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anonymous

  • 5 years ago

consider the function g(x) = x^3 - (5/2)x^2-2x+1. List any points of inflection. Please show work

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  1. anonymous
    • 5 years ago
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    To find the inflection points you have to derive the following function twice. After that, take 2 conditions for f''(x). (a) f''(x) = 0 (b) f''(x) DNE after that take f''(x) = 0 and find the zeros of the function, those zeros will be your inflection numbers, then substitute them in the function to get their y :) and your done ^_^ Give it a try :)

  2. anonymous
    • 5 years ago
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    what's DNE

  3. anonymous
    • 5 years ago
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    Doesn't exist, and you can use that case if and only if we have x in the denominator :)

  4. anonymous
    • 5 years ago
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    so in your case, you have a function that is considered a polynomial which means the range and domain is R, so you can factor out f''(x) and find the zeros ^_^

  5. anonymous
    • 5 years ago
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    clearer? :)

  6. anonymous
    • 5 years ago
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    i would factor out x out of the original function but what about the 1

  7. anonymous
    • 5 years ago
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    lol, no, to find the inflection points you have to take 2 conditions, like I have said before, but since for x, x exists , you must take the zeros of f''(x) and not f(x) :)

  8. anonymous
    • 5 years ago
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    oh you said factor out the second derivative. sry for the misunderstanding

  9. anonymous
    • 5 years ago
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    lol, no worries ^_^ yeah, the second derivative = f''(x) :) give it a try now.

  10. anonymous
    • 5 years ago
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    ok

  11. anonymous
    • 5 years ago
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    good luck :)

  12. anonymous
    • 5 years ago
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    thx

  13. anonymous
    • 5 years ago
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    np ^_^

  14. anonymous
    • 5 years ago
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    the 2nd deriv only gives me one number when = to zero

  15. anonymous
    • 5 years ago
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    so that's your inflection number :), I didn't calculate it yet, but if that's what you got, then that's the inflection number ^_^

  16. anonymous
    • 5 years ago
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    so i sub that in for all the Xs

  17. anonymous
    • 5 years ago
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    yes to get the y, and then you'll have an inflection point (x,y) ^_^

  18. anonymous
    • 5 years ago
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    if i substitute it in it will only give me one #, how is that a point? sry if i'm not understanding correctly

  19. anonymous
    • 5 years ago
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    no it's alright. Calc dear, when you found the inflection number by getting the zeros, then you have found point x. When you substitute x in the original function, you'll get point y , f(x) = y. Solve for y :)

  20. anonymous
    • 5 years ago
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    Did you understand it now? ^_^

  21. anonymous
    • 5 years ago
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    gotchya

  22. anonymous
    • 5 years ago
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    awesome :)

  23. anonymous
    • 5 years ago
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    thx alot that helped

  24. anonymous
    • 5 years ago
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    np ^_^ wish you all the best

  25. anonymous
    • 5 years ago
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    thnk u

  26. anonymous
    • 5 years ago
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    :) np

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