, the additive identity. You know these as negatives, like 3 and -3. 3 and -3 are additive inverses, since -3+3=0.

And we also have inverses for the operation of function composition. These are function pairs where, if we compose them, the result is the identity function y=x. So, for example, if we compose y=3x+1 and y=(x-1)/3, no matter which one is inside or outside, we get y=x. We call these "functional inverses" or "inverse functions."
• I think I am not grabbing the main concept. At , why do you need to express the function in terms of y, or function for x, again?
• If the function contains the point (a,b) the inverse of that function would then contain (b,a).

So if you swap the (Example) X and (Example) Y in the equation, we get the inverse relationship.
• Why the domains weren't covered in this video? The domain of g(x) is x != -3, and on the other hand the domain of the inverse function is y != 2. Does it mean that g(x) and g^-1(y) are inverse function of each other only when x != -3 and y != 2? Could someone explain this? Thanks.
• hi, sal have explained domains and range earlier, he focused on just getting the equation in this video, however it should be there of course, for g(x) it is x>-3 and for g^-1(x) it is x>2, in case of y it is y>2.
• When rearranging to solve for x, the answer I got was x = 3y+1 / 2-y

I worked out that this can be multiplied by -1/-1 so that the answer is written the same way as in the video. However, I can't make intuitive sense of why this little trick even works. It seems to be values must have changed somehow.

Does anyone have an intuitive explanation of this?
• I think that x = (3y+1) / (2-y) IS effectively identical to Sal's version of the correct answer, which is x = (-3y-1) / (-2+y) (ignoring his mistake of not having the 1 be negative). Multiplying both the numerator and the denominator by -1 makes no difference as the functions still output the same answer. Someone correct me on this if I am mistaken, but they are both correct, right?
• Hi all. when solving, i came up with an answer of (3X+1)/(2-X). can someone explain to me how this is equal to (-3X-1)/(X-2)?
• The best way to write the fraction would be: `- (3x+1)/(x-2)`
Your version has the minus distributed across the denominator.
The 2nd version has the minus distributed across the numerator.
Other than that, they are the same fraction.
• I was solving the exercise on this topic. I was asked to find an inverse of f(x)=7x+3/x-5. My answer was 5x+3/x-7, and it was marked wrong, with the correct answer being given as -5x-3/7-x.

Isn't that an equivalent solution? I simply factored a (-1) from the denominator, or is this something that cannot be done?
• You answer is equivalent. I would use the "report a problem" link with in the exercise to report the issue.
• i believe Sal made a mistake at
am i correct or did miss something?
• There is no error. Sal basically factored out a -1 from the denominator and then distributed it across the numerator.