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.
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.
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.
To find a value of d 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, we need to follow these steps:
Step 1: Find the average rate of change of F(s) from 0 to d.
To calculate the average rate of change, we use the formula:
Average rate of change = (F(d) - F(0)) / (d - 0)
Step 2: Find the instantaneous rate of change of F(s) at s = 1.
To find the instantaneous rate of change, we need to find the derivative of F(s) and evaluate it at s = 1.
Step 3: Set the average rate of change equal to the instantaneous rate of change and solve for d.
We can equate the average rate of change calculated in step 1 with the instantaneous rate of change calculated in step 2 and solve for d.
Let's perform these steps:
Step 1: Find the average rate of change of F(s) from 0 to d.
Average rate of change = (F(d) - F(0)) / (d - 0)
Inserting the equation for F(s) and simplifying:
Average rate of change = (5d^2 + d + 4 - (5(0)^2 + 0 + 4)) / (d - 0)
Average rate of change = (5d^2 + d + 4 - 4) / d
Average rate of change = 5d^2 + d / d
Average rate of change = 5d + 1
Step 2: Find the instantaneous rate of change of F(s) at s = 1.
To find the instantaneous rate of change, we need to find the derivative of F(s) and evaluate it at s = 1.
F'(s) = d/ds (5s^2 + s + 4)
= 10s + 1
To find the instantaneous rate of change at s = 1, evaluate F'(s) at s = 1:
F'(1) = 10(1) + 1
= 10 + 1
= 11
Step 3: Set the average rate of change equal to the instantaneous rate of change and solve for d.
Average rate of change = Instantaneous rate of change
5d + 1 = 11
5d = 10
d = 10/5
d = 2
Therefore, the value of d that satisfies the given conditions is d = 2.
To find a value of d that satisfies the given criteria, we need to calculate the average rate of change of F(s) from 0 to d and the instantaneous rate of change of F(s) at s=1, and then equate them.
Let's start by calculating the average rate of change of F(s) from 0 to d. The average rate of change is given by the formula:
Average Rate of Change = (F(d) - F(0)) / (d - 0)
We're given the function F(s) = 5s^2 + s + 4. So let's substitute the values into the formula:
Average Rate of Change = (F(d) - F(0)) / (d - 0)
Average Rate of Change = (5d^2 + d + 4 - 4) / d
Average Rate of Change = (5d^2 + d) / d
Now, let's find the instantaneous rate of change of F(s) at s=1. The instantaneous rate of change at a specific point can be found by taking the derivative of the function with respect to s and then substituting the value of s=1 into the derivative.
The derivative of F(s) = 5s^2 + s + 4 is obtained by differentiating each term of the function.
F'(s) = 10s + 1
Now we substitute s=1 into the derivative:
F'(1) = 10(1) + 1
F'(1) = 11
We want the average rate of change from 0 to d to equal the instantaneous rate of change at s=1. So we set them equal:
(5d^2 + d) / d = 11
To solve this equation, we can multiply both sides by d:
5d^2 + d = 11d
Now rearrange the equation:
5d^2 - 11d + d = 0
5d^2 - 10d = 0
Factor out the common factor, d:
d(5d - 10) = 0
From this equation, we have two possibilities:
1) d = 0
However, we're looking for a value of d greater than 0, so this solution is not valid.
2) 5d - 10 = 0
To solve for d, we isolate the variable:
5d = 10
d = 10/5
d = 2
So the value of d that satisfies the given criteria is d = 2.