Posted by t on Monday, February 1, 2010 at 5:31pm.
Let F(s)=5s2+1s+4. Find a value of d greater than 0 such that the average rate of change of F(s)from 0 to d equals the instantaneous rate of change of F(s)at s=1.

calculus  bobpursley, Monday, February 1, 2010 at 5:35pm
F'(1)=10+1=11
average= (finalF(s)initialF(s))/d
= (5d^2+d+44)/d= 5d+1
11=5d+1 solve for d.

calculus  drwls, Monday, February 1, 2010 at 5:41pm
I assume that your fucnction is
F(s) = 5s^2 +s + 4.
The instantaneous rate of change of F at s=1 is the value of the derivative at s=1. The derivative is F'(x) = 10s + 1. When s = 1, F' is 11.
The average rate of change of F from 0 to d is [F(d)  F(0)]/d. That equals
(5d^2 + d + 4  4)/d = 5d + 1
Set 5d +1 = 11 and solve for d.
I leave that part for you.
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