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mathcalculus

  • 3 years ago

find the critical points: s(t)= (t-1)^4 (t+5)^3

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  1. mathcalculus
    • 3 years ago
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    i found the derivative. 4(t-1)^3 3(t+5)^2

  2. kobesaurus
    • 3 years ago
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    is it just \[(t-1)^4(t+5)^2\] ?

  3. kobesaurus
    • 3 years ago
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    with the chain rule those are not the derivatives...

  4. mathcalculus
    • 3 years ago
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    yes but instead of 2 as an exponent its 3. i know im suppose to = 0 then im not sure what to do.

  5. kobesaurus
    • 3 years ago
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    wait that is right, t is single power, sorry.

  6. mathcalculus
    • 3 years ago
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    okay so i did it wrong can we review step by step

  7. kobesaurus
    • 3 years ago
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    just to be clear, it isnt the two equations plus one another, its them two multiplied by one another?

  8. mathcalculus
    • 3 years ago
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    yes

  9. kobesaurus
    • 3 years ago
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    kk

  10. mathcalculus
    • 3 years ago
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    so i get.... 4(t-1)^3 3(t+5)^2

  11. kobesaurus
    • 3 years ago
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    (t-1)^3 (t+5)^2 (17+7 t)

  12. mathcalculus
    • 3 years ago
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    how?

  13. kobesaurus
    • 3 years ago
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    I'm trying to figure that out, I used Wolframalpha to check my work and it got something different.

  14. Azteck
    • 3 years ago
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    Use the product rule!!! \[\huge s(t)= (t-1)^4 (t+5)^3\] \[\huge s'(t)=4(t-1)^3(t+5)^3+3(t+5)^2(t-1)^4\] \[\huge =(t-1)^3(t+5)^2[4t+20+3t-3]\] \[\huge =(t-1)^3(t+5)^2(7t+17)\] At S.P's s'(t)=0 \[\huge (t-1)^3(t+5)^2(7t+17)=0\] \[\huge t=1, t=-5, t=-\frac{17}{7}\]

  15. Azteck
    • 3 years ago
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    \[s(t)=y\] To find the values of y, sub in the values of t into the original equation.

  16. Azteck
    • 3 years ago
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    Don't think you need to describe the nature of the curve [ie. Maximum or minimum turning point, etc.]. But I think you still need to find the inflection points of the function.

  17. mathcalculus
    • 3 years ago
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    how did you get to this step: =(t−1)3(t+5)2[4t+20+3t−3]

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