y= x / (x^2+x)
what are the discontinuities on this graph ?
what happens at x=-1
more interesting, what happens when x = 0 ?
y = x/(x(x+1))
= 1/(x+1) , x ≠ 0
http://www.wolframalpha.com/input/?i=plot+y%3D+x+%2F+(x%5E2%2Bx)
so you also have a "hole" at (0,1)
e.g. let x = .0001
y = .0001/(.0001(1.0001))
= appr .99900999...
To find the discontinuities of the function y = x / (x^2+x), we need to identify the values of x at which the function is not defined or where there are vertical asymptotes.
Start by analyzing the denominator of the fraction, which is x^2 + x. This denominator equals 0 when x = 0 or x = -1, since both values cause division by zero.
So, the potential discontinuities occur at x = 0 and x = -1.
To confirm if these values are true discontinuities, you should check if the function has a limit as x approaches each of these points. If the limit exists, there is a removable discontinuity; if the limit is infinite, there is a vertical asymptote.
To find the limits as x approaches 0 and x approaches -1, you can substitute these values into the original function.
1. As x approaches 0:
lim(x->0) [x / (x^2 + x)]
Substitute x = 0:
lim(x->0) [0 / (0^2 + 0)]
Simplify:
lim(x->0) [0 / 0]
This indeterminate form indicates that further simplification is required to determine the limit. You can apply the L'Hôpital's Rule or factorize the numerator and denominator to simplify further.
2. As x approaches -1:
lim(x->-1) [x / (x^2 + x)]
Substitute x = -1:
-1 / (-1^2 - 1)
-1 / (1 - 1)
-1 / 0
Here, we also encounter an indeterminate form, which requires simplification.
Further simplification using appropriate techniques like factoring or L'Hôpital's Rule is essential to determine if these points are indeed discontinuities on the graph.