Question

Can you get the domain and range of an inverse function without graphing it?

Answer 1

you can, but it depends on your level of math.

Answer 2

domain you can im sure.. probably not the range.

Answer 3

what do you mean by level of math?

Answer 4

for the domain, just see what values of x work for your problem. example.. if it is a linear equation.. y = 2x+1 then you know it can be all real values of x.

Answer 5

I dont think they teach you how to find the range of an inverse function until calculus without making a graph and looking at it.

Answer 6

is what i mean.

Answer 7

oh. so you always need a graph to look for the range?

Answer 8

yes, you have a graphic calculator? makes it much easier than drawing :)

Answer 9

we can't use that. :|

Answer 10

eww

Answer 11

Can I ask a new question? Since you have answered my previous one.

Answer 12

though once u find the domain and range for the inverse, the domain and range for f(x) is the opposite.

Answer 13

sure

Answer 14

There are some cases were the inverse of a function is not a function.

Answer 15

well in our class my teacher gave a random domain in which he did not tell us how he got the domain, so can you explain how to give the right domain in those cases where the inverse is not a function?

Answer 16

do you mean the inverse is not a one to one function?

Answer 17

Yes.

Answer 18

For example

Answer 19

my teacher used this function: f(x)=(x-1)^2 -4

Answer 20

then we got the inverse which is +- sqrt of x+4 +1 = y

Answer 21

so what he did was to give a domain of x is greater than or equal to 1

Answer 22

for the inverse

Answer 23

so my question is, how did he get that domain? what was his basis for that domain?

Answer 24

the domain has of \[\pm \sqrt{x+4+1}\] has to be greater than 0 because you cannot have a negative number under a radical sign

Answer 25

when you are dealing with domain and functions.. you have to make sure that your answer does not break the rules... such as.. if you have a fraction.. x cannot = 0 in the denominator.. so you would leave that out of the domain.

Answer 26

The original fxn was (x-1)^2 -4 and my teacher gave a domain of X is greater than or equal to 1.

Answer 27

\[f(x)=(x-1)^2-4\] is a parabola that opens up, and has a vertex at \((1,-4)\)

Answer 28

yay satellite is here! he is much smarter than I..

Answer 29

if you make the domain
\[x\geq 1\] then it is no longer a parabola, it is half of one, the part to the right of the vertex, and so it is increasing and one to one

Answer 30

haha. :D so why is it confined into \[X \ge1\] only?

Answer 31

so now it is a one to one function, and you can find the inverse

Answer 32

why can't it be less than or equal to one?

Answer 33

you get to pick the domain, so long as on that domain the function is one to one

Answer 34

it could be
lets solve that one

Answer 35

so that means we can pick any domain we want? but how do we know if it is the right one?

Answer 36

\[x=(y-1)^2-4\]
\[x+4=(y-1)^2\]
\[\pm\sqrt{x+4}=y-1\]
\[y=\pm\sqrt{x+4}+1\] but since we know \(y\leq 1\) we get rid of the \(\pm\) and put
\[y=-\sqrt{x+4}+1\]

Answer 37

I'm kinda getting it now.

Answer 38

there is no "right" domain
you can pick any domain for which the original function is one to one
in your case you could say domain is \(x\geq 1\) or you could say \(x\leq 1\) or for that matter you could say \(1<x<5\) because it is one to one there too

Answer 39

sorry i hacked it up so much :)

Answer 40

It's okay Nameless :D

Answer 41

lets look at a picture of
\[y=(x-1)^2-4\]
http://www.wolframalpha.com/input/?i=y%3D%28x-1%29^2-4

Answer 42

so when you say pick the domain... do you mean pick the numbers in the intervals between key numbers?

Answer 43

you see it is not one to one, because it does not pass the horizontal line test
but you can make it one to one by restricting the domain, in other words by taking a piece of the function

Answer 44

So basically for this function:
f(x)=(x-1)^2 -4 we can say that it's domain is:

\[x \ge5 \]?

Answer 45

you could say 'let \(f(x)=(x-1)^2-4\) for \(x\geq 1\) and then it will be one to one

Answer 46

hey it is your function, you can do anything you like with it

Answer 47

so basically that fxn can have lots of domain?

Answer 48

you can restrict the domain in any way that you want

Answer 49

that's cool :D

Answer 50

yes

Answer 51

Thanks! Getting so confused there. :D

Answer 52

hope that helped. the restriction of \(x\geq 1\) was not arbitrary, it was made as large as possible so that the function was one to one
but \(x\leq1\) would work just as well, you would be taking the left side instead of the right side

Answer 53

i am confused by the statement you can pick your domain

Answer 54

Thank you so much satellite :D

Answer 55

i can define a function any way i like

Answer 56

yw

Answer 57

i can say "let \(f(x)=x^2+2x\) for \(3<x<5\)

Answer 58

so you can define to fit what you need?

Answer 59

thus in this case only half of the parabola was needed to make this an inverse? so this is not an inverse function until you restrict the domain

Answer 60

I guess that's what he's trying to say :D

Answer 61

Thanks for your help as well Nameless :D

Answer 62

np