Question

Can someone dumb this down for me?
--> Two functions are inverse functions if, for the functions f and g, f(g(x)) = x for every value of x in the domain of g, and g(f(x)) = x for every value of x in the domain of f.

Answer 1

http://www.youtube.com/watch?v=bBBUMHe900U

Answer 2

That video is talking about sine, consine, and tangent.

Answer 3

A function maps an input value (x) to an output value (y).
The inverse function, maps the output of the function back to its input.

If f(x) and g(x) are inverse functions, then if you find f(p) to be a number, let's say q, then using the inverse function, g(q) will give you back p.

Answer 4

hmm

Answer 5

Thanks though. :P

Answer 6

Okay, so, a peice just told me to find the inverse of something, I could just swap the Y and X?

Answer 7

Here is an example.

f(x) = 2x
That means that for any value of x you use, f(x) will be twice x.
For example, f(1) = 2, f(3) = 6, etc.

The inverse function of f(x) = 2x is g(x) = (1/2)x.
Function g takes any input number, and gives you back half of it.
For example, g(6) = 3, g(4) = 2.

Now try to use an x value in f then use the result in g.
Let's use x = 5. If you input 5 into f(x) you get f(5) = 2*5 = 10
Now use 10 in g. g(10) = (1/2)10 = 5. As you see, you got back to the original value of 5 you had used in f.

For any value of x used in inverse functions f(x) and g(x), f(g(x)) = g(f(x)) = x.
In other words, doing f(g(x)) cancels out what you did to x giving you x back. The same is true of g(f(x)).

Answer 8

\(\bf f(x)=y = 2x+1\qquad \textit{so it's inverse will be}\\ \quad \\
x = 2y+1\implies x-1=2y\implies \cfrac{x-1}{2}=y=f^{-1}\\ \quad \\
\textit{if indeed that's so, then } f(\quad f^{-1}(x)\quad )=x\quad and\quad f^{-1}(\quad f(x)\quad )=x\)

so let's test that
\(\bf f(\quad f^{-1}(x)\quad ) =

\cancel{2}\left(\cfrac{x-1}{\cancel{2}}\right)+1\implies (x-1)+1\implies x\\ \quad \\
f^{-1}(\quad f(x)\quad ) = \cfrac{(2x+1)-1}{2}\implies

\cfrac{\cancel{2}x}{\cancel{2}}=x\)

Answer 9

whatever the domain of the original function is, the inverse function will show it up in the RANGE

whatever the range of the original function is, the inverse function will show it up in the DOMAIN

Answer 10

Okay, I think it's a little better now. I'm about to take a quiz. Hang on. :P

Answer 11

What is a one-to-one function?

Answer 12

one that yields for every "unique x value" and "unique y value"

so the relation is one-to-one, exclusive to each unique "x" and "y"

Answer 13

And would the inverse of f(x)=x^3-3 be f^-1(x)=the cubed root of x+3?

Answer 14

So, like this?

Answer 15

ok... simple test.... a one-to-one function has to pass a "vertical line test" AND the "horizontal line test"

Answer 16

I think so because it is a horizontal line. Right? Okay.

Answer 17

And would the inverse of f(x)=x^3-3 be f^-1(x)=the cubed root of x+3?

Answer 18

hmm looks slanted

Answer 19

yea

Answer 20

Actually, I don't think the f(x)=x^3-3 is a one-to-one function. It looks like a weird s.

Answer 21

hmmm.... think so too... you're right... I was observing only the obtained expression, which is

Answer 22

Well, the only reason I say that is because one of my choices is *Not a one-to-one function.

Answer 23

Okay, the next one is the inverse of f(x)=5x+6. I think it's f^-1(x)=x-6/5. Sound right?

Answer 24

well.... what did you get when you swap the variables?

Answer 25

A relation is a function if for each x value there is only one y value. It passes the vertical line test.
A function is one-to-one if for each y value there is only one x value. It passes the horizontal line test.
Therefore, a one-to-one function passes both the vertical and horizontal line tests.

Answer 26

That. What I put. I just put up a new question, though.

Answer 27

I did really bad on that quiz. Apparentl the one we thought was not a one-to-one function, was. :(