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anonymous

  • 5 years ago

A(t)=t^5-5t^4+20t^3-17 I am having trouble setting the derivative equal to zero. Please help!

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  1. anonymous
    • 5 years ago
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    What did you get for the derivative?

  2. anonymous
    • 5 years ago
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    I've got \[A' = 5t^4-20t^3 + 60t^2 = 5t^2(t^2-4t + 12)\]

  3. anonymous
    • 5 years ago
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    THat's exactly what I've got

  4. myininaya
    • 5 years ago
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    a*b=0 implies either a=0 or b=0 or both=0

  5. anonymous
    • 5 years ago
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    Well I know that 5t^2=0 but I when I solve the parentheses for 0 (using quadratic formula) it comes out negative. I can't fit that into the rest of my problem!

  6. anonymous
    • 5 years ago
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    (I am solving for the intervals of increasing and decreasing in the function)

  7. myininaya
    • 5 years ago
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    you mean the number under the square root is negative?

  8. anonymous
    • 5 years ago
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    Yes

  9. myininaya
    • 5 years ago
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    so it is imaginary that means your only critical number is 0

  10. anonymous
    • 5 years ago
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    That just means that that part of the equation doesn't have any real roots.

  11. anonymous
    • 5 years ago
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    Okay. So if I am solving for the intervals, do I just divide my number in half?

  12. myininaya
    • 5 years ago
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    I like using test points before and after each critical number

  13. anonymous
    • 5 years ago
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    You're looking for intervals for what? Where it's increasing and decreasing?

  14. anonymous
    • 5 years ago
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    Yes, @polpak

  15. myininaya
    • 5 years ago
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    the derivative tells us the slope therefore tells us if is increasing or decreasing so choose a number before and after 0 to see what if it is increasing or decreasing

  16. anonymous
    • 5 years ago
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    So it's increasing from negative infinity to infinity, correct?

  17. anonymous
    • 5 years ago
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    The derivative is always positive except at 0. So it is increasing for \(x \in (-\infty,0) \bigcup (0,\infty)\)

  18. anonymous
    • 5 years ago
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    It is not increasing at 0.

  19. myininaya
    • 5 years ago
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    it has a horizontal tangent at x=0 thats what we found when we set A'=0

  20. myininaya
    • 5 years ago
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    A'=0 means we are trying to find all x that has slope 0

  21. anonymous
    • 5 years ago
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    Okay, thanks guys!

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