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) ?
To find the approximation of f′(0
To find the approximation of f′(0.5) using the given table, we need to calculate the difference quotient at x = 0.5. The difference quotient can be defined as:
f′(x) ≈ (f(x + h) – f(x)) / h
In this case, let’s take h to be a small positive number, such as 0.1. Therefore, x + h = 0.5 + 0.1 = 0.6.
Now let’s calculate the approximate value of f′(0.5) using the difference quotient for the given table values:
f′(0.5) ≈ (f(0.6) – f(0.5)) / 0.1
Substituting the function values from the table, we get:
f′(0.5) ≈ (2(0.6)^3 – 2(0.5)^3) / 0.1
Calculating further:
f′(0.5) ≈ (2(0.216) – 2(0.125)) / 0.1
f′(0.5) ≈ (0.432 – 0.25) / 0.1
f′(0.5) ≈ 0.182 / 0.1
f′(0.5) ≈ 1.82
Therefore, the approximation of f′(0.5) using the given table values is 1.82.
To find the difference between the approximation and the actual value of f′(0.5), we need to calculate the actual value of f′(0.5). Taking the derivative of f(x) = 2x^3, we get:
f′(x) = 6x^2
Substituting x = 0.5, we get:
f′(0.5) = 6(0.5)^2
f′(0.5) = 6(0.25)
f′(0.5) = 1.5
Therefore, the actual value of f′(0.5) is 1.5.
The difference between the approximation (1.82) and the actual value (1.5) of f′(0.5) is:
Difference = Approximation – Actual value
Difference = 1.82 – 1.5
Difference ≈ 0.32
Thus, the difference between the approximation and the actual value of f′(0.5) is approximately 0.32.
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