If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Algebra (all content)

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

Lesson 20: Finding inverse functions (Algebra 2 level)

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? •  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.
• what are the inverses of logarithmic functions ?
how do they look like, when represented on cartesian plane ? • At the very beginning of the video Sal said that it would not be possible to find the inverse of the function if the function had not been constrained to "x is greater than or equal to -2". I was wondering why this is. What is this constraint necessary?

Thank you! Keep up the great work. This site is saving me in Honors Algebra 2 :) • Can someone elaborate why at Sal chose to take the positive square root. • At , Sal said that if x was greater than or equal to -2 then the underlined expression would be positive, but zero isn't positive. • 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? • 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.
• 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. • 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!
• At , why is the range of y always grater or equal to 1? If i solve the equation for -2 I get 0.. • At I 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? • How are you defining the restrictions? I found this unclear. • 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.