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???
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
--------
(x-3)/x
(1+5/X)
--------
(1-3/x)
i forgot were to go from here
(x+5)/(x - 3)
is just
(x+5)
-----
(x-3)
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)
--------
(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 .