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If f(x) = x^3 - 3x + 2, find the critical points where f'(x) = 0.
If f(x) = x^3 - 3x + 2, find the critical points where f'(x) = 0.
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If f(x) = x^3 - 3x + 2, find the critical points where f'(x) = 0.
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Set f'(x) = 3x^2 - 3 = 0 and solve for x.
Questions & Step-by-step Solutions
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Q: If f(x) = x^3 - 3x + 2, find the critical points where f'(x) = 0.
Solution:
Set f'(x) = 3x^2 - 3 = 0 and solve for x.
Steps: 6
Show Steps
Step 1: Start with the function f(x) = x^3 - 3x + 2.
Step 2: Find the derivative of the function, which is f'(x). The derivative of x^3 is 3x^2, and the derivative of -3x is -3. So, f'(x) = 3x^2 - 3.
Step 3: Set the derivative equal to zero: 3x^2 - 3 = 0.
Step 4: Solve the equation 3x^2 - 3 = 0. First, add 3 to both sides: 3x^2 = 3.
Step 5: Divide both sides by 3: x^2 = 1.
Step 6: Take the square root of both sides: x = ±1. This gives us two critical points: x = 1 and x = -1.
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