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soccergal12

  • 2 years ago

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

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

  2. soccergal12
    • 2 years ago
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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?

  3. dpaInc
    • 2 years ago
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    right... that's what that means...

  4. dpaInc
    • 2 years ago
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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...

  5. dpaInc
    • 2 years ago
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    or vice versa....

  6. soccergal12
    • 2 years ago
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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?

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

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

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

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

  11. dpaInc
    • 2 years ago
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    so what's your critical number(s) ?

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

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

  14. soccergal12
    • 2 years ago
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    i did that and found the relative max

  15. dpaInc
    • 2 years ago
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    oh... sorry... i guess you're done...:)

  16. soccergal12
    • 2 years ago
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    perfect. thank you :)

  17. dpaInc
    • 2 years ago
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    yw...:)

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