Posted by **Calculus ** on Saturday, October 31, 2009 at 10:05pm.

I'm asked to find the domain of fuction

(x+5)/(x-3)

saying that x is all reals except for x can not equal three is wrong correct? Because i can evalue function when x = 3 as follows

3 + 5 = 8

so saying that the function has range x is a reals except x can not equal three is wrong because it can???

- Calculus -
**jim**, Saturday, October 31, 2009 at 10:13pm
No, it's not wrong.

When x=3, we have:

(x + 5) / ( x - 3)

= (3 + 5) / ( 3 - 3)

= 8 / 0

which is undefined, because of division by zero.

- Calculus -
**Calculus **, Saturday, October 31, 2009 at 10:15pm
no its 8

- HELP -
**HELP**, Saturday, October 31, 2009 at 10:16pm
I'm sure i did it right

- Calculus -
**jim**, Saturday, October 31, 2009 at 10:25pm
What is 8 divided by zero?

- Calculus -
**HELP**, Saturday, October 31, 2009 at 10:34pm
no...

also all reals is wrong because

i know you can never get a value greater than -5/3

- Calculus -
**HELP**, Saturday, October 31, 2009 at 10:45pm
wait what am I doing wrong here

(X+5)/(X - 3)

(x+5)/x

--------

(x-3)/x

(1+5/X)

--------

(1-3/x)

i forgot were to go from here

- Calculus -
**jim**, Saturday, October 31, 2009 at 10:47pm
(x+5)/(x - 3)

is just

(x+5)

-----

(x-3)

- Calculus -
**HELP**, Saturday, October 31, 2009 at 10:51pm
1 + 0

-------------

1 - 0

I'm just getting 1????

- Calculus -
**HELP**, Saturday, October 31, 2009 at 10:53pm
oh...

I did it right

graph it if you don't believe me

- Calculus -
**HELP**, Saturday, October 31, 2009 at 10:55pm
so the range is...

y is...

arg confused

- Calculus -
**HELP**, Saturday, October 31, 2009 at 10:57pm
...every number except 1?

- Calculus -
**HELP**, Saturday, October 31, 2009 at 11:11pm
So how come we can not say that x can not equal 3 because I'm getting 8 when x = 3 also how come we say range is all reals when its not because y will never equal 1

- Calculus -
**Reiny**, Sunday, November 1, 2009 at 7:28am
To the person posting as "Help"

The reply by Jim is correct, division by zero is undefined and causes a vertical asymptote in the corresponding graph for the given function.

Changing it to the form

(1+5/X)

--------

(1-3/x)

serves no purpose, since when x = 3 you are still dividing by zero.

Changing it to that form would be useful if we wanted to find the horizontal asymptote, by letting x become "very large". We can see that the above would approach 1 and the H.A. would be y = 1

As stated by jim, the domain is the set of real numbers, except x cannot be 3 .

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