What are the removable and non-removable discontinuities, if any, of #f(x)=(x^3 - x^2 - 72 x)/ (x - 9) #?

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Answer 1 · 2016-12-20T20:50:27

It has a removable singularity at #x=9#

We have:

# f(x)=(x^3 - x^2 - 72 x)/ (x - 9) #

Clearly when the denominator #x-9=0 => x=0# is the only singularity.

Let's see if we can remove it:

# f(x)=(x(x^2 - x - 72 ))/ (x - 9) #
# " " =(x(x+8)(x-9))/ (x - 9) #

And so we can "cancel" the #(x-9)# factor and remove the singularity to give:

# f(x) =x(x+8) #