Consider the function....?

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1 Answer
Jan 2, 2018

(a) Just #a=-1/2#

Explanation:

#f(x)# is discontinuous at #a#, if #f(a)# is not defined i.e #(x-a)# appears in denominator.

But we also want #lim(x->a)f(x)# to exist. This means that although #(x-a)# exists in denominator, it is also present in numertor and hence the two cancel each other out, in other words #(x-a)# is common factor between numerator and denominator.

The only linear factor common to both numerator and denominator is #(2x+1)# or #2(x-(-1/2))#, Hence for #a=-1/2# #f(x)# is discontinuous at #a# but #lim(x->a)f(x)# exists.

This is also apparent from the graph of the functionn, which shows that although function is ddiscontinuous at #x=2# and #x=-3#, it is not so at #x=-1/2#. Further at #x=1/2# and #x=3#, value of function is zero.

graph{(4x^3-12x^2-x+3)/(2x^3+3x^2-11x-6) [-10, 10, -5, 5]}