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Is f(x) = (x² - 2x - 1)²/³? Find f'(0).

a. f'(0) = 0
b. f'(0) = 1
c. f'(0) = -1
d. f'(0) = 2

User Mdewit
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1 Answer

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Final answer:

To find f'(0), we need to find the derivative of the given function f(x) = (x² - 2x - 1)²/³ and evaluate it at x = 0. The derivative is (2/3)(x⁴ - 4x³ + 4x² + 4x + 1)²/³ * (4x³ - 12x² + 8x + 4). When x = 0, f'(0) = 0. Option A is correct.

Step-by-step explanation:

To find f'(0), we need to find the derivative of the given function f(x) = (x² - 2x - 1)²/³ and evaluate it at x = 0. Let's go step by step:

Step 1: Rewrite the function using exponent properties:
f(x) = (x⁴ - 4x³ + 4x² + 4x + 1)²/³

Step 2: Use the chain rule to find the derivative:
f'(x) = (2/3)(x⁴ - 4x³ + 4x² + 4x + 1)²/³ * (4x³ - 12x² + 8x + 4)

Step 3: Substitute x = 0 into f'(x):
f'(0) = (2/3)(0⁴ - 4(0)³ + 4(0)² + 4(0) + 1)²/³ * (4(0)³ - 12(0)² + 8(0) + 4) = (2/3)(1)²/³ * (0) = 0

Therefore, f'(0) = 0. The correct answer choice is a) f'(0) = 0.

User Kvdb
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