Calculus
posted by Melissa on .
Let F(s)=5s2+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 the instantaneous rate of change of F(s) at s=1. d=

ignore the "d=". Sorry!

the instantaneous rate of change is given by
F'(x) which is 10s + 5
which, when s=1, becomes F'(1) = 15
for the "average" rate of change from 0 to d
F(0) = 4
F(d) = 5d^2 + 5d + 4
average rate of change = (5d^2 + 5d + 4  4)/(d0)
so solve for d in
(5d^2 + 5d)/d = 15
d^2 + d = 3d
d^2  2d = 0
d(d2) = 0
d = 0 or d = 2 ,but you want d > 0,
so d = 2