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arifelgon94 Group Title

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

  • 2 years ago
  • 2 years ago

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  1. AERONIK Group Title
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    at x=2 function is not defind.

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

    • 2 years ago
  3. zepdrix Group Title
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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

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

    • 2 years ago
  5. zepdrix Group Title
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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

    • 2 years ago
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