Calculus
posted by Calculus on .
I'm asked to find the domain of fuction
(x+5)/(x3)
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???

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. 
no its 8

I'm sure i did it right

What is 8 divided by zero?

no...
also all reals is wrong because
i know you can never get a value greater than 5/3 
wait what am I doing wrong here
(X+5)/(X  3)
(x+5)/x

(x3)/x
(1+5/X)

(13/x)
i forgot were to go from here 
(x+5)/(x  3)
is just
(x+5)

(x3) 
1 + 0

1  0
I'm just getting 1???? 
oh...
I did it right
graph it if you don't believe me 
so the range is...
y is...
arg confused 
...every number except 1?

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

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)

(13/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 .