Determine the point at which the function f(x) = x^3 - 3x^2 + 4 has a local mini

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Question: Determine the point at which the function f(x) = x^3 - 3x^2 + 4 has a local minimum.

Options:

Correct Answer: (1, 2)

Solution:

Find f\'(x) = 3x^2 - 6x. Setting f\'(x) = 0 gives x(x - 2) = 0, so x = 0 or x = 2. f\'\'(2) = 6 > 0, so (2, 1) is a local minimum.

Determine the point at which the function f(x) = x^3 - 3x^2 + 4 has a local minimum.

Determine the point at which the function f(x) = x^3 - 3x^2 + 4 has a local minimum.

Correct Answer: (2, 1)

Step 1: Write down the function f(x) = x^3 - 3x^2 + 4.
Step 2: Find the first derivative f'(x) to determine the slope of the function. The first derivative is f'(x) = 3x^2 - 6x.
Step 3: Set the first 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: This gives us two solutions, x = 0 and x = 2.
Step 6: To determine if these points are local minima or maxima, find the second derivative f''(x). The second derivative is f''(x) = 6x - 6.
Step 7: Evaluate the second derivative at the critical points. First, check x = 0: f''(0) = 6(0) - 6 = -6 (which indicates a local maximum).
Step 8: Now check x = 2: f''(2) = 6(2) - 6 = 6 (which indicates a local minimum).
Step 9: The local minimum occurs at x = 2. To find the corresponding y-value, substitute x = 2 back into the original function: f(2) = 2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0.
Step 10: Therefore, the point at which the function has a local minimum is (2, 0).

Critical Points – Identifying points where the derivative is zero to find potential local extrema.
Second Derivative Test – Using the second derivative to determine the concavity at critical points to classify them as local minima or maxima.