Thursday, June 12, 2014


In some cases, a function will have discontinuity, which is a value for x where the function is undefined. It could be removable discontinuity, in which case there is a "hole" in the graph, or non-removable discontinuity, which in many cases will be an asymptote.

Take for example the problem  f(x) = (x^2 - 1)/(x - 1).  This is undefined for x = 1.

Factored this becomes (x - 1)(x + 1)/(x - 1).  When simplifying we are in essence "removing" the x -1 from the numerator and the denominator and are left with x + 1.

The graph of x + 1 is a line with slope 1 and y-intercept of 1. But since we know x is undefined at 1, we have a hole at the coordinate (1,2) represented by an open circle.

Here's another example which has some discontinuity.

f(x) = (x^2 + 6x + 8)/(x^2 -2x - 8)

When factoring we get (x + 4)(x + 2)/[(x -4 )(x + 2)]

This function is undefined at x = 4 and x = -2.  Notice the removable discontinuity at x = -2. The other discontinuity at x = 4 is non-removable.

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