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Determine the critical points of f(x) = x^3 - 3x^2 + 4.
Determine the critical points of f(x) = x^3 - 3x^2 + 4.
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Practice Questions
1 question
Q1
Determine the critical points of f(x) = x^3 - 3x^2 + 4.
(0, 4)
(1, 2)
(2, 1)
(3, 0)
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f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x = 0 and x = 2. Critical points are (0, 4) and (2, 1).
Questions & Step-by-step Solutions
1 item
Q
Q: Determine the critical points of f(x) = x^3 - 3x^2 + 4.
Solution:
f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives x = 0 and x = 2. Critical points are (0, 4) and (2, 1).
Steps: 9
Show Steps
Step 1: Write down the function f(x) = x^3 - 3x^2 + 4.
Step 2: Find the derivative of the function, f'(x). The derivative of f(x) is f'(x) = 3x^2 - 6x.
Step 3: Set the derivative equal to zero to find critical points: 3x^2 - 6x = 0.
Step 4: Factor the equation: 3x(x - 2) = 0.
Step 5: Solve for x by setting each factor to zero: 3x = 0 gives x = 0, and x - 2 = 0 gives x = 2.
Step 6: Now we have the x-values of the critical points: x = 0 and x = 2.
Step 7: To find the corresponding y-values, substitute x = 0 into the original function: f(0) = 0^3 - 3(0)^2 + 4 = 4, so the point is (0, 4).
Step 8: Substitute x = 2 into the original function: f(2) = 2^3 - 3(2)^2 + 4 = 8 - 12 + 4 = 0, so the point is (2, 0).
Step 9: The critical points are (0, 4) and (2, 0).
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