What is the critical point of f(x) = x^3 - 3x^2 + 4?

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Question: What is the critical point of f(x) = x^3 - 3x^2 + 4?

Options:

Correct Answer: 2

Solution:

Setting f\'(x) = 3x^2 - 6x = 0 gives x(x - 2) = 0, so critical points are x = 0 and x = 2.

What is the critical point of f(x) = x^3 - 3x^2 + 4?

What is the critical point of f(x) = x^3 - 3x^2 + 4?

Step 1: Start with the function f(x) = x^3 - 3x^2 + 4.
Step 2: Find the derivative of the function, which is f'(x). The derivative of f(x) is f'(x) = 3x^2 - 6x.
Step 3: Set the derivative equal to zero to find critical points: 3x^2 - 6x = 0.
Step 4: Factor the equation: 3x(x - 2) = 0.
Step 5: Solve for x by setting each factor equal to zero: 3x = 0 or x - 2 = 0.
Step 6: From 3x = 0, we get x = 0. From x - 2 = 0, we get x = 2.
Step 7: The critical points are x = 0 and x = 2.

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