Question

Find the points of discontinuity ,if any.
F(x)=(x+3) / |x^(2) +3x|

My answer is at x=0 only cuz if we factored the Denom we will get |x (x+3)|
Then |x| |x+3|
Am i right ?

Answer 1

is the denominator in absolute values?

Answer 2

actually it doesn't make any difference, the discontinuities are the zeros of the denominator
that would include \(-3\) as well as \(0\)

Answer 3

The way i thought about is that if we want to get rid of the absolute value lines of x+3 we will have x+3 and 3-x right ? And both of them will cancel out with the numerator leaving us with 1/|x| and -1/|x| ....

Answer 4

What i mean is that for x<-3 the function will be -1/|x| and for x>-3 the function would be 1/|x| is't it ?

Answer 5

I 'm so freakin confused !

Answer 6

Sorry satellite 73 i didnt answer ur fist question yes the denominator is in absolute value im typing from my ipad i don't know how u see it but yes it is and that what confused me infact ..

Answer 7

discontinuity at \(x=0\) and also at \(x=-3\)

Answer 8

My answer is at x=0 only cuz if we factored the Denom we will get |x (x+3)|
Then |x| |x+3|

if you "cancel" the (x+3) the "simplified" expression is not *exactly* equivalent to the original fraction... even if you cancel common terms in the top and bottom you must take note of any "excluded" values... in other words, you must exclude both 0 and -3

Answer 9

I see that u agree with me phi that it's only discontinious at x=0 that's how i solved it but i wasn't sure ,however, i just downloaded a ghraphing program and i did graph the equation and there was an infinite discontinuity at zero only u can see that i once draw it as (x+3)/x^2+3x and as x+3/ -x^2-3x
Though i didn't understand what u mean by excluding both !! Cuz there's an infinite discontinuity at zero ...
Any way thanks for helping me out :)

Answer 10

***I see that u agree with me phi that it's only discontinious at x=0***

No, you are misinterpreting what I posted. There are 2 discontinuities: at x=0 and at x=-3

Answer 11

Are you saying that the way i draw the function is wrong? cuz the graph i had showed me an infinite discontinuity only at zero ...can you tell me how to graph such function please ?

Answer 12

See
http://www.wolframalpha.com/input/?i=F%28x%29%3D%28x%2B3%29+%2F+%7Cx%5E%282%29+%2B3x%7C+

There is a "hole" at x= -3

Answer 13

Aha!! So it's a removable discontinuity! okyyy now I get it ! the App i use is horrible I'm going to download WolframAlpha to graph functions..
so Thank you very much for making it so clear for me .I really appreciate it :)