Can an absolute value have a discontinuity f(x)= |x-9| / (x-9)?

1 Answer
LM
Jun 7, 2018

yes, at x = 9

Explanation:

when x = 9, x - 9 = 0

when x = 9, f(x) = (9-9)/(9-9)

= 0/0

any number divided by 0 is undefined, so there is a discontinuity in the graph of f(x) = (|x-9|)/(x-9), at x = 9.

this is what the entire graph looks like:
![desmos.com/calculator](useruploads.socratic.org)

when x < 9, x-9 < 0
|x - 9| is always greater than 0 since it is an absolute value, meaning that dividing it by a negative number will give a negative number.
when x - 9 is negative, |x - 9| is the additive inverse, and so dividing the two will give 1/-1, or -1.
that is why, for all values of x below 9, f(x) = -1.

when x - 9 is positive, |x - 9| is the same number, and so dividing the two will give 1/1, or 1.
that is why, for all values of x above 9, f(x) = 1.

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