Consider the interval I=[6,7.6]. Break I into four subintervals of length 0.4, namely the four subintervals
[6,6.4],[6.4,6.8],[6.8,7.2],[7.2,7.6].
Suppose that f(6)=19, f'(6)=0, f'(6.4)=−0.5, f'(6.8)=−0.1, and f'(7.2)=−0.1. What is the approximate value of f(7.6)?
I bet it is (using the left side) is
first figure the values a the left side of each interval.
f(6)=given 19
f(6.4)=f(6)+.4*f'(6)=19
f(6.8)=f(6.4)+f'(6.4)*.4=19-.2= 18.8
f(7.2)=f(6.8)+f'(6.8)*.4=18.8-.04=18.76
f(7.6)=18.76-.4*.1=18.72
so check all those, it is easy to make an error when doing it on a keyboard.
To approximate the value of f(7.6), we can use the Mean Value Theorem for Derivatives.
The Mean Value Theorem states that if a function f(x) is continuous on the interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
In this case, we can apply the Mean Value Theorem to the subinterval [6, 7.6].
Let's calculate the approximate value of f(7.6) using the Mean Value Theorem:
First, we calculate the differences for each subinterval:
Δx1 = 6.4 - 6 = 0.4
Δy1 = f(6.4) - f(6) = -0.5
Δx2 = 6.8 - 6.4 = 0.4
Δy2 = f(6.8) - f(6.4) = -0.1
Δx3 = 7.2 - 6.8 = 0.4
Δy3 = f(7.2) - f(6.8) = -0.1
Δx4 = 7.6 - 7.2 = 0.4
Now, let's use the Mean Value Theorem formula to find the approximate value of f(7.6):
f'(c1) = Δy1/Δx1 = -0.5 / 0.4 = -1.25
f'(c2) = Δy2/Δx2 = -0.1 / 0.4 = -0.25
f'(c3) = Δy3/Δx3 = -0.1 / 0.4 = -0.25
Since we are given that f'(6) = 0, we don't need to calculate f'(c4). Instead, we can directly use f'(6) = 0 as an approximation.
Now we have the following values of f'(c):
f'(c1) = -1.25
f'(c2) = -0.25
f'(c3) = -0.25
f'(c4) = 0
To approximate f(7.6), we can sum up these values multiplied by their corresponding Δx and add it to f(6):
f(7.6) ≈ f(6) + (f'(c1) * Δx1) + (f'(c2) * Δx2) + (f'(c3) * Δx3) + (f'(c4) * Δx4)
≈ 19 + (-1.25 * 0.4) + (-0.25 * 0.4) + (-0.25 * 0.4) + (0 * 0.4)
Calculating the above expression:
f(7.6) ≈ 19 - 0.5 - 0.1 - 0.1 + 0
≈ 18.3
Therefore, the approximate value of f(7.6) is approximately 18.3.