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the third-degree taylor polynomial for a function f about x=4 is (x−4)3512−(x−4)264 (x−4)4 2. what is the value of f′′′(4)?

User Spierala
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Answer: the value of f′′′(4) is 3/256.

Explanation:

Given the third-degree Taylor polynomial:

f(x) = (x−4)³/512 − (x−4)²/64 + (x−4)⁴/2

To find the value of f′′′(4), we need to differentiate the polynomial three times and evaluate it at x = 4.

First derivative:

f'(x) = 3(x−4)²/512 − 2(x−4)/64 + 4(x−4)³/2

Second derivative:

f''(x) = 6(x−4)/512 − 2/64 + 12(x−4)²/2

Third derivative:

f'''(x) = 6/512 + 24(x−4)/2

Now, substitute x = 4 into f'''(x):

f'''(4) = 6/512 + 24(4−4)/2

= 6/512 + 0

= 6/512

= 3/256

Therefore, the value of f′′′(4) is 3/256.

User Priyankchoudhary
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