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anonymous

  • 4 years ago

Let F(s)=5s^2+5s+4. Find a value of d greater than 0 such that the average rate of change of F(s) from 0 to d equals instantaneous rate of change of F(s) at s=1.

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  1. anonymous
    • 4 years ago
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    this is a similar prob to what i posted earlier but im getting the wrong answer so im doing something wrong

  2. JamesJ
    • 4 years ago
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    The average rate of change of F(s) from 0 to d is the average of F'(s) from 0 to d, which is \[ \frac{1}{d} \int_0^d dF/ds \ ds = \frac{F(d) - F(0)}{d} \] Hence you want to find a d such that \[ \frac{F(d) - F(0)}{d} = F'(1) \] So now you just need to evaluate.

  3. JamesJ
    • 4 years ago
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    talk to me

  4. anonymous
    • 4 years ago
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    hold on. im processing this information. lol

  5. anonymous
    • 4 years ago
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    so 5d+5=-2/5 ???

  6. JamesJ
    • 4 years ago
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    5d + 5 I agree with. Calculate F'(1) again.

  7. anonymous
    • 4 years ago
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    idk what im doing wrong

  8. anonymous
    • 4 years ago
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    10s+5=1? no?

  9. JamesJ
    • 4 years ago
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    No, you want the value of dF/ds when s = 1. That's what F'(1) means.

  10. JamesJ
    • 4 years ago
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    F'(s) = 10s + 5. Hence F'(1) = ...?

  11. anonymous
    • 4 years ago
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    15

  12. JamesJ
    • 4 years ago
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    yes

  13. JamesJ
    • 4 years ago
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    Thus 5d + 5 = 15

  14. anonymous
    • 4 years ago
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    i wasnt plugging in i was setting equal to 1

  15. anonymous
    • 4 years ago
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    so the answer is 2

  16. JamesJ
    • 4 years ago
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    Would seem so!

  17. anonymous
    • 4 years ago
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    yeah, it is

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