Question

Is there any algebraic method for finding the range of a rational function such as: f(x)=(x-7)/(x-2)

Answer 1

Find the inverse of the function.

Answer 2

y = (x-7)/(x-2)

(x-2-5)/(x-2)

1 - 5/(x-2)

x < 2 and 2 < x

x-2 < 0 and 0 < x-2

1/(x-2) < 0 and 0 < 1/(x-2)

-5/(x-2) < 0 and 0 < -5/(x-2)

1-5/(x-2) < 1 and 1 < 1-5/(x-2)

y < 1 and 1 < y

Answer 3

might have an error abounding :)

Answer 4

the horizontal asymptote is 1, so im satisfied

Answer 5

\[y = \frac{x - 7}{x - 2}\]

Swap x and y

\[x = \frac{y - 7}{y - 2}\]

Then isolate y:
\[x(y - 2) = y - 7\]
\[xy - 2x = y - 7\]
\[xy - y = 2x - 7\]
\[y(x - 1) = 2x - 7\]
\[y = \frac{2x - 7}{x - 1}\]

Thus \[y \ne 1\]

Answer 6

@amistre64 what are you hinting at?

Answer 7

With your "error aboundings". Cool it with that stuff

Answer 8

i was sure i made a mistake in typing :)

Answer 9

To find the range, find the inverse of the function. The domain of the inverse is the range of the original function.

Answer 10

and your write up is fine, but there was no need to swap x for y; in the end your result was
x\(\ne\) 1

Answer 11

I just addressed that just a second ago

Answer 12

there is more than 1 algebraic way to find the range. I simply presented one of them.

Answer 13

Is there a specific name for that method? I know you can find the inverse of the function or you can take the limit as x goes to infinity.

Answer 14

lol, im not good with names

Answer 15

I can't keep track of methods if they don't have a name.

Answer 16

@amistre64, just out of curiousity, what method would you use to find the range of this:

\[f(x) = \frac{\sqrt{2x - 5}}{x - 3}\]

Answer 17

i cant find a name for what i did :)

and asymptote is a thought; bottom is a higher degree than the top, so it goes to 0.

Answer 18

yeah, theres no easy way to do an inequality run with that one

Answer 19

There's a way to find the inverse, but it is quite tedious.

Answer 20

For \(\large f(x) = \dfrac{\sqrt{2x - 5}}{x - 3}\), can't we just use the facts that:

1. The domain is \(x \ge\dfrac{5}{2}\)
2. there is a vert asymptote at x=3
3. \(y\rightarrow\infty\) as \(x\rightarrow3^{+}\) and \(y\rightarrow -\infty\) as \(x\rightarrow 3^{-}\)
4. \(f(\dfrac{5}{2})=0\)
5. as \(x\rightarrow\infty\), \(y\rightarrow0\), because the degree of the num'r> degree of den'r.
6. The quotient of 2 continuous functions is continuous everywhere that the den'r is nonzero.

So putting that alllll together, we get range = \((-\infty,\infty)\), right?

I'm sure I'm missing something, lol.

Answer 21

domain is\[\{\frac52\le x<3\}\cup \{3<x\} \]

Answer 22

oh, of course it is.... sorry, I did know that but was sloppy above. lol

Answer 23

Should have said \(x \ge\dfrac{5}{2},~~x \neq3\)

Answer 24

hmm, y=0 for x=5/2

and then it would run up to x=3 in the y=-inf
then run from x=3 from y=+inf
and then horizontal out at y=0 which was already accounted for at the start of it

so i like the -inf to inf rande :) just not very algebraic

Answer 25

Yeah, that's basically what I was getting at... we can show that on [5/2,3) the range is \((-\infty,0]\) and on \((3,\infty)\) the range is \((0,\infty)\), so the union gives the whole range then as \((-\infty,\infty)\). But I'm not sure if it is rigorous enough for what Hero was looking for...? :)

Answer 26

... and it has to have a name :)

Answer 27

LOL right, that too! :)