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

Practice Questions

Q1
The critical points of the function f(x) = x^3 - 6x^2 + 9x + 1 are:
  1. x = 1, 3
  2. x = 0, 2
  3. x = 2, 4
  4. x = 1, 2

Questions & Step-by-Step Solutions

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.
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