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)?
To approximate the value of f(7.6) based on the given information, we will use the concept of the tangent line approximation or the linear approximation. This method relies on the linear behavior of the function that can be described by its derivative (slope). Here's how we can proceed step-by-step:
1. Identify the subinterval that contains the point 7.6, which in this case is [7.2, 7.6]. Since the given information does not provide the value of f′(7.6), we will assume that the derivative remains constant within this subinterval and use the value of f′(7.2) as an approximation.
2. Apply the linear approximation formula: Δy = f(x) - f(a) ≈ f′(a) * Δx, where Δy is the change in the function value, f(x) is the actual value at x, f(a) is the known value at a, f′(a) is the derivative at a, and Δx is the difference in x-values.
3. Calculate Δx, which is the difference between the upper bound of the targeted subinterval and the lower bound of the same subinterval: Δx = 7.6 - 7.2 = 0.4.
4. Substitute the values into the formula: Δy ≈ f′(7.2) * 0.4.
5. Compute the approximation: Δy = -0.1 * 0.4 = -0.04.
6. Finally, add the approximation to the known value at the lower bound of the targeted subinterval, f(7.2): f(7.6) ≈ f(7.2) + Δy = f(7.2) - 0.04.
Given that f(7.2) is not provided in the question, we are unable to compute the exact value of f(7.6). But with the given information, we can provide the approximate value of f(7.6) by adding -0.04 to f(7.2).