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 find the approximation of f'(0.5), we will use the formula for the derivative of a function:
f'(x) = lim(h->0) [f(x+h) – f(x)]/h
We want to find f'(0.5), so we can use the values from the table to approximate this derivative:
f'(0.5) ≈ [f(0.55) – f(0.45)]/(0.1)
f(0.55) = 2(0.55)^3 = 0.166375
f(0.45) = 2(0.45)^3 = 0.091125
Therefore, f'(0.5) ≈ (0.166375 – 0.091125)/(0.1) = 0.7535
To find the actual value of f'(0.5), we can take the derivative of f(x) and evaluate it at x=0.5:
f(x) = 2x^3
f'(x) = 6x^2
f'(0.5) = 6(0.5)^2 = 1.5
The difference between the approximation and the actual value of f'(0.5) is:
|f'(0.5) – f'(0.5) approx.| = |1.5 – 0.7535| = 0.7465
Therefore, the difference between the approximation and the actual value of f'(0.5) is 0.7465.
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