Posted by t on .
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,
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,
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.