What is the value of f''(x) for f(x) = 4x^3 - 6x^2 + 2 at x = 1? (2022)

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Question: What is the value of f\'\'(x) for f(x) = 4x^3 - 6x^2 + 2 at x = 1? (2022)

Options:

Correct Answer: 6

Exam Year: 2022

Solution:

f\'\'(x) = 24x - 12. At x = 1, f\'\'(1) = 24(1) - 12 = 12.

What is the value of f''(x) for f(x) = 4x^3 - 6x^2 + 2 at x = 1? (2022)

What is the value of f''(x) for f(x) = 4x^3 - 6x^2 + 2 at x = 1? (2022)

Step 1: Start with the function f(x) = 4x^3 - 6x^2 + 2.
Step 2: Find the first derivative f'(x) by differentiating f(x).
Step 3: The derivative of 4x^3 is 12x^2, and the derivative of -6x^2 is -12x. The derivative of a constant (2) is 0.
Step 4: So, f'(x) = 12x^2 - 12x.
Step 5: Now, find the second derivative f''(x) by differentiating f'(x).
Step 6: The derivative of 12x^2 is 24x, and the derivative of -12x is -12.
Step 7: So, f''(x) = 24x - 12.
Step 8: Now, substitute x = 1 into f''(x).
Step 9: Calculate f''(1) = 24(1) - 12.
Step 10: Simplify the expression: 24 - 12 = 12.

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