Question

is a function one to one if it has a verticle asymptote?

Answer 1

one to one meaning it has an inverse.

Answer 2

its one-to-one if it follows:
\[f(x_1) = f(x_2) \Longrightarrow x_1 = x_2\]
In other words, if it passes a Vertical Line test.

Answer 3

it doesnt matter if it has a vertical asymptote or not.

Answer 4

Yes even though the function approaches the vertical asymptote it never actually becomes it, even though it looks identical to it at extremely high values. So yes it would still have an inverse

Answer 5

wait, isn't that just test if its a function
to test if its one-to-one wouldn't you have to do like a Horizontal line test
Ex: y=x^2

Answer 6

it really depends on the function itself. a graph like:
\[y = \frac{1}{x^2}\]
has a vertical asymptote at x = 0, but does not have an inverse over the real numbers, while the graph:
\[y = \frac{1}{x}\]also has a vertical asymptote at x = 0, but does have an inverse over the real numbers.

Answer 7

I have a graph, and I'm suppose to determine whether it's one to one, and then graph the inverse of it.

Answer 8

the asymptote verticle lies on x=4

Answer 9

and it goes up and towards the left, from quadrant IV, through I, into II

Answer 10

can you take a picture an upload it?

Answer 11

ok ill get my cell phone one sec

Answer 12

do a vertical line test. if the graph hits each vertical line only once, its one to one. as far as graphing the inverse, like dichiaraj is saying, we would need a picture to see it better.

Answer 13

okay uploading

Answer 14

it passes a vertical line test, its one-to-one. assuming the pic is sideways lolol

Answer 15

it passes the verticle line test, so it's one to one. how do i figure out the equation?

Answer 16

I thought the horizontal line check was to see if it was a one-to-one function?

Answer 17

horizontal is just to check if its a function or not.

Answer 18

http://www.mathwords.com/h/horizontal_line_test.htm

Answer 19

it passes the horizontal test, then the verticle test.

Answer 20

so it's a function, with an inverse :P

Answer 21

vertical line test to determine if its a function
horizontal line test to determine if it has an inverse

Answer 22

wait wait...im getting them backwards lol

Answer 23

right, i stand corrected.

Answer 24

its vertical for function, horizontal for one-to-one.

Answer 25

so verticle test shows function, horizontal test shows if it has an inverse. okay

Answer 26

what does algebraic definition of f(x) mean on that picture i uploaded? I don't understand the question it's asking for.

Answer 27

right do the vertical to make sure each x has a y then the horizontal to check for one-to-one

Answer 28

function of x

Answer 29

ohhhh i thought it was asking for an equation to the graph, lmao

Answer 30

well their are ways to figure that out but no it's not asking that of you as of yet

Answer 31

how do I do part B of the second part?

Answer 32

"The inverse of f of x is one to one with f of x?"

Answer 33

i feel like these questions are vague!

Answer 34

I can't see what your asking on the pic you uploaded

Answer 35

B. If f(x) one to one? If so, determine it's inverse both algebraically, as well as graphically.

Answer 36

Heres a new pic

Answer 37

it's not letting me post another picture?

Answer 38

Does it give you a function?

Answer 39

or just the graph?

Answer 40

nope, just a graph. The teacher hinted that it has base e for it's function

Answer 41

with a verticle asymptote of 4

Answer 42

well graphicaly is going to be pretty easy, flip the paper over and draw the same graph, check the link for http://www.answers.com/topic/hyperbola an example

Answer 43

ok so i flip it and rotate it 90 degrees?

Answer 44

http://content.answcdn.com/main/content/img/ahd4/A4hypola.jpg here is a picture of a hyperbole and it's inverse graphed on the same paper

Answer 45

hmm i get log base 3 for this graph
notice point
(1,1)
starting with log(4-x)
log(4-1) = 1
log(3) = 1
base = 3

\[y = \log_{3} (4-x)\]

Answer 46

maybe he said "base 3" and not " base e"

Answer 47

no i could be wrong

Answer 48

i found a point (4,2) that contradicts my theory :(

Answer 49

lol =]

Answer 50

im over thinking it,

\[y = \ln (4-x)\]

Answer 51

im figuring out the graph right now, i think i got that part. :P

Answer 52

then solve for x to get inverse

nice :)
yeah you just flip the axis like zbay was saying

Answer 53

my graph turned out perfect. i think everything is good now. ill give u a medal =D

Answer 54

I like medals too!

Answer 55

lol joe

Answer 56

>.>

Answer 57

lolol

Answer 58

you got backwards :)