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 approximate value of f'(0
To find the approximate value of f'(0.5), we can use the values given in the table to estimate the derivative of the function at that point.
The function f(x) = 2x^3, where ^ denotes exponentiation, represents a cubic function. To find the derivative of this function, we can apply the power rule, which states that for a function of the form f(x) = ax^n, the derivative is given by f'(x) = nax^(n-1).
In this case, the function f(x) = 2x^3 has an exponent of 3, and the coefficient is 2. Using the power rule, the derivative of f(x) is f'(x) = 3 * 2 * x^(3-1) = 6x^2.
To find the approximation of f'(0.5), we need to substitute x = 0.5 into the derivative function. Plugging this value in, we get f'(0.5) = 6 * (0.5)^2 = 6 * 0.25 = 1.5.
Now, we need to compare this approximate value with the actual value of f'(0.5). Since we are not given the function or its derivative explicitly, we cannot calculate the actual value. Hence, we cannot determine the difference between the approximation and the actual value of f'(0.5) without additional information.
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