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:

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

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