What are the removable discontinuities of #f(x) = (x^3 - 3x^2 - x + 3) / (x+1)#?

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Answer 1 · 2015-06-26T19:02:54

The only discontinuity is at #x=-1# and it is removable.

Rational functions are continuous on their domains. (They are continuous except where the denominator is zero.)

#f(x) = (x^3 - 3x^2 - x + 3) / (x+1)# Is continuous except at #x=-1#

Not the when #x=-1#, the numerator is #0#, so we know that #x-(-1) = x+1# is a factor of the numerator.

We'll need the factoring in a minute, so either do the division or factor by grouping:

#x^3 - 3x^2 - x + 3 = (x^3 - 3x^2)+( - x + 3)#

# = x^2(x-3)-1(x-3)#

# = (x^2-1)(x-3) = (x+1)(x-1)(x-3)#

#lim_(xrarr-1)(x^3 - 3x^2 - x + 3) / (x+1) = lim_(xrarr-1)((x+1)(x-1)(x-3)) / (x+1) #

# = lim_(xrarr-1)((x-1)(x-3))#

# = (-2)(-4)=8#

Because this limit exists, the discontinuity is removable.

Not asked for, but

To remove the discontinuity, define #g# by:

#g(x) = { (f(x), "if", x != -1), (8, "if", x=-1) :}#

Note that #g(x) = f(x)# for all #x != -1# but #g# is continuous at #-1#.