Question

Find the inverse of each function. Is the inverse a function?
1) f(x)=(x+2)^2-4
2) f(x)=4x^3-1
3) f(x)=square root of x+4

Answer 1

Okay so for the first one, you want to set:

y=(x+2)^2-4

then change the y to x, and change the x-terms to y.

x=(y+2)^2-4

Then solve for y.

You'll have y = something.

Then just replace y with f(x).

Answer 2

sqrt(x+4) = y+2

y=sqrt(x+4)-2

f(x) = y=sqrt(x+4)-2

Answer 3

how do i get y by itself

Answer 4

I meant to say: f(x) = sqrt(x+4)-2

Answer 5

"how do i get y by itself"

You have to square root both sides.

Answer 6

x=(y+2)^2-4

x+4 = (y+2)^2

sqrt(x+4) = sqrt((y+2)^2)

sqrt (x+4) = y+2

y = sqrt(x+4) -2

Answer 7

the -2 is outside the square root?

Answer 8

yep

Answer 9

so that means its not a function right, since it has a square root?

Answer 10

uh... not quite. It's just not continuous, where the stuff inside the square root equals a negative number.

It's a "discontinuous" function.

Answer 11

So anything where x is less than -4 (like... x<-4, -5,... infinity), f(x) is undefined.

Answer 12

wait so the inverse is a function or not?

Answer 13

I'm sorry, I should have referred to the inverse as f^-1(x).

So anything where x is less than -4 (like... x<-4, -5,... infinity), f^-1(x) is undefined.

f^-1(x) =sqrt(x+4)-2

This is the inverse function.

Answer 14

so it is a function?

Answer 15

Yes, it is a function.

Inverse functions are functions... Just like functions are functions :-)

Answer 16

Here's a link that should help you more

http://tutorial.math.lamar.edu/Classes/CalcI/InverseFunctions.aspx

Answer 17

ok for number 2 you cube root?

Answer 18

Yup!

Show me your process and I'll check it.

Answer 19

i got cube root of x=1/4=y

Answer 20

x+1

Answer 21

first i added 1, cube rooted, and divided by 4

Answer 22

x=4y^3-1

x+1 = 4y^3

(x+1)/4 = y^3

(x/4+1/4) = y^3

cubedroot(x/4+1/4) = y

Answer 23

is cube root of x+1/4 ok?

Answer 24

They both need to be over 4. B/c when you divide on one side, then you divide everything by that number on the other side.

Answer 25

\[\sqrt[3]{x+1}/4\]

Answer 26

the 4 needs to be inside of the cubed root. cubed root of (x/4 +1/4)

Answer 27

can you do it using the equation thing please

Answer 28

\[\sqrt[3]{(x/4)+(1/4)}\]

Answer 29

so its not the same thing as everything over 4?

Answer 30

(x/4) + (1/4) = (x+1)/4

Answer 31

So what I did in the equation thing would be the same thing as saying:

\[\sqrt[3]{((X+1)/4)}\]

Answer 32

that wouldn't be a function right

Answer 33

It would. Just not any place where stuff inside the cubed root would equal a negative number.

Answer 34

how do you determine if its a function or no?

Answer 35

Where x=-1
(-1+1)/4 = 0/4 (this IS defined, but anything less than -1 is undefined)

Where x=-1.5

(-1.5+1)/4 = -.5/4

The cubed root of -.5/4 is undefined.

Answer 36

undefined would mean not a function?

Answer 37

I'll research that, I'm sorry for not answering the second part of the question very well. One sec.

Answer 38

for the last one i got x^2-4 = y

Answer 39

As far as I know, if the inverse of a function only has one output per each input, then it is a function. Try going onto google and graph x^(1/2) and x^(1/3).

Those are similar functions to what you have in your answers for a. and b. so with those, you'll see that are functions.

A problem you usually run in to is when you see something like y=sqrt(x^2 +2). That is a circle, so it is not a function. You don't have anything to worry about at least with the first and second ones.

Answer 40

Paul Dawkins give a really good set of instructions on how to find an inverse and then how to PROVE that what you found is truly the inverse of your function:
http://tutorial.math.lamar.edu/Classes/Alg/InverseFunctions.aspx

Answer 41

Let's see... x^2 = y+4

y = x^2 -4

Yep, you are right on that last one. And it is a function.

Answer 42

the first 2 don't look like functions from what i remember so im not sure

Answer 43

sometimes google graphs things wrong. I will go bust out my calculator

Answer 44

Try graphing the first and second ones. got a graphing calculator?

Answer 45

http://mpmath.googlecode.com/svn/gallery/gallery.html

Answer 46

This is an example of something that is NOT a function.

Answer 47

http://www.dummies.com/how-to/content/how-to-graph-a-circle.html

Answer 48

so im guessing they're not funtions

Answer 49

I keep getting conflicing answers. Must be that anything with a radical is not a function. Sorry for steering you wrong :-(

Answer 50

it's okay, how would something with a radical's graph look anyways?

Answer 51

I have 3 more questions if you dont mind helping me

Answer 52

Well I guess you would graph the f(x)= +sqrt(x+4)-2 and f(x)= -sqrt(x+4)-2

http://www.ehow.com/how_4515289_square-root-functions-fx-x.html

Answer 53

I'll do the rest tomorrow, thanks for helping!

Answer 54

k :-)

I'll look into the one about the cubed root.. I'm a little unsure on that one

Answer 55

Looks like with anything that has a radical which is an odd number... like the cubed root, the 5th root, 7th root, whatever...that those are functions. Here's what I'm thinking.

If you have the sqrt(4), then what are the possible answers?

Well (-2)(-2) =4 AND (2)(2) =4.

So square roots or any root that is an even number can't possibly be a function.

Okay well if you have the CUBED root(8), then what is/are the possible solution(s)?

(2)(2)(2) = 8 BUT!

You can't have: (-2)(-2)(-2) does not equal 8.

Therefore, cubed roots (or any root that is an odd number) only have one solution and is a function. Square roots(or any root that is an even number) always has two solutions and is NOT a function.

This was a good question, it helped refresh my memory on what is/isn't a function :-)

Answer 56

Oh and here's a video... he clarifies what I just said really well:

http://www.youtube.com/watch?v=oJLLQ2WeT5M