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### Course: Algebra (all content) > Unit 7

Lesson 20: Finding inverse functions (Algebra 2 level)- Finding inverse functions: linear
- Finding inverse functions: quadratic
- Finding inverse functions: quadratic (example 2)
- Finding inverse functions: radical
- Finding inverses of rational functions
- Finding inverse functions
- Finding inverses of linear functions

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# Finding inverse functions: quadratic

Sal finds the inverse of f(x)=(x+2)^2+1. Created by Sal Khan.

## Want to join the conversation?

- Why is the minimum negative two? If it were to be, let's say, negative 3, than it would add to be a negative, but once it's squared it will turn positive anyway. Shouldn't the domain be all real numbers?(5 votes)
- Madison,

As Sal said, the original function was defined to constrain x ≥ -2.

While he did not have to define the function in this manner, it was necessary to make it possible to find an inverse function.

The inverse of y=(x-2)+1 would look like this: http://www.khanacademy.org/cs/inverse-of-yx21/1789753003

But it would not be a function. because it has two y values for every one x value. A function can only have one y value for any x value.

By constraining the domain of the first function to x≥-2, then the inverse becomes a function because you only use the principal (positive) square root in the inverse function.

I hope that helps.(29 votes)

- what are the inverses of logarithmic functions ?

how do they look like, when represented on cartesian plane ?(4 votes)- The inverse of the logarithmic function is called the
**exponential**function. If you draw a graph of the exponential function, it looks like the logarithmic function but reflected in a line from bottom-left to top-right of the graph. Here's a graph of the logarithmic (red line) and exponential (blue line) functions together: http://www.regentsprep.org/Regents/math/algtrig/ATP8b/inversegraph.gif(13 votes)

- Can someone elaborate why at1:50Sal chose to take the positive square root.(3 votes)
- The values here will be positive. The thing squared is (x+2) and the minimum value possible for x is -2. Therefore, 0 (-2+2) and up is the set of possible values. Since the values are positive (well, technically nonnegative) the principal/positive square root is sufficient.(6 votes)

- At1:42, Sal said that if x was greater than or equal to -2 then the underlined expression would be positive, but zero isn't positive.(5 votes)
- Sal may have wanted to say non negative number.You should have got the point(0 votes)

- Is it really necessary in this instance to write y>=1 for the inverse function? If you try any values less than one, you would have the square root of a negative number and since the plane that the graph is on is limited to only real numbers, wouldn't you be able to leave the restriction off entirely?(2 votes)
- Nick,

This is true if you are limited to the "real" numbers, but many applications in mathematics allow "imaginary" or "complex" numbers. If you haven't learned about those yet, they allow the square root of a negative number… and they wind up becoming important when you look at the mathematics behind things that oscillate (like the pendulum on a grandfather clock, for example). Imaginary and complex numbers also are quite important in describing the way electricity moves through circuits, and many other problems.

Since such things as imaginary and complex numbers exist, you actually CAN get answers to the function at y's less than -1. They'll be answers that use imaginary or complex numbers, but they'll be answers nonetheless. However, those answers that you get won't be the inverse of the function we were given. So to make everything explicitly clear we include the restriction.

If all of that seems a little bit crazy and weird, you're right. Imaginary numbers come up usually in the first year of most student's algebra classes, towards the end, or in the second year of algebra. And most students usually have some trouble understanding the concept because it is a little strange. If you haven't encountered them yet, don't worry. You will eventually.(4 votes)

- Can someone explain why the value for x must be greater than or equal to -2? Why can't the output have a negative value?

And also, why does x, if greater than or equal to -2 means that y is greater than or equal to one? For I thought that y should be greater than or equal to one.(2 votes)- The constraint of x equaling or being greater than -2 is added because if you take the inverse of the original function, the inverse function wouldn't give you a real number for any value of x below -2. The product of any number squared is a positive number ( -2^2 = 2^2). Therefore, it is impossible (unless working with imaginary numbers) to get the square root of a negative number - because there is no situation where squaring a real number gets you a negative one! As for your other question - if you put in -2 into the original equation, you get 1. And if you put in any other x value greater than -2, the y value is greater than 1. So saying y must be greater than or equal to one is just saying the same thing in a different way. Hopefully I answered your questions completely!(5 votes)

- At2:30, why is the range of y always grater or equal to 1? If i solve the equation for -2 I get 0..(2 votes)
- The function is f(x)=(x+2)^2+1, so if you solve f(-2) you get:

f(-2) = [(-2)+2]^2+1

f(-2) = [0]^2+1

f(-2) = 1

You may be looking at when Sal writes the equation solved for the exponential expression as y-1=(x+2)^2 and forgetting to finish solving for y.

y-1 = (x+2)^2

y-1 = [(-2)+2]^2

y-1 = 0^2

y-1 = 0 :But we haven't solved for y yet!

y=1(4 votes)

- At1:17I still don't understand why Sal took the positive square root. Even though the domain of f is will make the expression (x+2) greater than or equal to 0, I don't see why Sal couldn't have taken the negative square root. It would have made logical sense right?(3 votes)
- Yes but how would you be able to plot the points on the graph with the negative square root(2 votes)

- How are you defining the restrictions? I found this unclear.(2 votes)
- If you look at the graph you will notice that what is plotted starts from the coordinates
`(-2, 1)`

. This is where we get the restraint of`x>=-2`

because when we enter any value above`-2`

into the function we will get a point on the curve.

The restraint couldn't be`x>=-3`

because there's no definition, according to the function that is graphed`for any value below`

-2`. Hence, it couldn't be`

-3`. The restraint is simply stating the minimum value that can be entered into the function for it to find a corresponding y value. This is called the domain.

Does that make sense?(3 votes)

- I have the whole entire video, but the problem i still have is that how do you do this inverse function with the Coordinate Geometry and if i'm in the wrong video then can someone point me to the right one please? I'm new to khanacademy. Thanks in advance.(3 votes)

## Video transcript

We've got the function f of x
is equal to x plus 2 squared plus 1, and we've constrained
our domain that x has to be greater than or equal
to negative 2. That's where we've
defined our function. And we want to
find its inverse. And I'll leave you to think
about why we had to constrain it to x being a greater than
or equal to negative 2. Wouldn't it have been possible
to find the inverse if we had just left it as the
full parabola? I'll leave you -- or maybe I'll
make a future video about that. But let's just figure
out the inverse here. So, like we've said in
the first video, in the introduction to inverses, we're
trying to find a mapping. Or, if we were to say that y --
if we were to say that y is equal to x plus 2
squared plus 1. This is the function you give
me an x and it maps to y. We want to go the other way. We want to take, I'll give you
a y and then map it to an x. So what we do is, we
essentially just solve for x in terms of y. So let's do that one
step at a time. So, the first thing to do, we
could subtract 1 from both sides of this equation. y minus 1 is equal to
x plus 2 squared. And now to solve here,
you might want to take the square root. And that actually will be
the correct thing to do. But it's very important to
think about whether you want to take the positive or
the negative square root at this step. So we've constrained our domain
to x is greater than or equal to negative 2. So this value right here, x
plus 2, if x is always greater than or equal to negative
2, x plus 2 will always be greater than or equal to 0. So this expression right here,
this right here is positive. This is positive. So we have a positive squared. So if we really want to get to
the x plus 2 in the appropriate domain, we want to take
the positive square root. And in the next video or the
video after that, we'll solve an example where you want to
take the negative square root. So we're going to take
theundefined positive square root, or just the principal
root, which is just the square root sign, of both sides. So you get the square root of y
minus 1 is equal to x plus 2. And one thing I should have
remembered to do is, from the beginning we had
a constraint on x. We had for x is greater than
or equal to negative 2. But what constraint
could we have on y? If you look at the graph right
here, x is greater than equal to negative 2. But what's why? What is the range of y-values
that we can get here? Well, if you just look at
the graph, y will always be greater than or equal to 1. And that just comes from the
fact that this term right here is always going to be
greater than or equal to 0. So the minimum value that the
function could take on is 1. So we could say for x is
greater than or equal to negative 2, and we could add
that y is always going to be greater than or equal to 1. y is always greater
than or equal to 1. The function is always
greater than or equal to that right there. To 1. And the reason why I want to
write it at the stage is because, you know, later
on, we're going to swap the the x's and y's. So let's just leave that there. So here we haven't explicitly
solved for x and y. But we can write for y is
greater than or equal to 1, this is going to be the domain
for our inverse, so to speak. And so here we can keep
it for y is greater than or equal to 1. This y constraint's
going to matter more. Because over here,
the domain is x. But for the inverse, the domain
is going to be the y-value. And then, let's see. We have the square root of y
minus 1 is equal to x plus 2. Now we can subtract
2 from both sides. We get the square root of y
minus 1 minus 2, is equal to x for y is greater
than or equal to 1. And so we've solved
for x in terms of y. Or, we could say, let me just
write it the other way. We could say, x is equal to,
I'm just swapping this. x is equal to the square root of y
minus one minus 2, for y is greater than or equal to one. So you see, now, the way
we've written it out. y is the input into the
function, which is going to be the inverse of that function. x the output. x is
now the range. So we could even rewrite
this as f inverse of y. That's what x is, is equal to
the square root of y minus 1 minus 2, for y is greater
than or equal to 1. And this is the
inverse function. We could say this
is our answer. But many times, people want
the answer in terms of x. And we know we could
put anything in here. If we put an a here, we
take f inverse of a. It'll become the square root
of a minus 1 minus 2, 4. Well, assuming a is greater
than or equal to 1. But we could put an x in here. So we can just rename
the the y for x. So we could just do
a renaming here. So we can just rename y for x. And then we would get -- let
me scroll down a little bit. We would f inverse of x. I'll highlight it here. Just to show you, we're
renaming y with x. You could rename it with
anything really, is equal to the square root of x minus 1. Of x minus 1. Minus 2 for, we have to
rename this to, for x being greater than or equal to 1. And so we now have our inverse
function as a function of x. And if we were to graph it,
let's try our best to graph it. Maybe the easiest thing to do
is to draw some points here. So the smallest value
x can take on is 1. If you put a 1 here,
you get a 0 here. So the point 1, negative 2,
is on our inverse graph. So 1, negative 2
is right there. And then if we go to 2,
let's see, 2 minus 1 is 1. The principle root
of that is 1. Minus 2. So it's negative 1, so
the point, 2, negative 1 is right there. And let's think about it. Let's see. If we did 5, I'm trying
take perfect squares. 5 minus 1 is 4, minus 2. So the point 5, 2 is,
let me make sure. 5 minus 1 is 4. Square root is 2. Minus 2 is 0. So the point 5, 0 is here. And so the inverse graph, it's
only defined for x greater than or equal to negative 1. So the inverse graph is going
to look something like this. It's going to look something
like, I started off well, and it got messy. So it's going to look
something like that. Just like that. And just like we saw, in the
first, the introduction to function inverses, these are
mirror images around the line y equals x. Let me graph y equals
x. y equals x. y equals x is that
line right there. Notice, they're mirror
images around that line. Over here, we map
the value 0 to 5. If x is 0, it gets mapped to 5. Here we go the other way. We're mapping 5 to -- we're
mapping 5 to the value 0. So that's why they're
mirror images. We've essentially
swapped the x and y.