The critical points of the function f(x) = x^3 - 6x^2 + 9x + 1 are:

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Question: The critical points of the function f(x) = x^3 - 6x^2 + 9x + 1 are:

Options:

Correct Answer: x = 1, 3

Solution:

Finding f\'(x) = 3x^2 - 12x + 9 and solving gives critical points at x = 1 and x = 3.

The critical points of the function f(x) = x^3 - 6x^2 + 9x + 1 are:

The critical points of the function f(x) = x^3 - 6x^2 + 9x + 1 are:

Step 1: Start with the function f(x) = x^3 - 6x^2 + 9x + 1.
Step 2: Find the derivative of the function, which is f'(x). The derivative tells us the slope of the function.
Step 3: Calculate the derivative: f'(x) = 3x^2 - 12x + 9.
Step 4: Set the derivative equal to zero to find critical points: 3x^2 - 12x + 9 = 0.
Step 5: Simplify the equation by dividing everything by 3: x^2 - 4x + 3 = 0.
Step 6: Factor the quadratic equation: (x - 1)(x - 3) = 0.
Step 7: Solve for x by setting each factor equal to zero: x - 1 = 0 or x - 3 = 0.
Step 8: This gives us the critical points: x = 1 and x = 3.

Finding Critical Points – This involves taking the derivative of a function and setting it to zero to find points where the function's slope is zero.
Derivative Calculation – Understanding how to differentiate polynomial functions correctly.
Solving Quadratic Equations – Applying methods to solve the resulting quadratic equation from the derivative.