What is the critical point of the function f(x) = x^2 - 4x + 4? (2022)

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Question: What is the critical point of the function f(x) = x^2 - 4x + 4? (2022)

Options:

Correct Answer: 2

Exam Year: 2022

Solution:

Find f\'(x) = 2x - 4. Set f\'(x) = 0, giving 2x - 4 = 0, hence x = 2.

What is the critical point of the function f(x) = x^2 - 4x + 4? (2022)

What is the critical point of the function f(x) = x^2 - 4x + 4? (2022)

Step 1: Start with the function f(x) = x^2 - 4x + 4.
Step 2: Find the derivative of the function, which is f'(x). The derivative of x^2 is 2x, and the derivative of -4x is -4. So, f'(x) = 2x - 4.
Step 3: Set the derivative equal to zero to find the critical points. This means we solve the equation 2x - 4 = 0.
Step 4: Solve the equation 2x - 4 = 0. Add 4 to both sides to get 2x = 4.
Step 5: Divide both sides by 2 to find x. So, x = 4 / 2 = 2.
Step 6: The critical point of the function is at x = 2.

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