Question

Give the domain and range of each of the following functions.

Answer 1

y = (x-2)/(3x-9)

Answer 2

For the domain, I got all real numbers except x cannot equal 3. For the range, how am I supposed to find that?

Answer 3

You can use a graph to visually see what the range is

Or you can swap x and y, then solve for y

Let's use the second method

y = (x-2)/(3x-9)
x = (y-2)/(3y-9) ... swap x and y; now isolate y
x(3y-9) = y-2
3x*y - 9x = y - 2
3x*y = y - 2 + 9x
3x*y - y = 9x - 2
y*(3x - 1) = 9x - 2
y = (9x - 2)/(3x - 1)

What I've done is shown that the inverse of
\[\Large f(x) = \frac{x-2}{3x-9}\]
is
\[\Large f^{-1}(x) = \frac{9x - 2}{3x - 1}\]
Finding the domain of the inverse will give you the range of the original function

Answer 4

So y is all real numbers except 1/3?

Answer 5

yep, it's impossible to plug in some value of x to get y = 1/3

so y = 1/3 is the horizontal asymptote

Answer 6

notice in y = (x-2)/(3x-9) the leading terms are x up top and 3x down below

x/(3x) = 1/3 where x is nonzero

so as x heads off to infinity, y is getting closer and closer to 1/3

Answer 7

Another example: y = 8/(x+1) + 5

I've already found that x cannot equal -1. For the y, I switched the x and y values (is there a reason for this? why can't I just solve for x?) and got

Answer 8

sure you can solve for x. What do you get when you do?

Answer 9

Is that true for all/most problems like this? the leading coefficients are the vertical asymptotes?

Answer 10

the swapping of x and y happens because I'm finding the inverse

Answer 11

I got it to this: yx + y - 5x - 12 and I'm kind of stuck

Answer 12

`Is that true for all/most problems like this? the leading coefficients are the vertical asymptotes?`

only if the degrees are the same and only if you have a rational function. See rule 2

http://2.bp.blogspot.com/-9uxn3U5jEz8/VC-TK4qZudI/AAAAAAAAAGc/zt-dgnMHBfw/s1600/Screen+Shot+2014-10-03+at+11.26.10+PM.png

Answer 13

In the case of the first example, the horizontal asymptote just happened to be the same as the range?

Answer 14

The horizontal asymptote is a visual indicator of what is left out of the range

Answer 15

Oh okay.

Answer 16

Back to the second example, what should I do from here?

yx + y - 5x - 12

Answer 17

Let's solve for x
y = 8/(x+1) + 5
y-5 = 8/(x+1)
(y-5)*(x+1) = 8
x*y + y - 5x - 5 = 8
x*y - 5x = 8 - y + 5
x*y - 5x = 13 - y
x*(y - 5) = 13 - y
x = (13 - y)/(y - 5)

Answer 18

The range is all real numbers except y = 5?

Answer 19

yes

Answer 20

And that would also be a vertical asymptote?

Answer 21

horizontal

Answer 22

Right

Answer 23

this is not a coincidence to the fact that there's a +5 at the end of y = 8/(x+1) + 5

as x gets bigger and bigger, y will slowly get closer to 5 but never get there

Answer 24

8/(x+1) will get smaller and smaller as x gets bigger

8/(x+1) will get closer and closer to 0

Answer 25

:) thanks so much. The world needs more math helpers like you!

Answer 26

glad to be of help