Question

How can you find the inverse of a function?

Answer 1

Give me an example. I will teach you.

Answer 2

Do you want to know what a function inverse is, exactly?

Answer 3

Nope. I totally forgot how it work.

Answer 4

Okay, the functions have two parts. The input(domain) and the output(range). In a function, you input and get the output.

Answer 5

So you get the range from the domain.
In the function inverse - you go from the range back to the domain.

Answer 6

So take it like this:

\( \color{Black}{\Rightarrow f(1) = 5 \Longrightarrow f^{-1}(5) = 1 }\)

You see what I did here?

Answer 7

a. {(-4,2),)(1,2), (6,-3),(5,3),(2,1)
b. \[y = \sqrt{x+1}\]
c. \[{ (x,y) |y = 1-x /x-2}\]

Answer 8

huh.

Answer 9

What is the question here? lol

Answer 10

How did -1 got there?

Answer 11

a b and c . The paper asked what was the the inverse of the following function. which I don't have any idea.

Answer 12

\( \color{Black}{\Rightarrow f^{-1} \textbf{ means f inverse, not f to the negative one power.} }\)

Answer 13

Ohhh

Answer 14

So you just switch the places of the range?

Answer 15

But what is the function here?

Answer 16

In the inverse, the range of the function becomes the domain, and the domain of the function becomes the range.

Answer 17

I think I have to make it a function first. -.=

Answer 18

Can I give you an example here?

Answer 19

Fell free.

Answer 20

*feel lol

Answer 21

Okay, think that a function is \(f(x) = x + 4\), then its inverse would be found by finding x in terms of y.

Answer 22

We first say that \(f(x) = y\), then:

\( \color{Black}{\Rightarrow y = x + 4 }\)

\( \color{Black}{\Rightarrow x = y - 4 }\)

Just replace \(x\) with \(f^{-1}(x) \) and \(y\) with \(x\).

Answer 23

\( \color{Black}{\Rightarrow f^{-1}(x) = x - 4 }\)

Answer 24

It just became negative. But how can you rewrite those ^ functions up there to be able to know the inverse of it?

Answer 25

hmm? for example?

Answer 26

y = square root of x +1

Answer 27

\( \color{Black}{\Rightarrow f(x) = \sqrt{x} + 1 }\)
This?

Answer 28

Yep

Answer 29

So let's take \(f(x) = y\)

\( \color{Black}{\Rightarrow y = \sqrt{x} + 1 }\)

\( \color{Black}{\Rightarrow y - 1= \sqrt{x} }\)

\( \color{Black}{\Rightarrow x = (y - 1)^2 }\)

\( \color{Black}{\Rightarrow x = y^2 - 2y + 1}\)

Replace \(x\) with \(f^{-1}(x) \) and \(y\) with \(x\).

\( \color{Black}{\Rightarrow f^{-1}(x) = x^2 - 2x + 1 }\)

Answer 30

So you always have to isolate x?

Answer 31

Exactly :D

Answer 32

Now I get it! But what if you were given a set of ordered pairs? :?

Answer 33

You just flip them :P
If you're given (3,2) then you just replace the domain with the range and range with domain(remember what I told you right :D) and you get (2,3)

Answer 34

:DDDD A million thank! xD

Answer 35

:P :D