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Algebra 2
Course: Algebra 2 > Unit 10
Lesson 1: Rational equationsFinding inverses of rational functions
CCSS.Math: , ,
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.
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Is the 1 not negative? The answer isn't...
g^-1(x) = (-1 - 3x)/(x - 2) ?(41 votes)
You are correct... At time stamp3:50, 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)"(25 votes)
At4:00in the video, it should be -1? Correct?(11 votes)
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.(11 votes)
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!(6 votes)
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."(16 votes)
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.(7 votes)
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.(3 votes)
- I think I am not grabbing the main concept. At1:47, why do you need to express the function in terms of y, or function for x, again?(4 votes)
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.(1 vote)
- 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?(3 votes)- 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?(6 votes)
- 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)?(3 votes)
- 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.(5 votes)
- 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?(3 votes)- You answer is equivalent. I would use the "report a problem" link with in the exercise to report the issue.(2 votes)
- At2:35, do you have to distribute the y, could you just divide the whole equation by x?(2 votes)
Keep in mind that we want to solve the equation for x.
To do that we have to "unlock" the parentheses and then gather all the x terms on one side of the equation and everything else on the other side.
That's why the y must be distributed.
Dividing by x would severely complicate things, as you would end up with
[ y ( x + 3 ) ] / x = 2 - 1/x and we would still not have only the x terms on one side of the equation.(3 votes)
I got f^-1(x) = 3x/2-x + 1/2-x
am I wrong? Why I got different result?(3 votes)- Your version should be: f^-1(x) = 3x/2-x - 1/2-x
You need a minus in front of the 2nd term.
Other than that, your version matches Sal's version. The differences are:
1) A factor of -1 applied to both numerators and denominators
2) You split the fraction into 2 terms rather than keeping your answer as one fraction.
These differences just make the result look different, but it is equivalent to Sal's.
Hope this helps.(1 vote)
3:50Video transcript
- [Instructor] All right,
let's say that we have the function f of x and it's
equal to two x plus five, over four minus three x. And what we wanna do is figure out what is the inverse of our function. Pause this video and try to figure that out before we work on that together. All right, now let's work on it together. Just as a reminder of what a function and an inverse even does, if this is the domain of a function and that's the set of all
values that you could input into the function for x
and get a valid output. And so let's say you have an x here, it's a member of the domain. And if I were to apply the function to it, or if I were to input
that x into that function. Then the function is
going to output a value in the range of the function
and we call that value f of x. Now an inverse, that goes the other way. If you were to input the f
of x value into the function that's going to take us back to x. So that's exactly what f inverse does. Now how do we actually
figure out the inverse of a function especially a function that's defined with a
rational expression like this. Well the way that I think about it is, let's say that y is equal
to our function of x or y is a function of x so we could say that y is equal to two x plus five, over four minus three x. For our inverse the relationship between x and y is going to be swapped. And so in our inverse it's going to be true that x is going to
be equal to two y plus five, over four minus three y. And then to be able to express this as a function of x, to say
that what is y as a function of x for our inverse we
now have to solve for y. So it's just a little bit of algebra here. So let's see if we can do that. So the first thing that I would do is multiply both sides of this
equation by four minus 3 y. If we do that, on the left
hand side we are going to get x times each of these terms. So we're going to get
four x minus three yx and then that's going to be equal to on the right hand side,
since we multiplied by the denominator here we're just going to be left with the numerator. It's going to be equal to two y plus five. And this could be a
little bit intimidating 'cause we're seeing xs and
ys, what are we trying to do, remember we're trying to solve for y. So let's gather all
the y terms on one side and all the non-y terms on the other side. So let's get rid of this two y here. Actually, well I could go either way. Let's get rid of this two y here, so let's subtract two y from both sides. And let's get rid of this four
x from the left hand side, so let's subtract four x from both sides. And then what're we going to be left with. On the left hand side
we're left with minus or negative five, or actually
it would be this way, it would be negative three yx minus two y. And you might say hey where is this going, but I'll show you in a second, is equal to, those cancel out and we're gonna have five minus four x. Now once again we are
trying to solve for y. So let's factor out a y here, and then we are going to have y, times negative three x minus two is equal to five minus four x. And now this is the homestretch. We can just divide both
sides of this equation by negative three x minus two and we're going to get y is equal to five minus four x, over
negative three x minus two. Now another way that
you could express this is you could multiply both the numerator and the denominator by negative one, that won't change the value. And then you would get, you would get in the numerator four x minus five, and in the denominator you
would get a three x plus two. So there you have it. Our f inverse as a function of x, which we could say is equal to this y is equal to this right over there.