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For the function f(x) = x^3 - 6x^2 + 9x, find the critical points.
For the function f(x) = x^3 - 6x^2 + 9x, find the critical points.
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Practice Questions
1 question
Q1
For the function f(x) = x^3 - 6x^2 + 9x, find the critical points.
x = 0, 3
x = 1, 2
x = 2, 3
x = 3, 4
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First, find f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 3)(x - 1) = 0, so critical points are x = 1 and x = 3.
Questions & Step-by-step Solutions
1 item
Q
Q: For the function f(x) = x^3 - 6x^2 + 9x, find the critical points.
Solution:
First, find f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 3)(x - 1) = 0, so critical points are x = 1 and x = 3.
Steps: 8
Show Steps
Step 1: Start with the function f(x) = x^3 - 6x^2 + 9x.
Step 2: Find the derivative of the function, which is f'(x). The derivative tells us the rate of change of the function.
Step 3: Calculate 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 - 3)(x - 1) = 0.
Step 7: Solve for x by setting each factor equal to zero: x - 3 = 0 gives x = 3, and x - 1 = 0 gives x = 1.
Step 8: The critical points are x = 1 and x = 3.
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