anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
this is a similar prob to what i posted earlier but im getting the wrong answer so im doing something wrong
JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
talk to me

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anonymous
  • anonymous
hold on. im processing this information. lol
anonymous
  • anonymous
so 5d+5=-2/5 ???
JamesJ
  • JamesJ
5d + 5 I agree with. Calculate F'(1) again.
anonymous
  • anonymous
idk what im doing wrong
anonymous
  • anonymous
10s+5=1? no?
JamesJ
  • JamesJ
No, you want the value of dF/ds when s = 1. That's what F'(1) means.
JamesJ
  • JamesJ
F'(s) = 10s + 5. Hence F'(1) = ...?
anonymous
  • anonymous
15
JamesJ
  • JamesJ
yes
JamesJ
  • JamesJ
Thus 5d + 5 = 15
anonymous
  • anonymous
i wasnt plugging in i was setting equal to 1
anonymous
  • anonymous
so the answer is 2
JamesJ
  • JamesJ
Would seem so!
anonymous
  • anonymous
yeah, it is

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