If the function f(x) = x^3 - 3x + 2 has a local maximum, what is the value of x?

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Q: If the function f(x) = x^3 - 3x + 2 has a local maximum, what is the value of x?

Solution: Setting f'(x) = 0 gives x = 1 as a local maximum.

Steps: 12

Step 1: Write down the function f(x) = x^3 - 3x + 2.
Step 2: Find the derivative of the function, f'(x). This tells us the slope of the function.
Step 3: Calculate the derivative: f'(x) = 3x^2 - 3.
Step 4: Set the derivative equal to zero to find critical points: 3x^2 - 3 = 0.
Step 5: Solve the equation 3x^2 - 3 = 0. This simplifies to x^2 = 1.
Step 6: Take the square root of both sides to find x: x = 1 or x = -1.
Step 7: To determine if x = 1 is a local maximum, check the second derivative, f''(x).
Step 8: Calculate the second derivative: f''(x) = 6x.
Step 9: Evaluate the second derivative at x = 1: f''(1) = 6(1) = 6, which is positive.
Step 10: Since f''(1) is positive, x = 1 is a local minimum, not a maximum. Check x = -1.
Step 11: Evaluate the second derivative at x = -1: f''(-1) = 6(-1) = -6, which is negative.
Step 12: Since f''(-1) is negative, x = -1 is a local maximum.