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What are the points of discontinuity? Are they removable? y = (x-5) / x^2 - 6x + 5

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Answer:

The points of discontinuity are: x=5 and x=1. The point of discontinuity x=5 can be removable.

Explanation:

The points of discontinuity are those points where the function is not defined. To find such points, we should factorize the denominator of the function needs to be factorized.


y= (x-5)/(x^2 - 6x + 5)

Considering the quadratic equation of the form ax^2+bx+c =0, then using the quadratic (see attached image), where a=, b=- and c=5, we have that the roots are:

x=5 and x=1.

If we simplify the fraction, by removing the term (x-5) from both the numerator and the denominator, we get:
y=1/(x-1), so we removed the point of discontinuity x=5.

What are the points of discontinuity? Are they removable? y = (x-5) / x^2 - 6x + 5-example-1
User Ramesh Srirangan
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