Approximating f'(0.5) and Calculating the Difference | A Math Tutorial

1. X | 0 | 1F(x) | 1 | 2Let f be the function given by f(x) = 2^(x^2). 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)?A. 0B. 0.176C. 0.824D. 1

To approximate f'(0

To approximate f'(0.5), we can use the formula for the derivative:

f'(x) = lim(h->0) (f(x+h) – f(x))/h

In this case, we want to approximate f'(0.5), so we substitute x = 0.5 into the formula:

f'(0.5) = lim(h->0) (f(0.5+h) – f(0.5))/h

Now let’s calculate the approximation using the values given in the table:

f(0) = 2^(0^2) = 1
f(1) = 2^(1^2) = 2
f(2) = 2^(2^2) = 16

Using these values, we can approximate f'(0.5) as:

f'(0.5) ≈ (f(1) – f(0))/1 = (2 – 1)/1 = 1

So the approximation for f'(0.5) is 1.

Now let’s find the actual value of f'(0.5) using calculus. We have the function f(x) = 2^(x^2), so we can find f'(x) by applying the chain rule:

f'(x) = d/dx (2^(x^2)) = (ln 2) * (2^x^2) * (2x)

Substituting x = 0.5:

f'(0.5) = (ln 2) * (2^(0.5^2)) * (2*0.5)
= (ln 2) * (2^(0.25)) * 1
≈ 0.4736

The actual value of f'(0.5) is approximately 0.4736.

Now let’s calculate the difference between the approximation and the actual value:

Difference = |approximation – actual value|
= |1 – 0.4736|
≈ 0.5264

Therefore, the difference between the approximation and the actual value of f'(0.5) is approximately 0.5264.

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