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If f(x) = x^3 - 3x^2 + 4, find the critical points of f.
If f(x) = x^3 - 3x^2 + 4, find the critical points of f.
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Practice Questions
1 question
Q1
If f(x) = x^3 - 3x^2 + 4, find the critical points of f.
x = 0, 1, 2
x = 1, 2
x = 0, 2
x = 1
Show Solution
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f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives 3x(x - 2) = 0, so x = 0 and x = 2 are critical points.
Questions & Step-by-step Solutions
1 item
Q
Q: If f(x) = x^3 - 3x^2 + 4, find the critical points of f.
Solution:
f'(x) = 3x^2 - 6x. Setting f'(x) = 0 gives 3x(x - 2) = 0, so x = 0 and x = 2 are critical points.
Steps: 7
Show Steps
Step 1: Start with the function f(x) = x^3 - 3x^2 + 4.
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 - 6x.
Step 4: Set the derivative equal to zero to find critical points: 3x^2 - 6x = 0.
Step 5: Factor the equation: 3x(x - 2) = 0.
Step 6: Solve for x by setting each factor equal to zero: 3x = 0 gives x = 0, and x - 2 = 0 gives x = 2.
Step 7: The critical points are x = 0 and x = 2.
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