# Calculus

posted by .

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 -

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 -

no its 8

• HELP -

I'm sure i did it right

• Calculus -

What is 8 divided by zero?

• Calculus -

no...
also all reals is wrong because
i know you can never get a value greater than -5/3

• Calculus -

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 -

(x+5)/(x - 3)

is just

(x+5)
-----
(x-3)

• Calculus -

1 + 0
-------------
1 - 0

I'm just getting 1????

• Calculus -

oh...
I did it right

graph it if you don't believe me

• Calculus -

so the range is...
y is...
arg confused

• Calculus -

...every number except 1?

• Calculus -

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 -

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 .