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.