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
Dec 22, 2016

All discontinuities: set the denominator equal to 0 and solve for #x#.

Removable ones: #x#-values that make numerator and denominator 0 simultaneously.

Explanation:

Any function #f(x)# will be discontinuous at #x#-values that make the function undefined. Here, that will simply be any #x#-value that creates a "division by zero".

What values of #x# create division by zero? In this case, it will be any #x# that satisfies

#x^2-9=0#

which we can factor to get

#(x+3)(x-3)=0#

So division by zero occurs when #x=+-3#.

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 #0/0# situation.

For this function, it is easy to see that when #x=3#, we get #0/0#, because #x-3# is a factor of both sides of the fraction. Thus, #x=3# is a removable discontinuity.

graph{(y-(x-3)/(x^2-9))((x-3)^2+(y-0.167)^2-.02)=0 [-11.5, 8.5, -4.046, 5.95]}