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roselin

  • 2 years ago

f(x) x^3-8/x^2-4 , x not equal to 2, X not equal to -2 3, x=2 4, x= -2 at which point is the function continuous?

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  1. tcarroll010
    • 2 years ago
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    2 questions 1) is your equation\[f(x) = \frac{ x ^{3} - 8 }{ x ^{2} - 4 }\] with x not = 2 or -2? And 2) what do the second 2 lines mean? (3, x=2 and 4, x = -2)?

  2. roselin
    • 2 years ago
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    |dw:1351367741430:dw|

  3. roselin
    • 2 years ago
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    |dw:1351367907016:dw|

  4. tcarroll010
    • 2 years ago
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    Start by factoring the numerator and denominator. Hint: you will then be able to get the common factor of (x-2) to cancel.

  5. roselin
    • 2 years ago
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    okay

  6. tcarroll010
    • 2 years ago
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    What do you get after factoring? Show your work.

  7. roselin
    • 2 years ago
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    (x-2) ( x^2+2x+4)/ (x+2)(x-2)

  8. roselin
    • 2 years ago
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    (x^2+2x+4)/(x+2)

  9. tcarroll010
    • 2 years ago
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    So, the first line has the restriction of x not = 2 or -2 because the first line is not defined for those 2 values of x. So far, we have a max of 2 points of discontinuity (before going to lines 2 and 3). For line 2, looking at how x-2 cancelled out (you did that correctly! Good job!) We DEFINE f(2) = 3 and that is what the "factored out" equation would have given, so the overall function IS continuous at x=2. This leaves line 3 and x = -2 as the only possible point of discontinuity. The "factored out" equation would give 4/0, so the overall function is discontinuous only at x = -2.

  10. roselin
    • 2 years ago
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    okay,thanks a lot.

  11. tcarroll010
    • 2 years ago
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    You're welcome! Medals are nice.

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