How do you find the x values at which #f(x)=(x-3)/(x^2-9)# is not continuous, which of the discontinuities are removable?
1 Answer
All discontinuities: set the denominator equal to 0 and solve for
Removable ones:
Explanation:
Any function
What values of
#x^2-9=0#
which we can factor to get
#(x+3)(x-3)=0#
So division by zero occurs when
To find out if either of these values is a removable discontinuity, we check them one at a time to see if they also make the numerator equal to zero, thus creating a
For this function, it is easy to see that when
graph{(y-(x-3)/(x^2-9))((x-3)^2+(y-0.167)^2-.02)=0 [-11.5, 8.5, -4.046, 5.95]}