infinite discontinuities are not removable.
(x^2+4x+3) = (x+3)(x+1)
x^2-9 = (x+3)(x-3)
So, for all values of x except x = -3,
f(x) = (x+1)/(x-3)
At x = -3, f(x) = 0/0, which can be any value; it is undefined.
The discontinuity there is removable, if f(-3) is defined to be 1/3.
Are infinite discontinuities removable? Also, please help me with this question:
f(x)=x^2+4x+3 / x^2-9 has one removable discontinuity and one vertical asymptote. Find and identify the x-value for each. I found the asymptote at x=3, but please help for the discontinuity.
Thanks!
1 answer
1 answer