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 find the derivative of the function f(x) = 2x^3.
The derivative of f(x) can be found using the power rule for derivatives. According to the power rule, for any real number n, the derivative of x^n is nx^(n-1).
In this case, we have f(x) = 2x^3, so applying the power rule, we get:
f′(x) = 3 * 2x^3-1 = 6x^2
Now, let’s find the actual value of f′(0.5) by plugging in x = 0.5 into the derivative:
f′(0.5) = 6 * (0.5)^2 = 6 * 0.25 = 1.5
To approximate f′(0.5) using the table values, we can calculate the slope between the points with x = 0 and x = 1:
Slope = (f(1) – f(0)) / (1 – 0) = (2(1)^3 – 2(0)^3) / (1 – 0) = 2 – 0 = 2
Since the function f(x) = 2x^3 is a continuously differentiable function, we can assume that the derivative is constant between x = 0 and x = 1. Therefore, we can approximate f′(0.5) by taking the slope obtained above: f′(0.5) ≈ 2.
Now, let’s calculate the difference between the approximation and the actual value:
Difference = Actual value – Approximation = 1.5 – 2 = -0.5
Therefore, the difference between the approximation and the actual value of f′(0.5) is -0.5.
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