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anonymous

  • 5 years ago

Find critical points of: A(t)=t^5-5t^4+20t^3-17 I have an answer, but I think it's incorrect. Help, please!

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  1. anonymous
    • 5 years ago
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    What answer did you get?

  2. anonymous
    • 5 years ago
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    First I found derivative: A'(t)=5t^4-20t^3+60t^2 Then factored out 5t^2 for: 5t^2(t^2-4t+12) So I got x=0, but used quadratic equation and got 4+-sqrt(-32) all over 2

  3. anonymous
    • 5 years ago
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    I am concerned because I don't think there is supposed to be a negative under the sqrt sign.

  4. anonymous
    • 5 years ago
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    That just means that you won't have any real roots from that factor.

  5. anonymous
    • 5 years ago
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    Okay. So If I'm finding absolute extrema, what do I do in this case? And my work looks correct to you?

  6. anonymous
    • 5 years ago
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    So your only (real) critical points will be when t = 0.

  7. anonymous
    • 5 years ago
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    Yes it is correct.

  8. anonymous
    • 5 years ago
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    Okay.

  9. anonymous
    • 5 years ago
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    Does that mean that the quadratic part is where it DNE?

  10. anonymous
    • 5 years ago
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    DNE?

  11. anonymous
    • 5 years ago
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    does not exist

  12. anonymous
    • 5 years ago
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    The function exists for any real input. You just won't ever have a zero from that quadratic over the real numbers.

  13. anonymous
    • 5 years ago
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    So look at what your derivative is doing about 0 and see whether you have a min, a max, or an inflection point.

  14. anonymous
    • 5 years ago
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    Okay. Thank you! So is t=0 my only critical point?

  15. anonymous
    • 5 years ago
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    For real inputs, yes.

  16. anonymous
    • 5 years ago
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    So I'm on the closed interval [-4,3] I evaluate A(0), A(-4), and A(3), correct?

  17. anonymous
    • 5 years ago
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    Sure, but you don't really have to bother with 0 since it's pretty easy to see that it's an inflection point and not a local min/max.

  18. anonymous
    • 5 years ago
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    Hmm...How do I know that?

  19. anonymous
    • 5 years ago
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    Well, if t is some number close to 0, but negative what would your derivative be?

  20. anonymous
    • 5 years ago
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    Wait - do I evaluate the derivative or the original equation?

  21. anonymous
    • 5 years ago
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    We're looking at the derivative to find if 0 is a local min max.

  22. anonymous
    • 5 years ago
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    I thought I evaluate the function. Sorry I don't understand what I evaluate the derivative for.

  23. anonymous
    • 5 years ago
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    And the derivative would be positive because t^2 is positive and (12 +/1 a very small number) is positive.

  24. anonymous
    • 5 years ago
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    err +/- rather.

  25. anonymous
    • 5 years ago
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    So my absolute max is 361 @ t=3 and absolute min is -3601 @ t=-4 right? (I'm only looking for max extrema.)

  26. anonymous
    • 5 years ago
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    Ok, you know that if the derivative is positive the curve is sloping upward. If the derivative is negatve it slopes down. If you have a local max then the derivative will be 0 at that point, but just before that point it will be positive, and after it will be negative. If you have a local min then the derivative will be 0, but just before the point it will be negative, and just after that it will be positive. If you have an inflection point (think y=x^3 around 0) you will have the derivative is 0, but it doesn't change signs from one side of the point to the other. So when looking for extrema you want to take into consideration local mins/maxes and the values at the end points.

  27. anonymous
    • 5 years ago
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    Since in our case the derivative is positive just before 0 and positive just after 0 we can see that it's just an inflection point and not a possible min/max.

  28. anonymous
    • 5 years ago
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    Oh. Okay! I get the y=x^3 part. My professor talked about that.

  29. anonymous
    • 5 years ago
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    Have a nice day. I'm off to the store! =)

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