Let f be the function given by f(x)=2×3. Selected values of f are given in the table above. If the values in the table are used to approximate f′(0.5), what is the difference between the approximation and the actual value of f′(0.5) ?
0.433The numerical value of the derivative at x=0.5 obtained from the calculator is f′(0.5)=0.567. A difference quotient can be used with the values in the table to estimate the derivative as [f(1)−f(0)]/(1−0) = (2−1)/1 = 1. The error between the actual derivative value and this approximation is 0.433.
To approximate f'(0.5), we can use the forward difference formula:
f'(x) ≈ [f(x+h) – f(x)] / h
where h is the step size, which is given by the difference between the x-values in the table. In this case, h = 0.2.
So we have:
f'(0.5) ≈ [f(0.7) – f(0.5)] / 0.2
f(0.7) = 2(0.7)^3 = 0.686
f(0.5) = 2(0.5)^3 = 0.5
So the approximation is:
f'(0.5) ≈ (0.686 – 0.5) / 0.2 = 0.93
To find the actual value of f'(0.5), we can take the derivative of f(x):
f'(x) = 6x^2
So:
f'(0.5) = 6(0.5)^2 = 1.5
The difference between the approximation and the actual value is:
1.5 – 0.93 = 0.57
Therefore, the difference between the approximation and the actual value of f'(0.5) is 0.57.
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