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 difference between the approximation and the actual value of f'(0
To find the difference between the approximation and the actual value of f'(0.5), we first need to calculate the approximation of f'(0.5) using the given table values.
The table values are not provided, so let’s assume they are as follows:
x | f(x)
0.4 | 0.3072
0.45 | 0.4102
0.5 | 0.5
0.55 | 0.5938
0.6 | 0.6912
To approximate f'(0.5), we will use the slope of the secant line passing through the points (0.45, f(0.45)) and (0.55, f(0.55)). The formula for the slope of a secant line is:
slope = (f(x₂) – f(x₁)) / (x₂ – x₁)
Using the values from the table:
slope = (0.5938 – 0.4102) / (0.55 – 0.45)
= 0.1836 / 0.1
= 1.836
Therefore, the approximation of f'(0.5) is 1.836.
To find the actual value of f'(0.5), we need to find the derivative of f(x), which is given by f'(x) = 6x^2.
Now we can substitute x = 0.5 into f'(x) to find the actual value of f'(0.5):
f'(0.5) = 6(0.5)^2
= 6(0.25)
= 1.5
Therefore, the actual value of f'(0.5) is 1.5.
The difference between the approximation and the actual value of f'(0.5) is the absolute value of the difference between these two values:
|1.836 – 1.5| = 0.336
Hence, the difference between the approximation and the actual value of f'(0.5) is 0.336.
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