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What is the critical point of f(x) = x^3 - 6x^2 + 9x?
What is the critical point of f(x) = x^3 - 6x^2 + 9x?
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What is the critical point of f(x) = x^3 - 6x^2 + 9x?
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Setting f'(x) = 0 gives critical points at x = 1, 2, and 3.
Questions & Step-by-step Solutions
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Q: What is the critical point of f(x) = x^3 - 6x^2 + 9x?
Solution:
Setting f'(x) = 0 gives critical points at x = 1, 2, and 3.
Steps: 10
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Step 1: Start with the function f(x) = x^3 - 6x^2 + 9x.
Step 2: Find the derivative of the function, f'(x).
Step 3: The derivative f'(x) is calculated as f'(x) = 3x^2 - 12x + 9.
Step 4: Set the derivative equal to zero: 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 to zero: x - 1 = 0 or x - 3 = 0.
Step 8: This gives the solutions x = 1 and x = 3.
Step 9: Check for any other critical points by finding the second derivative, but in this case, we only have x = 1 and x = 3.
Step 10: The critical points are x = 1, 2, and 3.
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