Find the critical points of the function f(x) = x^3 - 6x^2 + 9x.

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Q: Find the critical points of the function f(x) = x^3 - 6x^2 + 9x.

Solution: f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1 and x = 3. Critical points are (1, f(1)) and (3, f(3)).

Steps: 10

Step 1: Write down the function f(x) = x^3 - 6x^2 + 9x.
Step 2: Find the derivative of the function, which is f'(x). The derivative of f(x) is f'(x) = 3x^2 - 12x + 9.
Step 3: Set the derivative equal to zero to find critical points: 3x^2 - 12x + 9 = 0.
Step 4: Simplify the equation by dividing everything by 3: x^2 - 4x + 3 = 0.
Step 5: Factor the quadratic equation: (x - 1)(x - 3) = 0.
Step 6: Solve for x by setting each factor to zero: x - 1 = 0 gives x = 1, and x - 3 = 0 gives x = 3.
Step 7: Now, find the y-values for the critical points by substituting x back into the original function f(x).
Step 8: Calculate f(1) = 1^3 - 6(1^2) + 9(1) = 1 - 6 + 9 = 4, so the first critical point is (1, 4).
Step 9: Calculate f(3) = 3^3 - 6(3^2) + 9(3) = 27 - 54 + 27 = 0, so the second critical point is (3, 0).
Step 10: The critical points are (1, 4) and (3, 0).