Asked by t
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.
Answers
Answered by
bobpursley
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.
average= (finalF(s)-initialF(s))/d
= (5d^2+d+4-4)/d= 5d+1
11=5d+1 solve for d.
Answered by
drwls
I assume that your function 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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.