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+4-4)/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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