How do you find the discontinuities for #y=(4x)/(x^3-9x)#?

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Answer 1 · 2015-06-17T07:12:42

Factor the denominator to get:

#y = (4x)/(x(x-3)(x+3)) = 4/((x-3)(x+3))#

with exclusion #x != 0#

This has a removable discontinuity at #x=0# and simple poles at #x=+-3#

#y = f(x) = (4x)/(x(x-3)(x+3)) = 4/((x-3)(x+3))*x/x#

#= 4/((x-3)(x+3))# with exclusion #x != 0#

#f(0)# is undefined since #0/0# is undefined,

So #f(x)# has a discontinuity at #x=0#.

This is a removable discontinuity as we can simply define:

#g(0) = lim_(x->0) f(x) = -4/9#

#g(x) = f(x)# when #x != 0#

#f(-3)# and #f(3)# are simple pole discontinuities, where the denominator is zero and the numerator is non-zero.

graph{4x/(x(x-3)(x+3)) [-10, 10, -5, 5]}