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)/(d-0)
so solve for d in
(5d^2 + 5d)/d = 15
d^2 + d = 3d
d^2 - 2d = 0
d(d-2) = 0
d = 0 or d = 2 ,but you want d > 0,
so d = 2
To find the value of d that satisfies the given conditions, we need to follow these steps:
Step 1: Find the average rate of change of F(s) from 0 to d.
The average rate of change of F(s) from 0 to d is given by:
Average rate of change = [F(d) - F(0)] / (d - 0)
Step 2: Find the instantaneous rate of change of F(s) at s = 1.
The instantaneous rate of change of F(s) at s = 1 is given by the derivative of F(s) with respect to s:
F'(s) = 10s + 5
Step 3: Set the average rate of change equal to the instantaneous rate of change and solve for d:
[F(d) - F(0)] / (d - 0) = F'(1)
Step 4: Substitute the expressions for F(s) and F'(s) into the equation and solve for d:
[5d^2 + 5d + 4 - 4] / d = 10(1) + 5
Simplifying the equation:
[5d^2 + 5d] / d = 15
Multiplying both sides by d to eliminate the denominator:
5d^2 + 5d = 15d
Rearranging the equation:
5d^2 - 10d = 0
Factoring out common factor:
5d(d - 2) = 0
Using the zero-product property:
d = 0 or d - 2 = 0
Since we're looking for a positive value of d, we can disregard the solution d = 0. Therefore, the value of d that satisfies the given conditions is:
d = 2
To find the value of d that satisfies the given condition, we first need to calculate the average rate of change of F(s) from 0 to d.
The average rate of change of a function f(x) over an interval [a, b] is given by:
average rate of change = (f(b) - f(a)) / (b - a)
In this case, we want to find the average rate of change of F(s) from 0 to d. So, we substitute the values into the formula:
average rate of change = (F(d) - F(0)) / (d - 0)
To proceed, we need to calculate F(d) and F(0) using the given function F(s)=5s^2+5s+4:
F(d) = 5d^2 + 5d + 4
F(0) = 5(0)^2 + 5(0) + 4 = 4
Substituting these values into the formula for average rate of change, we have:
average rate of change = (5d^2 + 5d + 4 - 4) / (d - 0) = (5d^2 + 5d) / d = 5d + 5
Now, we want the average rate of change from 0 to d to be equal to the instantaneous rate of change at s=1.
The instantaneous rate of change of a function at a specific point is given by the derivative of the function evaluated at that point.
In this case, we need to find the derivative of F(s) and evaluate it at s=1. Let's find the derivative of F(s) first:
F'(s) = d/ds(5s^2 + 5s + 4) = 10s + 5
Now, let's evaluate the derivative at s=1:
F'(1) = 10(1) + 5 = 10 + 5 = 15
Therefore, we can set the average rate of change equal to the instantaneous rate of change and solve for d:
5d + 5 = 15
Subtracting 5 from both sides:
5d = 10
Dividing both sides by 5:
d = 2
Therefore, the value of d that satisfies the given condition is d = 2.