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 actual value of f′(0.5) using the given function f(x) = 2x^3.
The derivative of f(x) with respect to x, denoted as f'(x), gives us the rate of change of the function at any given point.
In this case, to find f′(x), we differentiate f(x) = 2x^3 with respect to x.
f'(x) = d/dx (2x^3)
To differentiate the function, we apply the power rule:
f'(x) = 3*2x^(3-1)
Simplifying the expression, we have:
f'(x) = 6x^2
Substituting x = 0.5 into f'(x), we find:
f'(0.5) = 6*(0.5)^2
= 6*(0.25)
= 1.5
Now, let’s look at the table provided and use the values to approximate f′(0.5).
+—–+—–+
| x | f |
+—–+—–+
| 0.4 | 0.32|
| 0.5 | 0.5 |
| 0.6 | 0.72|
+—–+—–+
To approximate f′(0.5), we can use the finite difference formula:
f′(0.5) ≈ (f(0.6) – f(0.4)) / (0.6 – 0.4)
Substituting the values from the table:
f′(0.5) ≈ (0.72 – 0.32) / (0.6 – 0.4)
≈ 0.4 / 0.2
≈ 2
Now we can find the difference between the approximation and the actual value of f′(0.5):
Difference = Approximation – Actual value
= 2 – 1.5
= 0.5
So, the difference between the approximation and the actual value of f′(0.5) is 0.5.
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