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.
The derivative of f(x) = 2x^3 is f'(x) = 6x^2.
To approximate f'(0.5) using the given table, we can use the formula:
f'(0.5) ≈ [f(0.6) – f(0.4)] / [0.6 – 0.4]
f'(0.5) ≈ [(2(0.6)^3) – (2(0.4)^3)] / 0.2
f'(0.5) ≈ (0.864 – 0.128) / 0.2
f'(0.5) ≈ 3.68
Therefore, the approximation of f'(0.5) using the table is 3.68.
To find the actual value of f'(0.5), we can simply evaluate f'(x) at x=0.5:
f'(0.5) = 6(0.5)^2
f'(0.5) = 1.5
The difference between the approximation and actual value of f'(0.5) is:
|3.68 – 1.5| = 2.18
Therefore, the difference between the approximation and actual value of f'(0.5) is 2.18.
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