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soccergal12

find all relative extrema and points of inflection: f(x) = 2 / (x^2 - 4) i got x^2 = -4/3 as an inflection point. but i think i did something wrong

  • one year ago
  • one year ago

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  1. dpaInc
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    no... nothing wrong there... what happens when you take the square root of a negative number?

    • one year ago
  2. soccergal12
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    well you can't take a sqaure root of a negative number. so i wouldn't have any inflection points right?

    • one year ago
  3. dpaInc
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    right... that's what that means...

    • one year ago
  4. dpaInc
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    but that does not mean the graph is all concave up or all concave down.... there can be places on your graph where in one interval its ccup and in another it's ccdown...

    • one year ago
  5. dpaInc
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    or vice versa....

    • one year ago
  6. soccergal12
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    okay, the question does not ask for concavity, but should i still determine if there is some type of concavity? just to be safe?

    • one year ago
  7. dpaInc
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    sorry... hate when chrome crashes... no we don't need concavity but the question did ask for points of inflection...

    • one year ago
  8. dpaInc
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    so according to what u got, there are no points of inflection...

    • one year ago
  9. dpaInc
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    now all you need are relative max/mins... do you have your first derivative?

    • one year ago
  10. soccergal12
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    yes , it was (-4x) / (x^2 -4)^2

    • one year ago
  11. dpaInc
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    so what's your critical number(s) ?

    • one year ago
  12. soccergal12
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    0, 2 and -2, i said there was a relatve max at (0, -1/2)

    • one year ago
  13. dpaInc
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    ok.... now do the first derivative test on the intervals: (-infinity, -2) (-2, 0) (0, 2) (2, infinity)

    • one year ago
  14. soccergal12
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    i did that and found the relative max

    • one year ago
  15. dpaInc
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    oh... sorry... i guess you're done...:)

    • one year ago
  16. soccergal12
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    perfect. thank you :)

    • one year ago
  17. dpaInc
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    yw...:)

    • one year ago
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