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arifelgon94

  • 2 years ago

y= (2x^2 + 5) / (x^2 - 2x) point of discontinuity

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  1. AERONIK
    • 2 years ago
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    at x=2 function is not defind.

  2. AERONIK
    • 2 years ago
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    so the function is discontinuous at that point.....

  3. zepdrix
    • 2 years ago
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    We have discontinuities whenever the DENOMINATOR is zero. So let's set the denominator equal to 0, and solve for x! Those values of X are the trouble makers! :O \[\large (x^2-2x)=0\]Factoring an x out of each term gives us: \[\large x(x-2)=0\] Know what to do from here to solve for x? :D Need to take advantage of the Zero Factor Property! :D

  4. arifelgon94
    • 2 years ago
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    thnk you... but how did you solve it?

  5. zepdrix
    • 2 years ago
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    Aeron mentioned one of the values, but there is also another one!! Make sure you don't miss that trouble maker! :3

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