How do you find the x values at which #f(x)=(x+2)/(x^2-3x-10)# is not continuous, which of the discontinuities are removable?

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Answer 1 · 2016-12-16T19:33:21

#x=5# is a non removable discontinuity, while #x=-2# can be removed by simplifying the function

As #f(x)# is a rational function the only values where it is not continuous are the values of #x# for which the denominator is null:

#x^2-3x-10 = 0#

#x = frac (3+-sqrt(9+40)) 2 = frac(3+-7) 2#

that is:

#x=-2# and #x=5#

We can therefore factorize the denominator as:

#x^2-3x-10 = (x+2)(x-5)#

and conclude that #x=5# is a non removable discontinuity, while #x=-2# can be removed by simplifying the function and writing it as:

#f(x) = frac (x+2) (x^2-3x-10) = frac (x+2) ((x+2)(x-5)) = 1/(x-5)#